Rd Sharma XII Vol 1 2020 for Class 12 Commerce Maths Chapter 4 - Algebra Of Matrices

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Page No 4.18:

Question 1:

Compute the following sums:
(i) $[\begin{matrix} 3 & - 2 \\ 1 & 4 \end{matrix}] + [\begin{matrix} - 2 & 4 \\ 1 & 3 \end{matrix}]$

(ii) $[\begin{matrix} 2 & 1 & 3 \\ 0 & 3 & 5 \\ - 1 & 2 & 5 \end{matrix}] + [\begin{matrix} 1 & - 2 & 3 \\ 2 & 6 & 1 \\ 0 & - 3 & 1 \end{matrix}]$

Answer:

$(i) [\begin{matrix} 3 & - 2 \\ 1 & 4 \end{matrix}] + [\begin{matrix} - 2 & 4 \\ 1 & 3 \end{matrix}] \Rightarrow [\begin{matrix} 3 - 2 & - 2 + 4 \\ 1 + 1 & 4 + 3 \end{matrix}] \Rightarrow [\begin{matrix} 1 & 2 \\ 2 & 7 \end{matrix}]$

$(i i) [\begin{matrix} 2 & 1 & 3 \\ 0 & 3 & 5 \\ - 1 & 2 & 5 \end{matrix}] + [\begin{matrix} 1 & - 2 & 3 \\ 2 & 6 & 1 \\ 0 & - 3 & 1 \end{matrix}] \Rightarrow [\begin{matrix} 2 + 1 & 1 - 2 & 3 + 3 \\ 0 + 2 & 3 + 6 & 5 + 1 \\ - 1 + 0 & 2 - 3 & 5 + 1 \end{matrix}] \Rightarrow [\begin{matrix} 3 & - 1 & 6 \\ 2 & 9 & 6 \\ - 1 & - 1 & 6 \end{matrix}]$

Page No 4.18:

Question 2:

Let A = $[\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}]$ , B = $[\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}]$ and C = $[\begin{matrix} - 2 & 5 \\ 3 & 4 \end{matrix}]$ . Find each of the following:
(i) 2A − 3B
(ii) B − 4C
(iii) 3A − C
(iv) 3A − 2B + 3C

Answer:

$(i) 2 A - 3 B \Rightarrow 2 A - 3 B = 2 [\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}] - 3 [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}] \Rightarrow 2 A - 3 B = [\begin{matrix} 4 & 8 \\ 6 & 4 \end{matrix}] - [\begin{matrix} 3 & 9 \\ - 6 & 15 \end{matrix}] \Rightarrow 2 A - 3 B = [\begin{matrix} 4 - 3 & 8 - 9 \\ 6 + 6 & 4 - 15 \end{matrix}] \Rightarrow 2 A - 3 B = [\begin{matrix} 1 & - 1 \\ 12 & - 11 \end{matrix}] (i i) B - 4 C \Rightarrow B - 4 C = [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}] - 4 [\begin{matrix} - 2 & 5 \\ 3 & 4 \end{matrix}] \Rightarrow B - 4 C = [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}] - [\begin{matrix} - 8 & 20 \\ 12 & 16 \end{matrix}] \Rightarrow B - 4 C = [\begin{matrix} 1 + 8 & 3 - 20 \\ - 2 - 12 & 5 - 16 \end{matrix}] \Rightarrow B - 4 C = [\begin{matrix} 9 & - 17 \\ - 14 & - 11 \end{matrix}] (i i i) 3 A - C \Rightarrow 3 A - C = 3 [\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}] - [\begin{matrix} - 2 & 5 \\ 3 & 4 \end{matrix}] \Rightarrow 3 A - C = [\begin{matrix} 6 & 12 \\ 9 & 6 \end{matrix}] - [\begin{matrix} - 2 & 5 \\ 3 & 4 \end{matrix}] \Rightarrow 3 A - C = [\begin{matrix} 6 + 2 & 12 - 5 \\ 9 - 3 & 6 - 4 \end{matrix}] \Rightarrow 3 A - C = [\begin{matrix} 8 & 7 \\ 6 & 2 \end{matrix}] (i v) 3 A - 2 B + 3 C \Rightarrow 3 A - 2 B + 3 C = 3 [\begin{matrix} 2 & 4 \\ 3 & 2 \end{matrix}] - 2 [\begin{matrix} 1 & 3 \\ - 2 & 5 \end{matrix}] + 3 [\begin{matrix} - 2 & 5 \\ 3 & 4 \end{matrix}] \Rightarrow 3 A - 2 B + 3 C = [\begin{matrix} 6 & 12 \\ 9 & 6 \end{matrix}] - [\begin{matrix} 2 & 6 \\ - 4 & 10 \end{matrix}] + [\begin{matrix} - 6 & 15 \\ 9 & 12 \end{matrix}] \Rightarrow 3 A - 2 B + 3 C = [\begin{matrix} 6 - 2 - 6 & 12 - 6 + 15 \\ 9 + 4 + 9 & 6 - 10 + 12 \end{matrix}] \Rightarrow 3 A - 2 B + 3 C = [\begin{matrix} - 2 & 21 \\ 22 & 8 \end{matrix}]$

Page No 4.18:

Question 3:

If A = $[\begin{matrix} 2 & 3 \\ 5 & 7 \end{matrix}]$ , B = $[\begin{matrix} - 1 & 0 & 2 \\ 3 & 4 & 1 \end{matrix}]$ , C = $[\begin{matrix} - 1 & 2 & 3 \\ 2 & 1 & 0 \end{matrix}]$ , find
(i) A + B and B + C
(ii) 2B + 3A and 3C − 4B.

Answer:

$(i) A + B = [\begin{matrix} 2 & 3 \\ 5 & 7 \end{matrix}] + [\begin{matrix} - 1 & 0 & 2 \\ 3 & 4 & 1 \end{matrix}]$

It is not possible to add these matrices because the number of elements in A are not equal to the
number of elements in B. So, A + B does not exist.

$\Rightarrow B + C = [\begin{matrix} - 1 & 0 & 2 \\ 3 & 4 & 1 \end{matrix}] + [\begin{matrix} - 1 & 2 & 3 \\ 2 & 1 & 0 \end{matrix}] \Rightarrow B + C = [\begin{matrix} - 1 - 1 & 0 + 2 & 2 + 3 \\ 3 + 2 & 4 + 1 & 1 + 0 \end{matrix}] \Rightarrow B + C = [\begin{matrix} - 2 & 2 & 5 \\ 5 & 5 & 1 \end{matrix}]$

$(ii) 2 B + 3 A = 2 [\begin{matrix} - 1 & 0 & 2 \\ 3 & 4 & 1 \end{matrix}] + 3 [\begin{matrix} 2 & 3 \\ 5 & 7 \end{matrix}]$

It is not possible to add these matrices because the number of elements in B are not equal to the
number of elements in A. So, 2B + 3A does not exist.

$\Rightarrow 3 C - 4 B = 3 [\begin{matrix} - 1 & 2 & 3 \\ 2 & 1 & 0 \end{matrix}] - 4 [\begin{matrix} - 1 & 0 & 2 \\ 3 & 4 & 1 \end{matrix}] \Rightarrow 3 C - 4 B = [\begin{matrix} - 3 & 6 & 9 \\ 6 & 3 & 0 \end{matrix}] - [\begin{matrix} - 4 & 0 & 8 \\ 12 & 16 & 4 \end{matrix}] \Rightarrow 3 C - 4 B = [\begin{matrix} - 3 + 4 & 6 - 0 & 9 - 8 \\ 6 - 12 & 3 - 16 & 0 - 4 \end{matrix}] \Rightarrow 3 C - 4 B = [\begin{matrix} 1 & 6 & 1 \\ - 6 & - 13 & - 4 \end{matrix}]$

Page No 4.18:

Question 4:

Let A = $[\begin{matrix} - 1 & 0 & 2 \\ 3 & 1 & 4 \end{matrix}],$ B = $[\begin{matrix} 0 & - 2 & 5 \\ 1 & - 3 & 1 \end{matrix}]$ and C = $[\begin{matrix} 1 & - 5 & 2 \\ 6 & 0 & - 4 \end{matrix}]$ . Compute 2A − 3B + 4C.

Answer:

$Here, 2 A - 3 B + 4 C = 2 [\begin{matrix} - 1 & 0 & 2 \\ 3 & 1 & 4 \end{matrix}] - 3 [\begin{matrix} 0 & - 2 & 5 \\ 1 & - 3 & 1 \end{matrix}] + 4 [\begin{matrix} 1 & - 5 & 2 \\ 6 & 0 & - 4 \end{matrix}] \Rightarrow 2 A - 3 B + 4 C = [\begin{matrix} - 2 & 0 & 4 \\ 6 & 2 & 8 \end{matrix}] - [\begin{matrix} 0 & - 6 & 15 \\ 3 & - 9 & 3 \end{matrix}] + [\begin{matrix} 4 & - 20 & 8 \\ 24 & 0 & - 16 \end{matrix}] \Rightarrow 2 A - 3 B + 4 C = [\begin{matrix} - 2 - 0 + 4 & 0 + 6 - 20 & 4 - 15 + 8 \\ 6 - 3 + 24 & 2 + 9 + 0 & 8 - 3 - 16 \end{matrix}] \Rightarrow 2 A - 3 B + 4 C = [\begin{matrix} 2 & - 14 & - 3 \\ 27 & 11 & - 11 \end{matrix}]$

Page No 4.18:

Question 5:

If A = diag (2 − 59), B = diag (11 − 4) and C = diag (−6 3 4), find
(i) A − 2B
(ii) B + C − 2A
(iii) 2A + 3B − 5C

Answer:

$Here, A = [\begin{matrix} 2 & 0 & 0 \\ 0 & - 5 & 0 \\ 0 & 0 & 9 \end{matrix}], B = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 4 \end{matrix}] and C = [\begin{matrix} - 6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{matrix}] (i) A - 2 B \Rightarrow A - 2 B = [\begin{matrix} 2 & 0 & 0 \\ 0 & - 5 & 0 \\ 0 & 0 & 9 \end{matrix}] - 2 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 4 \end{matrix}] \Rightarrow A - 2 B = [\begin{matrix} 2 & 0 & 0 \\ 0 & - 5 & 0 \\ 0 & 0 & 9 \end{matrix}] - [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & - 8 \end{matrix}] \Rightarrow A - 2 B = [\begin{matrix} 0 & 0 & 0 \\ 0 & - 7 & 0 \\ 0 & 0 & 17 \end{matrix}] = diag (0 - 7 17) (ii) B + C - 2 A \Rightarrow B + C - 2 A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 4 \end{matrix}] + [\begin{matrix} - 6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{matrix}] - 2 [\begin{matrix} 2 & 0 & 0 \\ 0 & - 5 & 0 \\ 0 & 0 & 9 \end{matrix}] \Rightarrow B + C - 2 A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 4 \end{matrix}] + [\begin{matrix} - 6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{matrix}] - [\begin{matrix} 4 & 0 & 0 \\ 0 & - 10 & 0 \\ 0 & 0 & 18 \end{matrix}] \Rightarrow B + C - 2 A = [\begin{matrix} 1 - 6 - 4 & 0 + 0 - 0 & 0 + 0 - 0 \\ 0 + 0 - 0 & 1 + 3 + 10 & 0 + 0 - 0 \\ 0 + 0 - 0 & 0 + 0 - 0 & - 4 + 4 - 18 \end{matrix}] \Rightarrow B + C - 2 A = [\begin{matrix} - 9 & 0 & 0 \\ 0 & 14 & 0 \\ 0 & 0 & - 18 \end{matrix}] = diag (- 9 14 - 18) (iii) 2 A + 3 B - 5 C \Rightarrow 2 A + 3 B - 5 C = 2 [\begin{matrix} 2 & 0 & 0 \\ 0 & - 5 & 0 \\ 0 & 0 & 9 \end{matrix}] + 3 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 4 \end{matrix}] - 5 [\begin{matrix} - 6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{matrix}] \Rightarrow 2 A + 3 B - 5 C = [\begin{matrix} 4 & 0 & 0 \\ 0 & - 10 & 0 \\ 0 & 0 & 18 \end{matrix}] + [\begin{matrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & - 12 \end{matrix}] - [\begin{matrix} - 30 & 0 & 0 \\ 0 & 15 & 0 \\ 0 & 0 & 20 \end{matrix}] \Rightarrow 2 A + 3 B - 5 C = [\begin{matrix} 4 + 3 + 30 & 0 + 0 - 0 & 0 + 0 - 0 \\ 0 + 0 - 0 & - 10 + 3 - 15 & 0 + 0 - 0 \\ 0 + 0 - 0 & 0 + 0 - 0 & 18 - 12 - 20 \end{matrix}] \Rightarrow 2 A + 3 B - 5 C = [\begin{matrix} 37 & 0 & 0 \\ 0 & - 22 & 0 \\ 0 & 0 & - 14 \end{matrix}] = diag (37 - 22 - 14)$

Page No 4.18:

Question 6:

Given the matrices
A = $[\begin{matrix} 2 & 1 & 1 \\ 3 & - 1 & 0 \\ 0 & 2 & 4 \end{matrix}]$ , B = $[\begin{matrix} 9 & 7 & - 1 \\ 3 & 5 & 4 \\ 2 & 1 & 6 \end{matrix}]$ and C = $[\begin{matrix} 2 & - 4 & 3 \\ 1 & - 1 & 0 \\ 9 & 4 & 5 \end{matrix}]$
Verify that (A + B) + C = A + (B + C).

Answer:

$Here, LHS = (A + B) + C = ([\begin{matrix} 2 & 1 & 1 \\ 3 & - 1 & 0 \\ 0 & 2 & 4 \end{matrix}] + [\begin{matrix} 9 & 7 & - 1 \\ 3 & 5 & 4 \\ 2 & 1 & 6 \end{matrix}]) + [\begin{matrix} 2 & - 4 & 3 \\ 1 & - 1 & 0 \\ 9 & 4 & 5 \end{matrix}] = ([\begin{matrix} 2 + 9 & 1 + 7 & 1 - 1 \\ 3 + 3 & - 1 + 5 & 0 + 4 \\ 0 + 2 & 2 + 1 & 4 + 6 \end{matrix}]) + [\begin{matrix} 2 & - 4 & 3 \\ 1 & - 1 & 0 \\ 9 & 4 & 5 \end{matrix}] = [\begin{matrix} 11 & 8 & 0 \\ 6 & 4 & 4 \\ 2 & 3 & 10 \end{matrix}] + [\begin{matrix} 2 & - 4 & 3 \\ 1 & - 1 & 0 \\ 9 & 4 & 5 \end{matrix}] = [\begin{matrix} 11 + 2 & 8 - 4 & 0 + 3 \\ 6 + 1 & 4 - 1 & 4 + 0 \\ 2 + 9 & 3 + 4 & 10 + 5 \end{matrix}] = [\begin{matrix} 13 & 4 & 3 \\ 7 & 3 & 4 \\ 11 & 7 & 15 \end{matrix}] RHS = A + (B + C) = [\begin{matrix} 2 & 1 & 1 \\ 3 & - 1 & 0 \\ 0 & 2 & 4 \end{matrix}] + ([\begin{matrix} 9 & 7 & - 1 \\ 3 & 5 & 4 \\ 2 & 1 & 6 \end{matrix}] + [\begin{matrix} 2 & - 4 & 3 \\ 1 & - 1 & 0 \\ 9 & 4 & 5 \end{matrix}]) = [\begin{matrix} 2 & 1 & 1 \\ 3 & - 1 & 0 \\ 0 & 2 & 4 \end{matrix}] + ([\begin{matrix} 9 + 2 & 7 - 4 & - 1 + 3 \\ 3 + 1 & 5 - 1 & 4 + 0 \\ 2 + 9 & 1 + 4 & 6 + 5 \end{matrix}]) = [\begin{matrix} 2 & 1 & 1 \\ 3 & - 1 & 0 \\ 0 & 2 & 4 \end{matrix}] + [\begin{matrix} 11 & 3 & 2 \\ 4 & 4 & 4 \\ 11 & 5 & 11 \end{matrix}] = [\begin{matrix} 2 + 11 & 1 + 3 & 1 + 2 \\ 3 + 4 & - 1 + 4 & 0 + 4 \\ 0 + 11 & 2 + 5 & 4 + 11 \end{matrix}] = [\begin{matrix} 13 & 4 & 3 \\ 7 & 3 & 4 \\ 11 & 7 & 15 \end{matrix}] ∴ LHS = RHS$

Hence proved.

Page No 4.18:

Question 7:

Find matrices X and Y, if X + Y = $[\begin{matrix} 5 & 2 \\ 0 & 9 \end{matrix}]$ and X − Y = $[\begin{matrix} 3 & 6 \\ 0 & - 1 \end{matrix}]$

Answer:

$Given : (X + Y) + (X - Y) = [\begin{matrix} 5 & 2 \\ 0 & 9 \end{matrix}] + [\begin{matrix} 3 & 6 \\ 0 & - 1 \end{matrix}] \Rightarrow 2 X = [\begin{matrix} 5 + 3 & 2 + 6 \\ 0 + 0 & 9 - 1 \end{matrix}] \Rightarrow 2 X = [\begin{matrix} 8 & 8 \\ 0 & 8 \end{matrix}] \Rightarrow X = \frac{1}{2} [\begin{matrix} 8 & 8 \\ 0 & 8 \end{matrix}] \Rightarrow X = [\begin{matrix} 4 & 4 \\ 0 & 4 \end{matrix}] Now, (X + Y) - (X - Y) = [\begin{matrix} 5 & 2 \\ 0 & 9 \end{matrix}] - [\begin{matrix} 3 & 6 \\ 0 & - 1 \end{matrix}] \Rightarrow X + Y - X + Y = [\begin{matrix} 5 - 3 & 2 - 6 \\ 0 - 0 & 9 + 1 \end{matrix}] \Rightarrow 2 Y = [\begin{matrix} 2 & - 4 \\ 0 & 10 \end{matrix}] \Rightarrow Y = \frac{1}{2} [\begin{matrix} 2 & - 4 \\ 0 & 10 \end{matrix}] \Rightarrow Y = [\begin{matrix} 1 & - 2 \\ 0 & 5 \end{matrix}] ∴ X = [\begin{matrix} 4 & 4 \\ 0 & 4 \end{matrix}] and Y = [\begin{matrix} 1 & - 2 \\ 0 & 5 \end{matrix}]$

Page No 4.18:

Question 8:

Find X if Y = $[\begin{matrix} 3 & 2 \\ 1 & 4 \end{matrix}]$ and 2X + Y = $[\begin{matrix} 1 & 0 \\ - 3 & 2 \end{matrix}]$

Answer:

$Given : 2 X + Y = [\begin{matrix} 1 & 0 \\ - 3 & 2 \end{matrix}] \Rightarrow 2 X + [\begin{matrix} 3 & 2 \\ 1 & 4 \end{matrix}] = [\begin{matrix} 1 & 0 \\ - 3 & 2 \end{matrix}] \Rightarrow 2 X = [\begin{matrix} 1 & 0 \\ - 3 & 2 \end{matrix}] - [\begin{matrix} 3 & 2 \\ 1 & 4 \end{matrix}] \Rightarrow 2 X = [\begin{matrix} 1 - 3 & 0 - 2 \\ - 3 - 1 & 2 - 4 \end{matrix}] \Rightarrow 2 X = [\begin{matrix} - 2 & - 2 \\ - 4 & - 2 \end{matrix}] \Rightarrow X = \frac{1}{2} [\begin{matrix} - 2 & - 2 \\ - 4 & - 2 \end{matrix}] \Rightarrow X = [\begin{matrix} - 1 & - 1 \\ - 2 & - 1 \end{matrix}]$

Page No 4.18:

Question 9:

Find matrices X and Y, if 2X − Y = $[\begin{matrix} 6 & - 6 & 0 \\ - 4 & 2 & 1 \end{matrix}]$ and X + 2Y = $[\begin{matrix} 3 & 2 & 5 \\ - 2 & 1 & - 7 \end{matrix}]$

Answer:

$Given : (2 X - Y) = [\begin{matrix} 6 & - 6 & 0 \\ - 4 & 2 & 1 \end{matrix}] . . . (1) (X + 2 Y) = [\begin{matrix} 3 & 2 & 5 \\ - 2 & 1 & - 7 \end{matrix}] . . . (2) Multiplying eq . (1) by eq . (2), we get 2 (2 X - Y) = 2 [\begin{matrix} 6 & - 6 & 0 \\ - 4 & 2 & 1 \end{matrix}] \Rightarrow 4 X - 2 Y = [\begin{matrix} 12 & - 12 & 0 \\ - 8 & 4 & 2 \end{matrix}] . . . (3) From eq . (3) and eq . (4), we get (4 X - 2 Y) + (X + 2 Y) = [\begin{matrix} 12 & - 12 & 0 \\ - 8 & 4 & 2 \end{matrix}] + [\begin{matrix} 3 & 2 & 5 \\ - 2 & 1 & - 7 \end{matrix}] \Rightarrow 5 X = [\begin{matrix} 12 + 3 & - 12 + 2 & 0 + 5 \\ - 8 - 2 & 4 + 1 & 2 - 7 \end{matrix}] \Rightarrow 5 X = [\begin{matrix} 15 & - 10 & 5 \\ - 10 & 5 & - 5 \end{matrix}] \Rightarrow X = \frac{1}{5} [\begin{matrix} 15 & - 10 & 5 \\ - 10 & 5 & - 5 \end{matrix}] \Rightarrow X = [\begin{matrix} 3 & - 2 & 1 \\ - 2 & 1 & - 1 \end{matrix}] Putting the value of X in eq . (2), we get (X + 2 Y) = [\begin{matrix} 3 & 2 & 5 \\ - 2 & 1 & - 7 \end{matrix}] \Rightarrow [\begin{matrix} 3 & - 2 & 1 \\ - 2 & 1 & - 1 \end{matrix}] + 2 Y = [\begin{matrix} 3 & 2 & 5 \\ - 2 & 1 & - 7 \end{matrix}] \Rightarrow 2 Y = [\begin{matrix} 3 & 2 & 5 \\ - 2 & 1 & - 7 \end{matrix}] - [\begin{matrix} 3 & - 2 & 1 \\ - 2 & 1 & - 1 \end{matrix}] \Rightarrow 2 Y = [\begin{matrix} 3 - 3 & 2 + 2 & 5 - 1 \\ - 2 + 2 & 1 - 1 & - 7 + 1 \end{matrix}] \Rightarrow Y = [\begin{matrix} 0 & 2 & 2 \\ 0 & 0 & - 3 \end{matrix}]$

Page No 4.18:

Question 10:

If X − Y = $[\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}]$ and X + Y = $[\begin{matrix} 3 & 5 & 1 \\ - 1 & 1 & 4 \\ 11 & 8 & 0 \end{matrix}]$ , find X and Y.

Answer:

$Here, X - Y + X + Y = [\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}] + [\begin{matrix} 3 & 5 & 1 \\ - 1 & 1 & 4 \\ 11 & 8 & 0 \end{matrix}] \Rightarrow 2 X = [\begin{matrix} 1 + 3 & 1 + 5 & 1 + 1 \\ 1 - 1 & 1 + 1 & 0 + 4 \\ 1 + 11 & 0 + 8 & 0 + 0 \end{matrix}] \Rightarrow 2 X = [\begin{matrix} 4 & 6 & 2 \\ 0 & 2 & 4 \\ 12 & 8 & 0 \end{matrix}] \Rightarrow X = \frac{1}{2} [\begin{matrix} 4 & 6 & 2 \\ 0 & 2 & 4 \\ 12 & 8 & 0 \end{matrix}] \Rightarrow X = [\begin{matrix} 2 & 3 & 1 \\ 0 & 1 & 2 \\ 6 & 4 & 0 \end{matrix}] Now, (X - Y) - (X + Y) = [\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}] - [\begin{matrix} 3 & 5 & 1 \\ - 1 & 1 & 4 \\ 11 & 8 & 0 \end{matrix}] \Rightarrow X - Y - X - Y = [\begin{matrix} 1 - 3 & 1 - 5 & 1 - 1 \\ 1 + 1 & 1 - 1 & 0 - 4 \\ 1 - 11 & 0 - 8 & 0 - 0 \end{matrix}] \Rightarrow - 2 Y = [\begin{matrix} - 2 & - 4 & 0 \\ 2 & 0 & - 4 \\ - 10 & - 8 & 0 \end{matrix}] \Rightarrow Y = - \frac{1}{2} [\begin{matrix} - 2 & - 4 & 0 \\ 2 & 0 & - 4 \\ - 10 & - 8 & 0 \end{matrix}] \Rightarrow Y = [\begin{matrix} 1 & 2 & 0 \\ - 1 & 0 & 2 \\ 5 & 4 & 0 \end{matrix}] ∴ X = [\begin{matrix} 2 & 3 & 1 \\ 0 & 1 & 2 \\ 6 & 4 & 0 \end{matrix}] and Y = [\begin{matrix} 1 & 2 & 0 \\ - 1 & 0 & 2 \\ 5 & 4 & 0 \end{matrix}]$

Page No 4.18:

Question 11:

Find matrix A, if $[\begin{matrix} 1 & 2 & - 1 \\ 0 & 4 & 9 \end{matrix}]$ + A = $[\begin{matrix} 9 & - 1 & 4 \\ - 2 & 1 & 3 \end{matrix}]$

Answer:

$Here, A = [\begin{matrix} 9 & - 1 & 4 \\ - 2 & 1 & 3 \end{matrix}] - [\begin{matrix} 1 & 2 & - 1 \\ 0 & 4 & 9 \end{matrix}] \Rightarrow A = [\begin{matrix} 9 - 1 & - 1 - 2 & 4 + 1 \\ - 2 - 0 & 1 - 4 & 3 - 9 \end{matrix}] \Rightarrow A = [\begin{matrix} 8 & - 3 & 5 \\ - 2 & - 3 & - 6 \end{matrix}]$

Page No 4.18:

Question 12:

If A = $[\begin{matrix} 9 & 1 \\ 7 & 8 \end{matrix}]$ , B = $[\begin{matrix} 1 & 5 \\ 7 & 12 \end{matrix}]$ , find matrix C such that 5A + 3B + 2C is a null matrix.

Answer:

$Given : 5 A + 3 B + 2 C = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \Rightarrow 5 [\begin{matrix} 9 & 1 \\ 7 & 8 \end{matrix}] + 3 [\begin{matrix} 1 & 5 \\ 7 & 12 \end{matrix}] + 2 C = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \Rightarrow [\begin{matrix} 45 & 5 \\ 35 & 40 \end{matrix}] + [\begin{matrix} 3 & 15 \\ 21 & 36 \end{matrix}] + 2 C = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \Rightarrow [\begin{matrix} 45 + 3 & 5 + 15 \\ 35 + 21 & 40 + 36 \end{matrix}] + 2 C = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \Rightarrow [\begin{matrix} 48 & 20 \\ 56 & 76 \end{matrix}] + 2 C = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \Rightarrow 2 C = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] - [\begin{matrix} 48 & 20 \\ 56 & 76 \end{matrix}] \Rightarrow C = \frac{1}{2} [\begin{matrix} - 48 & - 20 \\ - 56 & - 76 \end{matrix}] \Rightarrow C = [\begin{matrix} - 24 & - 10 \\ - 28 & - 38 \end{matrix}]$

Page No 4.18:

Question 13:

If A = $[\begin{matrix} 2 & - 2 \\ 4 & 2 \\ - 5 & 1 \end{matrix}]$ , B = $[\begin{matrix} 8 & 0 \\ 4 & - 2 \\ 3 & 6 \end{matrix}]$ , find matrix X such that 2A + 3X = 5B.

Answer:

$Given : 2 A + 3 X = 5 B \Rightarrow 2 [\begin{matrix} 2 & - 2 \\ 4 & 2 \\ - 5 & 1 \end{matrix}] + 3 X = 5 [\begin{matrix} 8 & 0 \\ 4 & - 2 \\ 3 & 6 \end{matrix}] \Rightarrow [\begin{matrix} 4 & - 4 \\ 8 & 4 \\ - 10 & 2 \end{matrix}] + 3 X = [\begin{matrix} 40 & 0 \\ 20 & - 10 \\ 15 & 30 \end{matrix}] \Rightarrow 3 X = [\begin{matrix} 40 & 0 \\ 20 & - 10 \\ 15 & 30 \end{matrix}] - [\begin{matrix} 4 & - 4 \\ 8 & 4 \\ - 10 & 2 \end{matrix}] \Rightarrow 3 X = [\begin{matrix} 40 - 4 & 0 + 4 \\ 20 - 8 & - 10 - 4 \\ 15 + 10 & 30 - 2 \end{matrix}] \Rightarrow 3 X = [\begin{matrix} 36 & 4 \\ 12 & - 14 \\ 25 & 28 \end{matrix}] \Rightarrow X = \frac{1}{3} [\begin{matrix} 36 & 4 \\ 12 & - 14 \\ 25 & 28 \end{matrix}] \Rightarrow X = [\begin{matrix} 12 & \frac{4}{3} \\ 4 & \frac{- 14}{3} \\ \frac{25}{3} & \frac{28}{3} \end{matrix}]$

Page No 4.18:

Question 14:

If A = $[\begin{matrix} 1 & - 3 & 2 \\ 2 & 0 & 2 \end{matrix}]$ and, B = $[\begin{matrix} 2 & - 1 & - 1 \\ 1 & 0 & - 1 \end{matrix}]$ , find the matrix C such that A + B + C is zero matrix.

Answer:

$Given : A + B + C = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] \Rightarrow [\begin{matrix} 1 & - 3 & 2 \\ 2 & 0 & 2 \end{matrix}] + [\begin{matrix} 2 & - 1 & - 1 \\ 1 & 0 & - 1 \end{matrix}] + C = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] \Rightarrow [\begin{matrix} 1 + 2 & - 3 - 1 & 2 - 1 \\ 2 + 1 & 0 + 0 & 2 - 1 \end{matrix}] + C = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] \Rightarrow [\begin{matrix} 3 & - 4 & 1 \\ 3 & 0 & 1 \end{matrix}] + C = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] \Rightarrow C = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] - [\begin{matrix} 3 & - 4 & 1 \\ 3 & 0 & 1 \end{matrix}] \Rightarrow C = [\begin{matrix} 0 - 3 & 0 + 4 & 0 - 1 \\ 0 - 3 & 0 - 0 & 0 - 1 \end{matrix}] \Rightarrow C = [\begin{matrix} - 3 & 4 & - 1 \\ - 3 & 0 & - 1 \end{matrix}]$

Page No 4.18:

Question 15:

Find x, y satisfying the matrix equations

(i) $[\begin{matrix} x - y & 2 & - 2 \\ 4 & x & 6 \end{matrix}] + [\begin{matrix} 3 & - 2 & 2 \\ 1 & 0 & - 1 \end{matrix}] = [\begin{matrix} 6 & 0 & 0 \\ 5 & 2 x + y & 5 \end{matrix}]$

(ii) $[\begin{matrix} x & y + 2 & z - 3 \end{matrix}] + [\begin{matrix} y & 4 & 5 \end{matrix}] = [\begin{matrix} 4 & 9 & 12 \end{matrix}]$

(iii) $x [\begin{matrix} 2 \\ 1 \end{matrix}] + y [\begin{matrix} 3 \\ 5 \end{matrix}] + [\begin{matrix} - 8 \\ - 11 \end{matrix}] = 0$

Answer:

$(i) Given : [\begin{matrix} x - y & 2 & - 2 \\ 4 & x & 6 \end{matrix}] + [\begin{matrix} 3 & - 2 & 2 \\ 1 & 0 & - 1 \end{matrix}] = [\begin{matrix} 6 & 0 & 0 \\ 5 & 2 x + y & 5 \end{matrix}] \Rightarrow [\begin{matrix} x - y + 3 & 2 - 2 & - 2 + 2 \\ 4 + 1 & x + 0 & 6 - 1 \end{matrix}] = [\begin{matrix} 6 & 0 & 0 \\ 5 & 2 x + y & 5 \end{matrix}] \Rightarrow [\begin{matrix} x - y + 3 & 0 & 0 \\ 5 & x & 5 \end{matrix}] = [\begin{matrix} 6 & 0 & 0 \\ 5 & 2 x + y & 5 \end{matrix}] \Rightarrow x - y + 3 = 6 \Rightarrow x - y = 6 - 3 \Rightarrow x - y = 3 . . . (1) Also, x = 2 x + y \Rightarrow - x = y . . . (2) Putting the value of y in eq . (1), we get x - (- x) = 3 \Rightarrow 2 x = 3 \Rightarrow x = \frac{3}{2} Putting the value of x in eq . (2), we get - (\frac{3}{2}) = y \Rightarrow y = - \frac{3}{2}$

$(ii) [\begin{matrix} x & y + 2 & z - 3 \end{matrix}] + [\begin{matrix} y & 4 & 5 \end{matrix}] = [\begin{matrix} 4 & 9 & 12 \end{matrix}] \Rightarrow [\begin{matrix} x + y & y + 2 + 4 & z - 3 + 5 \end{matrix}] = [\begin{matrix} 4 & 9 & 12 \end{matrix}] \Rightarrow [\begin{matrix} x + y & y + 6 & z + 2 \end{matrix}] = [\begin{matrix} 4 & 9 & 12 \end{matrix}] ∴ x + y = 4 . . . (1) Also, y + 6 = 9 \Rightarrow y = 3 z + 2 = 12 \Rightarrow z = 10 Putting the value of y in eq . (1), we get x + 3 = 4 \Rightarrow x = 4 - 3 \Rightarrow x = 1 ∴ x = 1, y = 3 and z = 10$

$(iii) Given : x [\begin{matrix} 2 \\ 1 \end{matrix}] + y [\begin{matrix} 3 \\ 5 \end{matrix}] + [\begin{matrix} - 8 \\ - 11 \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] \Rightarrow [\begin{matrix} 2 x + 3 y - 8 \\ x + 5 y - 11 \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] \Rightarrow 2 x + 3 y - 8 = 0 \Rightarrow 2 x + 3 y = 8 . . . (1) Also, x + 5 y - 11 = 0 \Rightarrow x + 5 y = 11 \Rightarrow x = 11 - 5 y . . . (2) Putting the value of x in (1), we get 2 (11 - 5 y) + 3 y = 8 \Rightarrow 22 - 10 y + 3 y = 8 \Rightarrow - 7 y = 8 - 22 \Rightarrow - 7 y = - 14 \Rightarrow y = 2 Putting the value of y in (2), we get x = 11 - 5 (2) \Rightarrow x = 11 - 10 \Rightarrow x = 1 ∴ x = 1 and y = 2$

Page No 4.19:

Question 16:

If 2 $[\begin{matrix} 3 & 4 \\ 5 & x \end{matrix}] + [\begin{matrix} 1 & y \\ 0 & 1 \end{matrix}] = [\begin{matrix} 7 & 0 \\ 10 & 5 \end{matrix}]$ , find x and y.

Answer:

$Given : 2 [\begin{matrix} 3 & 4 \\ 5 & x \end{matrix}] + [\begin{matrix} 1 & y \\ 0 & 1 \end{matrix}] = [\begin{matrix} 7 & 0 \\ 10 & 5 \end{matrix}] \Rightarrow [\begin{matrix} 6 & 8 \\ 10 & 2 x \end{matrix}] + [\begin{matrix} 1 & y \\ 0 & 1 \end{matrix}] = [\begin{matrix} 7 & 0 \\ 10 & 5 \end{matrix}] \Rightarrow [\begin{matrix} 6 + 1 & 8 + y \\ 10 + 0 & 2 x + 1 \end{matrix}] = [\begin{matrix} 7 & 0 \\ 10 & 5 \end{matrix}] \Rightarrow [\begin{matrix} 7 & 8 + y \\ 10 & 2 x + 1 \end{matrix}] = [\begin{matrix} 7 & 0 \\ 10 & 5 \end{matrix}] ∴ 8 + y = 0 \Rightarrow y = - 8 Also, 2 x + 1 = 5 \Rightarrow 2 x = 4 \Rightarrow x = \frac{4}{2} = 2 ∴ x = 2 and y = - 8$

Page No 4.19:

Question 17:

Find the value of λ, a non-zero scalar, if λ $[\begin{matrix} 1 & 0 & 2 \\ 3 & 4 & 5 \end{matrix}] + 2 [\begin{matrix} 1 & 2 & 3 \\ - 1 & - 3 & 2 \end{matrix}] = [\begin{matrix} 4 & 4 & 10 \\ 4 & 2 & 14 \end{matrix}]$

Answer:

$Given : λ [\begin{matrix} 1 & 0 & 2 \\ 3 & 4 & 5 \end{matrix}] + 2 [\begin{matrix} 1 & 2 & 3 \\ - 1 & - 3 & 2 \end{matrix}] = [\begin{matrix} 4 & 4 & 10 \\ 4 & 2 & 14 \end{matrix}] \Rightarrow [\begin{matrix} λ & 0 & 2 λ \\ 3 λ & 4 λ & 5 λ \end{matrix}] + [\begin{matrix} 2 & 4 & 6 \\ - 2 & - 6 & 4 \end{matrix}] = [\begin{matrix} 4 & 4 & 10 \\ 4 & 2 & 14 \end{matrix}] \Rightarrow [\begin{matrix} λ + 2 & 0 + 4 & 2 λ + 6 \\ 3 λ - 2 & 4 λ - 6 & 5 λ + 4 \end{matrix}] = [\begin{matrix} 4 & 4 & 10 \\ 4 & 2 & 14 \end{matrix}] \Rightarrow λ + 2 = 4 \Rightarrow λ = 4 - 2 ∴ λ = 2$

Page No 4.19:

Question 18:

(i) Find a matrix X such that 2A + B + X = O, where
A = $[\begin{matrix} - 1 & 2 \\ 3 & 4 \end{matrix}]$ , B = $[\begin{matrix} 3 & - 2 \\ 1 & 5 \end{matrix}]$
(ii) If A = $[\begin{matrix} 8 & 0 \\ 4 & - 2 \\ 3 & 6 \end{matrix}]$ and B = $[\begin{matrix} 2 & - 2 \\ 4 & 2 \\ - 5 & 1 \end{matrix}]$ , then find the matrix X of order 3 × 2 such that 2A + 3X = 5B.

Answer:

$(i) 2 A + B + X = 0 \Rightarrow 2 [\begin{matrix} - 1 & 2 \\ 3 & 4 \end{matrix}] + [\begin{matrix} 3 & - 2 \\ 1 & 5 \end{matrix}] + X = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \Rightarrow [\begin{matrix} - 2 & 4 \\ 6 & 8 \end{matrix}] + [\begin{matrix} 3 & - 2 \\ 1 & 5 \end{matrix}] + X = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \Rightarrow [\begin{matrix} - 2 + 3 & 4 - 2 \\ 6 + 1 & 8 + 5 \end{matrix}] + X = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \Rightarrow [\begin{matrix} 1 & 2 \\ 7 & 13 \end{matrix}] + X = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] ∴ X = [\begin{matrix} - 1 & - 2 \\ - 7 & - 13 \end{matrix}]$

$(ii) 2 A + 3 X = 5 B \Rightarrow 2 [\begin{matrix} 8 & 0 \\ 4 & - 2 \\ 3 & 6 \end{matrix}] + 3 X = 5 [\begin{matrix} 2 & - 2 \\ 4 & 2 \\ - 5 & 1 \end{matrix}] \Rightarrow [\begin{matrix} 16 & 0 \\ 8 & - 4 \\ 6 & 12 \end{matrix}] + 3 X = [\begin{matrix} 10 & - 10 \\ 20 & 10 \\ - 25 & 5 \end{matrix}] \Rightarrow 3 X = [\begin{matrix} 10 & - 10 \\ 20 & 10 \\ - 25 & 5 \end{matrix}] - [\begin{matrix} 16 & 0 \\ 8 & - 4 \\ 6 & 12 \end{matrix}] \Rightarrow 3 X = [\begin{matrix} 10 - 16 & - 10 - 0 \\ 20 - 8 & 10 + 4 \\ - 25 - 6 & 5 - 12 \end{matrix}] \Rightarrow 3 X = [\begin{matrix} - 6 & - 10 \\ 12 & 14 \\ - 31 & - 7 \end{matrix}] \Rightarrow X = \frac{1}{3} [\begin{matrix} - 6 & - 10 \\ 12 & 14 \\ - 31 & - 7 \end{matrix}] ∴ X = [\begin{matrix} - 2 & \frac{- 10}{3} \\ 4 & \frac{14}{3} \\ \frac{- 31}{3} & \frac{- 7}{3} \end{matrix}]$

Page No 4.19:

Question 19:

Find x, y, z and t, if
(i) $3 [\begin{matrix} x & y \\ z & t \end{matrix}] = [\begin{matrix} x & 6 \\ - 1 & 2 t \end{matrix}] + [\begin{matrix} 4 & x + y \\ z + t & 3 \end{matrix}]$

(ii) $2 [\begin{matrix} x & 5 \\ 7 & y - 3 \end{matrix}] + [\begin{matrix} 3 & 4 \\ 1 & 2 \end{matrix}] = [\begin{matrix} 7 & 14 \\ 15 & 14 \end{matrix}]$

Answer:

$(i) 3 [\begin{matrix} x & y \\ z & t \end{matrix}] = [\begin{matrix} x & 6 \\ - 1 & 2 t \end{matrix}] + [\begin{matrix} 4 & x + y \\ z + t & 3 \end{matrix}] \Rightarrow [\begin{matrix} 3 x & 3 y \\ 3 z & 3 t \end{matrix}] = [\begin{matrix} x + 4 & 6 + x + y \\ - 1 + z + t & 2 t + 3 \end{matrix}] ∴ 3 x = x + 4 \Rightarrow 3 x - x = 4 \Rightarrow 2 x = 4 \Rightarrow x = 2 Also, 3 y = 6 + x + y \Rightarrow 3 y - y = 6 + x \Rightarrow 2 y = 6 + x . . . (1) Putting the value of x in eq . (1), we get 2 y = 6 + 2 \Rightarrow 2 y = 8 \Rightarrow y = 4 Now, 3 t = 2 t + 3 \Rightarrow 3 t - 2 t = 3 \Rightarrow t = 3 3 z = - 1 + z + t \Rightarrow 3 z - z = - 1 + t \Rightarrow 2 z = - 1 + t . . . (2) Putting the value of t in eq . (2), we get 2 z = - 1 + 3 \Rightarrow 2 z = 2 \Rightarrow z = 1 ∴ x = 2, y = 4, z = 1 and t = 3$

$(ii) 2 [\begin{matrix} x & 5 \\ 7 & y - 3 \end{matrix}] + [\begin{matrix} 3 & 4 \\ 1 & 2 \end{matrix}] = [\begin{matrix} 7 & 14 \\ 15 & 14 \end{matrix}] \Rightarrow [\begin{matrix} 2 x & 10 \\ 14 & 2 y - 6 \end{matrix}] + [\begin{matrix} 3 & 4 \\ 1 & 2 \end{matrix}] = [\begin{matrix} 7 & 14 \\ 15 & 14 \end{matrix}] \Rightarrow [\begin{matrix} 2 x + 3 & 10 + 4 \\ 14 + 1 & 2 y - 6 + 2 \end{matrix}] = [\begin{matrix} 7 & 14 \\ 15 & 14 \end{matrix}] \Rightarrow [\begin{matrix} 2 x + 3 & 14 \\ 15 & 2 y - 4 \end{matrix}] = [\begin{matrix} 7 & 14 \\ 15 & 14 \end{matrix}] ∴ 2 x + 3 = 7 \Rightarrow 2 x = 4 \Rightarrow x = 2 Also, 2 y - 4 = 14 \Rightarrow 2 y = 18 \Rightarrow y = 9$

Page No 4.19:

Question 20:

If X and Y are 2 × 2 matrices, then solve the following matrix equations for X and Y.

$2 X + 3 Y = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}], 3 X + 2 Y = [\begin{matrix} - 2 & 2 \\ 1 & - 5 \end{matrix}]$

Answer:

We have,
$3 (2 X + 3 Y) - 2 (3 X + 2 Y) = 3 [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}] - 2 [\begin{matrix} - 2 & 2 \\ 1 & - 5 \end{matrix}] \Rightarrow 6 X + 9 Y - 6 X - 4 Y = [\begin{matrix} 6 & 9 \\ 12 & 0 \end{matrix}] + [\begin{matrix} 4 & - 4 \\ - 2 & 10 \end{matrix}] \Rightarrow 5 Y = [\begin{matrix} 6 + 4 & 9 - 4 \\ 12 - 2 & 0 + 10 \end{matrix}] \Rightarrow Y = \frac{1}{5} [\begin{matrix} 10 & 5 \\ 10 & 10 \end{matrix}] \Rightarrow Y = [\begin{matrix} 2 & 1 \\ 2 & 2 \end{matrix}] . . . (1)$

Also,
$2 (2 X + 3 Y) - 3 (3 X + 2 Y) = 2 [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}] - 3 [\begin{matrix} - 2 & 2 \\ 1 & - 5 \end{matrix}] \Rightarrow 4 X + 6 Y - 9 X - 6 Y = [\begin{matrix} 4 & 6 \\ 8 & 0 \end{matrix}] + [\begin{matrix} 6 & - 6 \\ - 3 & 15 \end{matrix}] \Rightarrow - 5 X = [\begin{matrix} 6 + 4 & 6 - 6 \\ 8 - 3 & 0 + 15 \end{matrix}] \Rightarrow X = \frac{1}{- 5} [\begin{matrix} 10 & 0 \\ 5 & 15 \end{matrix}] \Rightarrow X = [\begin{matrix} - 2 & 0 \\ - 1 & - 3 \end{matrix}] . . . (2)$

From (1) and (2), we get

$X = [\begin{matrix} - 2 & 0 \\ - 1 & - 3 \end{matrix}] and Y = [\begin{matrix} 2 & 1 \\ 2 & 2 \end{matrix}]$ .

Page No 4.19:

Question 21:

In a certain city there are 30 colleges. Each college has 15 peons, 6 clerks, 1 typist and 1 section officer. Express the given information as a column matrix. Using scalar multiplication, find the total number of posts of each kind in all the colleges.

Answer:

Number of different types of posts in any college is given by

X = $[\begin{matrix} 15 \\ 6 \\ 1 \\ 1 \end{matrix}]$

Total number of posts of each kind in all the colleges = 30X

= 30 $[\begin{matrix} 15 \\ 6 \\ 1 \\ 1 \end{matrix}]$

= $[\begin{matrix} 450 \\ 180 \\ 30 \\ 30 \end{matrix}]$

Page No 4.19:

Question 22:

The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves Rs 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value?

Answer:

Let the monthly incomes of Aryan and Babban be 3x and 4x, respectively.

Suppose their monthly expenditures are 5y and 7y, respectively.

Since each saves Rs 15,000 per month,

$Monthly saving of Aryan : 3 x - 5 y = 15, 000 Monthly saving of Babban : 4 x - 7 y = 15, 000$

The above system of equations can be written in the matrix form as follows:

$[\begin{matrix} 3 & - 5 \\ 4 & - 7 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} 15000 \\ 15000 \end{matrix}]$

or,
AX = B, where $A = [\begin{matrix} 3 & - 5 \\ 4 & - 7 \end{matrix}], X = [\begin{matrix} x \\ y \end{matrix}] and B = [\begin{matrix} 15000 \\ 15000 \end{matrix}]$

Now,

$|A| = |\begin{matrix} 3 & - 5 \\ 4 & - 7 \end{matrix}| = - 21 - (- 20) = - 1$

Adj A= ${[\begin{matrix} - 7 & - 4 \\ 5 & 3 \end{matrix}]}^{T} = [\begin{matrix} - 7 & 5 \\ - 4 & 3 \end{matrix}]$

So, $A^{- 1} = \frac{1}{|A|} a d j A = - 1 [\begin{matrix} - 7 & 5 \\ - 4 & 3 \end{matrix}] = [\begin{matrix} 7 & - 5 \\ 4 & - 3 \end{matrix}]$

$∴ X = A^{- 1} B \Rightarrow [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} 7 & - 5 \\ 4 & - 3 \end{matrix}] [\begin{matrix} 15000 \\ 15000 \end{matrix}] \Rightarrow [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} 105000 - 75000 \\ 60000 - 45000 \end{matrix}] \Rightarrow [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} 30000 \\ 15000 \end{matrix}] \Rightarrow x = 30, 000 and y = 15, 000$

Therefore,

Monthly income of Aryan = $3 \times Rs 30, 000 = Rs 90, 000$

Monthly income of Babban = $4 \times Rs 30, 000 = Rs 1, 20, 000$

From this problem, we are encouraged to understand the power of savings. We should save certain part of our monthly income for the future.

Page No 4.41:

Question 1:

Compute the indicated products:
(i) $[\begin{matrix} a & b \\ - b & a \end{matrix}] [\begin{matrix} a & - b \\ b & a \end{matrix}]$

(ii) $[\begin{matrix} 1 & - 2 \\ 2 & 3 \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ - 3 & 2 & - 1 \end{matrix}]$

(iii) $[\begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{matrix}] [\begin{matrix} 1 & - 3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{matrix}]$

Answer:

$(i) [\begin{matrix} a & b \\ - b & a \end{matrix}] [\begin{matrix} a & - b \\ b & a \end{matrix}] \Rightarrow [\begin{matrix} a \times a + b \times b & a \times (- b) + b \times a \\ (- b) \times a + a \times b & (- b) \times (- b) + a \times a \end{matrix}] \Rightarrow [\begin{matrix} a^{2} + b^{2} & - a b + a b \\ - a b + a b & b^{2} + a^{2} \end{matrix}] \Rightarrow [\begin{matrix} a^{2} + b^{2} & 0 \\ 0 & a^{2} + b^{2} \end{matrix}]$

$(i i) [\begin{matrix} 1 & - 2 \\ 2 & 3 \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ - 3 & 2 & - 1 \end{matrix}] \Rightarrow [\begin{matrix} 1 \times 1 + (- 2) \times (- 3) & 1 \times 2 + (- 2) \times 2 & 1 \times 3 + (- 2) \times (- 1) \\ 2 \times 1 + 3 \times (- 3) & 2 \times 2 + 3 \times 2 & 2 \times 3 + 3 \times (- 1) \end{matrix}] \Rightarrow [\begin{matrix} 1 + 6 & 2 - 4 & 3 + 2 \\ 2 - 9 & 4 + 6 & 6 - 3 \end{matrix}] \Rightarrow [\begin{matrix} 7 & - 2 & 5 \\ - 7 & 10 & 3 \end{matrix}]$

$(i i i) [\begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{matrix}] [\begin{matrix} 1 & - 3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{matrix}] \Rightarrow [\begin{matrix} 2 \times 1 + 3 \times 0 + 4 \times 3 & 2 \times (- 3) + 3 \times 2 + 4 \times 0 & 2 \times 5 + 3 \times 4 + 4 \times 5 \\ 3 \times 1 + 4 \times 0 + 5 \times 3 & 3 \times (- 3) + 4 \times 2 + 5 \times 0 & 3 \times 5 + 4 \times 4 + 5 \times 5 \\ 4 \times 1 + 5 \times 0 + 6 \times 3 & 4 \times (- 3) + 5 \times 2 + 6 \times 0 & 4 \times 5 + 5 \times 4 + 6 \times 5 \end{matrix}] \Rightarrow [\begin{matrix} 2 + 0 + 12 & - 6 + 6 + 0 & 10 + 12 + 20 \\ 3 + 0 + 15 & - 9 + 8 + 0 & 15 + 16 + 25 \\ 4 + 0 + 18 & - 12 + 10 + 0 & 20 + 20 + 30 \end{matrix}] \Rightarrow [\begin{matrix} 14 & 0 & 42 \\ 18 & - 1 & 56 \\ 22 & - 2 & 70 \end{matrix}]$

Page No 4.41:

Question 2:

Show that AB ≠ BA in each of the following cases:
(i) $A = [\begin{matrix} 5 & - 1 \\ 6 & 7 \end{matrix}] and B = [\begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix}]$

(ii) $A = [\begin{matrix} - 1 & 1 & 0 \\ 0 & - 1 & 1 \\ 2 & 3 & 4 \end{matrix}] and B = [\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{matrix}]$

(iii) $A = [\begin{matrix} 1 & 3 & 0 \\ 1 & 1 & 0 \\ 4 & 1 & 0 \end{matrix}] and B = [\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 5 & 1 \end{matrix}]$

Answer:

$(i)$
$A B = [\begin{matrix} 5 & - 1 \\ 6 & 7 \end{matrix}] [\begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix}] \Rightarrow A B = [\begin{matrix} 10 - 3 & 5 - 4 \\ 12 + 21 & 6 + 28 \end{matrix}] \Rightarrow A B = [\begin{matrix} 7 & 1 \\ 33 & 34 \end{matrix}] . . . (1) Also, B A = [\begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix}] [\begin{matrix} 5 & - 1 \\ 6 & 7 \end{matrix}] \Rightarrow B A = [\begin{matrix} 10 + 6 & - 2 + 7 \\ 15 + 24 & - 3 + 28 \end{matrix}] \Rightarrow B A = [\begin{matrix} 16 & 5 \\ 39 & 25 \end{matrix}] . . . (2) ∴ AB ≠ BA (From eqs . (1) and (2))$

$(ii) A B = [\begin{matrix} - 1 & 1 & 0 \\ 0 & - 1 & 1 \\ 2 & 3 & 4 \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{matrix}] \Rightarrow A B = [\begin{matrix} - 1 + 0 + 0 & - 2 + 1 + 0 & - 3 + 0 + 0 \\ 0 + 0 + 1 & 0 - 1 + 1 & 0 + 0 + 0 \\ 2 + 0 + 4 & 4 + 3 + 4 & 6 + 0 + 0 \end{matrix}] \Rightarrow A B = [\begin{matrix} - 1 & - 1 & - 3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{matrix}] . . . (1) Also, B A = [\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{matrix}] [\begin{matrix} - 1 & 1 & 0 \\ 0 & - 1 & 1 \\ 2 & 3 & 4 \end{matrix}] \Rightarrow B A = [\begin{matrix} - 1 + 0 + 6 & 1 - 2 + 9 & 0 + 2 + 12 \\ 0 + 0 + 0 & 0 - 1 + 0 & 0 + 1 + 0 \\ - 1 + 0 + 0 & 1 - 1 + 0 & 0 + 1 + 0 \end{matrix}] \Rightarrow B A = [\begin{matrix} 5 & 8 & 14 \\ 0 & - 1 & 1 \\ - 1 & 0 & 1 \end{matrix}] . . . (2) ∴ AB ≠ BA (From eqs . (1) and (2))$

$(iii) A B = [\begin{matrix} 1 & 3 & 0 \\ 1 & 1 & 0 \\ 4 & 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 5 & 1 \end{matrix}] \Rightarrow A B = [\begin{matrix} 0 + 3 + 0 & 1 + 0 + 0 & 0 + 0 + 0 \\ 0 + 1 + 0 & 1 + 0 + 0 & 0 + 0 + 0 \\ 0 + 1 + 0 & 4 + 0 + 0 & 0 + 0 + 0 \end{matrix}] \Rightarrow A B = [\begin{matrix} 3 & 1 & 0 \\ 1 & 1 & 0 \\ 1 & 4 & 0 \end{matrix}] . . . (1) Also, B A = [\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 5 & 1 \end{matrix}] [\begin{matrix} 1 & 3 & 0 \\ 1 & 1 & 0 \\ 4 & 1 & 0 \end{matrix}] \Rightarrow B A = [\begin{matrix} 0 + 1 + 0 & 0 + 1 + 0 & 0 + 0 + 0 \\ 1 + 0 + 0 & 3 + 0 + 0 & 0 + 0 + 0 \\ 0 + 5 + 4 & 0 + 5 + 1 & 0 + 0 + 0 \end{matrix}] \Rightarrow B A = [\begin{matrix} 1 & 1 & 0 \\ 1 & 3 & 0 \\ 9 & 6 & 0 \end{matrix}] . . . (2) ∴ AB ≠ BA (From eqs . (1) and (2))$

Page No 4.41:

Question 3:

Compute the products AB and BA whichever exists in each of the following cases:
(i) $A = [\begin{matrix} 1 & - 2 \\ 2 & 3 \end{matrix}] and B = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{matrix}]$

(ii) $A = [\begin{matrix} 3 & 2 \\ - 1 & 0 \\ - 1 & 1 \end{matrix}] and B = [\begin{matrix} 4 & 5 & 6 \\ 0 & 1 & 2 \end{matrix}]$
(iii) A = [1 −1 2 3] and $B = [\begin{matrix} 0 \\ 1 \\ 3 \\ 2 \end{matrix}]$

(iv) [a, b] $[\begin{matrix} c \\ d \end{matrix}]$ + [a, b, c, d] $[\begin{matrix} a \\ b \\ c \\ d \end{matrix}]$

Answer:

$(i) A B = [\begin{matrix} 1 & - 2 \\ 2 & 3 \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{matrix}] \Rightarrow A B = [\begin{matrix} 1 - 4 & 2 - 6 & 3 - 2 \\ 2 + 6 & 4 + 9 & 6 + 3 \end{matrix}] \Rightarrow A B = [\begin{matrix} - 3 & - 4 & 1 \\ 8 & 13 & 9 \end{matrix}]$

Since the number of columns in B is greater then the number of rows in A, BA does not exists.

$(ii) A B = [\begin{matrix} 3 & 2 \\ - 1 & 0 \\ - 1 & 1 \end{matrix}] [\begin{matrix} 4 & 5 & 6 \\ 0 & 1 & 2 \end{matrix}] \Rightarrow A B = [\begin{matrix} 12 + 0 & 15 + 2 & 18 + 4 \\ - 4 + 0 & - 5 + 0 & - 6 + 0 \\ - 4 + 0 & - 5 + 1 & - 6 + 2 \end{matrix}] \Rightarrow A B = [\begin{matrix} 12 & 17 & 22 \\ - 4 & - 5 & - 6 \\ - 4 & - 4 & - 4 \end{matrix}] Also, B A = [\begin{matrix} 4 & 5 & 6 \\ 0 & 1 & 2 \end{matrix}] [\begin{matrix} 3 & 2 \\ - 1 & 0 \\ - 1 & 1 \end{matrix}] \Rightarrow B A = [\begin{matrix} 12 - 5 - 6 & 8 + 0 + 6 \\ 0 - 1 - 2 & 0 + 0 + 2 \end{matrix}] \Rightarrow B A = [\begin{matrix} 1 & 14 \\ - 3 & 2 \end{matrix}]$

$(iii) A B = [\begin{matrix} 1 & - 1 & 2 & 3 \end{matrix}] [\begin{matrix} 0 \\ 1 \\ 3 \\ 2 \end{matrix}] \Rightarrow A B = [\begin{matrix} 0 + (- 1) + 6 + 6 \end{matrix}] \Rightarrow A B = [\begin{matrix} 11 \end{matrix}] Also, B A = [\begin{matrix} 0 \\ 1 \\ 3 \\ 2 \end{matrix}] [\begin{matrix} 1 & - 1 & 2 & 3 \end{matrix}] \Rightarrow B A = [\begin{matrix} 0 & 0 & 0 & 0 \\ 1 & - 1 & 2 & 3 \\ 3 & - 3 & 6 & 9 \\ 2 & - 2 & 4 & 6 \end{matrix}]$

$(iv) [\begin{matrix} a & b \end{matrix}] [\begin{matrix} c \\ d \end{matrix}] + [\begin{matrix} a & b & c & d \end{matrix}] [\begin{matrix} a \\ b \\ c \\ d \end{matrix}] \Rightarrow [\begin{matrix} a c + b d \end{matrix}] + [\begin{matrix} a^{2} + b^{2} + c^{2} + d^{2} \end{matrix}] [\begin{matrix} a^{2} + b^{2} + c^{2} + d^{2} + a c + b d \end{matrix}]$

Page No 4.41:

Question 4:

Show that AB ≠ BA in each of the following cases:
(i) $A = [\begin{matrix} 1 & 3 & - 1 \\ 2 & - 1 & - 1 \\ 3 & 0 & - 1 \end{matrix}] and B = [\begin{matrix} - 2 & 3 & - 1 \\ - 1 & 2 & - 1 \\ - 6 & 9 & - 4 \end{matrix}]$

(ii) $A = [\begin{matrix} 10 & - 4 & - 1 \\ - 11 & 5 & 0 \\ 9 & - 5 & 1 \end{matrix}] and B = [\begin{matrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{matrix}]$

Answer:

$(i) A B = [\begin{matrix} 1 & 3 & - 1 \\ 2 & - 1 & - 1 \\ 3 & 0 & - 1 \end{matrix}] [\begin{matrix} - 2 & 3 & - 1 \\ - 1 & 2 & - 1 \\ - 6 & 9 & - 4 \end{matrix}] \Rightarrow A B = [\begin{matrix} - 2 - 3 + 6 & 3 + 6 - 9 & - 1 - 3 + 4 \\ - 4 + 1 + 6 & 6 - 2 - 9 & - 2 + 1 + 4 \\ - 6 - 0 + 6 & 9 + 0 - 9 & - 3 - 0 + 4 \end{matrix}] \Rightarrow A B = [\begin{matrix} 1 & 0 & 0 \\ 3 & - 5 & 3 \\ 0 & 0 & 1 \end{matrix}] . . . (1) Also, B A = [\begin{matrix} - 2 & 3 & - 1 \\ - 1 & 2 & - 1 \\ - 6 & 9 & - 4 \end{matrix}] [\begin{matrix} 1 & 3 & - 1 \\ 2 & - 1 & - 1 \\ 3 & 0 & - 1 \end{matrix}] \Rightarrow B A = [\begin{matrix} - 2 + 6 - 3 & - 6 - 3 + 0 & 2 - 3 + 1 \\ - 1 + 4 - 3 & - 3 - 2 + 0 & 1 - 2 + 1 \\ - 6 + 18 - 12 & - 18 - 9 + 0 & 6 - 9 + 4 \end{matrix}] \Rightarrow B A = [\begin{matrix} 1 & - 9 & 0 \\ 0 & - 5 & 0 \\ 0 & - 27 & 1 \end{matrix}] . . . (2) ∴ AB ≠ BA (From eqs . (1) and (2))$

Page No 4.41:

Question 5:

Evaluate the following:
(i) $([\begin{matrix} 1 & 3 \\ - 1 & - 4 \end{matrix}] + [\begin{matrix} 3 & - 2 \\ - 1 & 1 \end{matrix}]) [\begin{matrix} 1 & 3 & 5 \\ 2 & 4 & 6 \end{matrix}]$

(ii) $[1 2 3] [\begin{matrix} 1 & 0 & 2 \\ 2 & 0 & 1 \\ 0 & 1 & 2 \end{matrix}] [\begin{matrix} 2 \\ 4 \\ 6 \end{matrix}]$

(iii) $[\begin{matrix} 1 & - 1 \\ 0 & 2 \\ 2 & 3 \end{matrix}] ([\begin{matrix} 1 & 0 & 2 \\ 2 & 0 & 1 \end{matrix}] - [\begin{matrix} 0 & 1 & 2 \\ 1 & 0 & 2 \end{matrix}])$

Answer:

$(i) ([\begin{matrix} 1 & 3 \\ - 1 & - 4 \end{matrix}] + [\begin{matrix} 3 & - 2 \\ - 1 & 1 \end{matrix}]) [\begin{matrix} 1 & 3 & 5 \\ 2 & 4 & 6 \end{matrix}] \Rightarrow ([\begin{matrix} 1 + 3 & 3 - 2 \\ - 1 - 1 & - 4 + 1 \end{matrix}]) [\begin{matrix} 1 & 3 & 5 \\ 2 & 4 & 6 \end{matrix}] \Rightarrow [\begin{matrix} 4 & 1 \\ - 2 & - 3 \end{matrix}] [\begin{matrix} 1 & 3 & 5 \\ 2 & 4 & 6 \end{matrix}] \Rightarrow [\begin{matrix} 4 + 2 & 12 + 4 & 20 + 6 \\ - 2 - 6 & - 6 - 12 & - 10 - 18 \end{matrix}] \Rightarrow [\begin{matrix} 6 & 16 & 26 \\ - 8 & - 18 & - 28 \end{matrix}]$

$(i i) [\begin{matrix} 1 & 2 & 3 \end{matrix}] [\begin{matrix} 1 & 0 & 2 \\ 2 & 0 & 1 \\ 0 & 1 & 2 \end{matrix}] [\begin{matrix} 2 \\ 4 \\ 6 \end{matrix}] \Rightarrow [\begin{matrix} 1 + 4 + 0 & 0 + 0 + 3 & 2 + 2 + 6 \end{matrix}] [\begin{matrix} 2 \\ 4 \\ 6 \end{matrix}] \Rightarrow [\begin{matrix} 5 & 3 & 10 \end{matrix}] [\begin{matrix} 2 \\ 4 \\ 6 \end{matrix}] \Rightarrow [\begin{matrix} 10 + 12 + 60 \end{matrix}] \Rightarrow [\begin{matrix} 82 \end{matrix}]$

$(i i i) [\begin{matrix} 1 & - 1 \\ 0 & 2 \\ 2 & 3 \end{matrix}] ([\begin{matrix} 1 & 0 & 2 \\ 2 & 0 & 1 \end{matrix}] - [\begin{matrix} 0 & 1 & 2 \\ 1 & 0 & 2 \end{matrix}]) \Rightarrow [\begin{matrix} 1 & - 1 \\ 0 & 2 \\ 2 & 3 \end{matrix}] [\begin{matrix} 1 - 0 & 0 - 1 & 2 - 2 \\ 2 - 1 & 0 - 0 & 1 - 2 \end{matrix}] \Rightarrow [\begin{matrix} 1 & - 1 \\ 0 & 2 \\ 2 & 3 \end{matrix}] [\begin{matrix} 1 & - 1 & 0 \\ 1 & 0 & - 1 \end{matrix}] \Rightarrow [\begin{matrix} 1 - 1 & - 1 + 0 & 0 + 1 \\ 0 + 2 & 0 + 0 & 0 - 2 \\ 2 + 3 & - 2 + 0 & 0 - 3 \end{matrix}] \Rightarrow [\begin{matrix} 0 & - 1 & 1 \\ 2 & 0 & - 2 \\ 5 & - 2 & - 3 \end{matrix}]$

Page No 4.41:

Question 6:

If A = $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ , B = $[\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}]$ and C = $[\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}]$ , then show that A² = B² = C² = I₂.

Answer:

$Here, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 + 0 & 0 + 0 \\ 0 + 0 & 0 + 1 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] . . . (1) B^{2} = B B \Rightarrow B^{2} = [\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}] [\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}] \Rightarrow B^{2} = [\begin{matrix} 1 + 0 & 0 - 0 \\ 0 - 0 & 0 + 1 \end{matrix}] \Rightarrow B^{2} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] . . . (2) C^{2} = C C \Rightarrow B^{2} = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] \Rightarrow B^{2} = [\begin{matrix} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{matrix}] \Rightarrow B^{2} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] . . . (3) We know, I_{2} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] . . . (4)$

$\Rightarrow A^{2} = B^{2} = C^{2} = I_{2} [From eqs . (1), (2), (3) and (4)]$

Page No 4.42:

Question 7:

If A = $[\begin{matrix} 2 & - 1 \\ 3 & 2 \end{matrix}]$ and B = $[\begin{matrix} 0 & 4 \\ - 1 & 7 \end{matrix}]$ , find 3A² − 2B + I

Answer:

$Given : A = [\begin{matrix} 2 & - 1 \\ 3 & 2 \end{matrix}] Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 2 & - 1 \\ 3 & 2 \end{matrix}] [\begin{matrix} 2 & - 1 \\ 3 & 2 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 4 - 3 & - 2 - 2 \\ 6 + 6 & - 3 + 4 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 & - 4 \\ 12 & 1 \end{matrix}] 3 A^{2} - 2 B + I \Rightarrow 3 A^{2} - 2 B + I = 3 [\begin{matrix} 1 & - 4 \\ 12 & 1 \end{matrix}] - 2 [\begin{matrix} 0 & 4 \\ - 1 & 7 \end{matrix}] + [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow 3 A^{2} - 2 B + I = [\begin{matrix} 3 & - 12 \\ 36 & 3 \end{matrix}] - [\begin{matrix} 0 & 8 \\ - 2 & 14 \end{matrix}] + [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow 3 A^{2} - 2 B + I = [\begin{matrix} 3 - 0 + 1 & - 12 - 8 + 0 \\ 36 + 2 + 0 & 3 - 14 + 1 \end{matrix}] \Rightarrow 3 A^{2} - 2 B + I = [\begin{matrix} 4 & - 20 \\ 38 & - 10 \end{matrix}]$

Page No 4.42:

Question 8:

If A = $[\begin{matrix} 4 & 2 \\ - 1 & 1 \end{matrix}]$ , prove that (A − 2I) (A − 3I) = O

Answer:

$Given : (A - 2 I) (A - 3 I) \Rightarrow (A - 2 I) (A - 3 I) = ([\begin{matrix} 4 & 2 \\ - 1 & 1 \end{matrix}] - 2 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]) ([\begin{matrix} 4 & 2 \\ - 1 & 1 \end{matrix}] - 3 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]) \Rightarrow (A - 2 I) (A - 3 I) = ([\begin{matrix} 4 & 2 \\ - 1 & 1 \end{matrix}] - [\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}]) ([\begin{matrix} 4 & 2 \\ - 1 & 1 \end{matrix}] - [\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}]) \Rightarrow (A - 2 I) (A - 3 I) = [\begin{matrix} 4 - 2 & 2 - 0 \\ - 1 - 0 & 1 - 2 \end{matrix}] [\begin{matrix} 4 - 3 & 2 - 0 \\ - 1 - 0 & 1 - 3 \end{matrix}] \Rightarrow (A - 2 I) (A - 3 I) = [\begin{matrix} 2 & 2 \\ - 1 & - 1 \end{matrix}] [\begin{matrix} 1 & 2 \\ - 1 & - 2 \end{matrix}] \Rightarrow (A - 2 I) (A - 3 I) = [\begin{matrix} 2 - 2 & 4 - 4 \\ - 1 + 1 & - 2 + 2 \end{matrix}] \Rightarrow (A - 2 I) (A - 3 I) = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \Rightarrow (A - 2 I) (A - 3 I) = 0 Hence proved .$

Page No 4.42:

Question 9:

If A = $[\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}]$ , show that A² = $[\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}]$ and A³ = $[\begin{matrix} 1 & 3 \\ 0 & 1 \end{matrix}]$ .

Answer:

$Given : A = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 + 0 & 1 + 1 \\ 0 + 0 & 0 + 1 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}] A^{3} = A^{2} A \Rightarrow A^{3} = [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] \Rightarrow A^{3} = [\begin{matrix} 1 + 0 & 1 + 2 \\ 0 + 0 & 0 + 1 \end{matrix}] \Rightarrow A^{3} = [\begin{matrix} 1 & 3 \\ 0 & 1 \end{matrix}]$

Hence proved.

Page No 4.42:

Question 10:

If A = $[\begin{matrix} a b & b^{2} \\ - a^{2} & - a b \end{matrix}]$ , show that A² = O

Answer:

$Given : A = [\begin{matrix} a b & b^{2} \\ - a^{2} & - a b \end{matrix}] Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} a b & b^{2} \\ - a^{2} & - a b \end{matrix}] [\begin{matrix} a b & b^{2} \\ - a^{2} & - a b \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} a^{2} b^{2} - a^{2} b^{2} & a b^{3} - a b^{3} \\ - a^{3} b + a^{3} b & - a^{2} b^{2} + a^{2} b^{2} \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \Rightarrow A^{2} = 0 Hence proved .$

Page No 4.42:

Question 11:

If A = $[\begin{matrix} \cos 2 θ & \sin 2 θ \\ - \sin 2 θ & \cos 2 θ \end{matrix}]$ , find A².

Answer:

$Given : A = [\begin{matrix} \cos 2 θ & \sin 2 θ \\ - \sin 2 θ & c o s 2 θ \end{matrix}] Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} \cos 2 θ & \sin 2 θ \\ - \sin 2 θ & \cos 2 θ \end{matrix}] [\begin{matrix} \cos 2 θ & \sin 2 θ \\ - \sin 2 θ & \cos 2 θ \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} \cos^{2} (2 θ) - \sin^{2} (2 θ) & \cos (2 θ) \sin 2 θ + \cos (2 θ) \sin 2 θ \\ - \cos (2 θ) \sin 2 θ - \sin 2 θ \cos 2 θ & - \sin^{2} (2 θ) + \cos^{2} (2 θ) \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} \cos (2 \times 2 θ) & 2 \sin 2 θ \cos 2 θ \\ - 2 \sin 2 θ \cos (2 θ) & \cos (2 \times 2 θ) \end{matrix}] [∵ \cos^{2} θ - \sin^{2} θ = \cos 2 θ] \Rightarrow A^{2} = [\begin{matrix} \cos 4 θ & \sin (2 \times 2 θ) \\ - \sin (2 \times 2 θ) & \cos 4 θ \end{matrix}] [∵ \sin 2 θ = 2 \sin θ \cos θ] \Rightarrow A^{2} = [\begin{matrix} \cos 4 θ & \sin 4 θ \\ - \sin 4 θ & \cos 4 θ \end{matrix}]$

Page No 4.42:

Question 12:

If A = $[\begin{matrix} 2 & - 3 & - 5 \\ - 1 & 4 & 5 \\ 1 & - 3 & - 4 \end{matrix}]$ and B = $[\begin{matrix} - 1 & 3 & 5 \\ 1 & - 3 & - 5 \\ - 1 & 3 & 5 \end{matrix}]$ , show that AB = BA = O₃_×3.

Answer:

$Here, A B = [\begin{matrix} 2 & - 3 & - 5 \\ - 1 & 4 & 5 \\ 1 & - 3 & - 4 \end{matrix}] [\begin{matrix} - 1 & 3 & 5 \\ 1 & - 3 & - 5 \\ - 1 & 3 & 5 \end{matrix}] \Rightarrow A B = [\begin{matrix} - 2 - 3 + 5 & 6 + 9 - 15 & 10 + 15 - 25 \\ 1 + 4 - 5 & - 3 - 12 + 15 & - 5 - 20 + 25 \\ - 1 - 3 + 4 & 3 + 9 - 12 & 5 + 15 - 20 \end{matrix}] \Rightarrow A B = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] \Rightarrow A B = 0_{3 \times 3} . . . (1) B A = [\begin{matrix} - 1 & 3 & 5 \\ 1 & - 3 & - 5 \\ - 1 & 3 & 5 \end{matrix}] [\begin{matrix} 2 & - 3 & - 5 \\ - 1 & 4 & 5 \\ 1 & - 3 & - 4 \end{matrix}] \Rightarrow B A = [\begin{matrix} - 2 - 3 + 5 & 3 + 12 - 15 & 5 + 15 - 20 \\ 2 + 3 - 5 & - 3 - 12 + 15 & - 5 - 15 + 20 \\ - 2 - 3 + 5 & 3 + 12 - 15 & 5 + 15 - 20 \end{matrix}] \Rightarrow B A = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] \Rightarrow B A = 0_{3 \times 3} . . . (2) \Rightarrow A B = B A = 0_{3 \times 3} [From eqs . (1) and (2)]$

Page No 4.42:

Question 13:

If A = $[\begin{matrix} 0 & c & - b \\ - c & 0 & a \\ b & - a & 0 \end{matrix}]$ and B = $[\begin{matrix} a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2} \end{matrix}]$ , show that AB = BA = O₃_×3.

Answer:

$Here, A B = [\begin{matrix} 0 & c & - b \\ - c & 0 & a \\ b & - a & 0 \end{matrix}] [\begin{matrix} a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2} \end{matrix}] \Rightarrow A B = [\begin{matrix} 0 + a b c - a b c & 0 + b^{2} c - b^{2} c & 0 + b c^{2} - b c^{2} \\ - a^{2} c + 0 + a^{2} c & - a b c + 0 + a b c & - a c^{2} + 0 + a c^{2} \\ a^{2} b - a^{2} b + 0 & a b^{2} - a b^{2} + 0 & a b c - a b c + 0 \end{matrix}] \Rightarrow A B = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] \Rightarrow A B = O_{3 \times 3} . . . (1) B A = [\begin{matrix} a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2} \end{matrix}] [\begin{matrix} 0 & c & - b \\ - c & 0 & a \\ b & - a & 0 \end{matrix}] \Rightarrow B A = [\begin{matrix} 0 - a b c + a b c & a^{2} c + 0 - a^{2} c & - a^{2} b + a^{2} b + 0 \\ 0 - b^{2} c + b^{2} c & a b c + 0 - a b c & - a b^{2} + a b^{2} + 0 \\ 0 - b c^{2} + b c^{2} & a c^{2} + 0 - a c^{2} & - a b c + a b c + 0 \end{matrix}] \Rightarrow B A = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] \Rightarrow B A = O_{3 \times 3} . . . (2) \Rightarrow A B = B A = O_{3 \times 3} [From eqs . (1) and (2)]$

Page No 4.42:

Question 14:

If A = $[\begin{matrix} 2 & - 3 & - 5 \\ - 1 & 4 & 5 \\ 1 & - 3 & - 4 \end{matrix}]$ and B = $[\begin{matrix} 2 & - 2 & - 4 \\ - 1 & 3 & 4 \\ 1 & - 2 & - 3 \end{matrix}]$ , show that AB = A and BA = B.

Answer:

$Given : A B = [\begin{matrix} 2 & - 3 & - 5 \\ - 1 & 4 & 5 \\ 1 & - 3 & - 4 \end{matrix}] [\begin{matrix} 2 & - 2 & - 4 \\ - 1 & 3 & 4 \\ 1 & - 2 & - 3 \end{matrix}] \Rightarrow A B = [\begin{matrix} 4 + 3 - 5 & - 4 - 9 + 10 & - 8 - 12 + 15 \\ - 2 - 4 + 5 & 2 + 12 - 10 & 4 + 16 - 15 \\ 2 + 3 - 4 & - 2 - 9 + 8 & - 4 - 12 + 12 \end{matrix}] \Rightarrow A B = [\begin{matrix} 2 & - 3 & - 5 \\ - 1 & 4 & 5 \\ 1 & - 3 & - 4 \end{matrix}] \Rightarrow A B = A B A = [\begin{matrix} 2 & - 2 & - 4 \\ - 1 & 3 & 4 \\ 1 & - 2 & - 3 \end{matrix}] [\begin{matrix} 2 & - 3 & - 5 \\ - 1 & 4 & 5 \\ 1 & - 3 & - 4 \end{matrix}] \Rightarrow B A = [\begin{matrix} 4 + 2 - 4 & - 6 - 8 + 12 & - 10 - 10 + 16 \\ - 2 - 3 + 4 & 3 + 12 - 12 & 5 + 15 - 16 \\ 2 + 2 - 3 & - 3 - 8 + 9 & - 5 - 10 + 12 \end{matrix}] \Rightarrow B A = [\begin{matrix} 2 & - 2 & - 4 \\ - 1 & 3 & 4 \\ 1 & - 2 & - 3 \end{matrix}] \Rightarrow B A = B$

Page No 4.42:

Question 15:

Let A = $[\begin{matrix} - 1 & 1 & - 1 \\ 3 & - 3 & 3 \\ 5 & 5 & 5 \end{matrix}]$ and B = $[\begin{matrix} 0 & 4 & 3 \\ 1 & - 3 & - 3 \\ - 1 & 4 & 4 \end{matrix}]$ , compute A² − B².

Answer:

$Given : A = [\begin{matrix} - 1 & 1 & - 1 \\ 3 & - 3 & 3 \\ 5 & 5 & 5 \end{matrix}] Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} - 1 & 1 & - 1 \\ 3 & - 3 & 3 \\ 5 & 5 & 5 \end{matrix}] [\begin{matrix} - 1 & 1 & - 1 \\ 3 & - 3 & 3 \\ 5 & 5 & 5 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 + 3 - 5 & - 1 - 3 - 5 & 1 + 3 - 5 \\ - 3 - 9 + 15 & 3 + 9 + 15 & - 3 - 9 + 15 \\ - 5 + 15 + 25 & 5 - 15 + 25 & - 5 + 15 + 25 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} - 1 & - 9 & - 1 \\ 3 & 27 & 3 \\ 35 & 15 & 35 \end{matrix}] B^{2} = B B \Rightarrow B^{2} = [\begin{matrix} 0 & 4 & 3 \\ 1 & - 3 & - 3 \\ - 1 & 4 & 4 \end{matrix}] [\begin{matrix} 0 & 4 & 3 \\ 1 & - 3 & - 3 \\ - 1 & 4 & 4 \end{matrix}] \Rightarrow B^{2} = [\begin{matrix} 0 + 4 - 3 & 0 - 12 + 12 & 0 - 12 + 12 \\ 0 - 3 + 3 & 4 + 9 - 12 & 3 + 9 - 12 \\ 0 + 4 - 4 & - 4 - 12 + 16 & - 3 - 12 + 16 \end{matrix}] \Rightarrow B^{2} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A^{2} - B^{2} \Rightarrow A^{2} - B^{2} = [\begin{matrix} - 1 & - 9 & - 1 \\ 3 & 27 & 3 \\ 35 & 15 & 35 \end{matrix}] - [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow A^{2} - B^{2} = [\begin{matrix} - 1 - 1 & - 9 - 0 & - 1 - 0 \\ 3 - 0 & 27 - 1 & 3 - 0 \\ 35 - 0 & 15 - 0 & 35 - 1 \end{matrix}] \Rightarrow A^{2} - B^{2} = [\begin{matrix} - 2 & - 9 & - 1 \\ 3 & 26 & 3 \\ 35 & 15 & 34 \end{matrix}]$

Page No 4.42:

Question 16:

For the following matrices verify the associativity of matrix multiplication i.e. (AB) C = A (BC):
(i) $A = [\begin{matrix} 1 & 2 & 0 \\ - 1 & 0 & 1 \end{matrix}], B = [\begin{matrix} 1 & 0 \\ - 1 & 2 \\ 0 & 3 \end{matrix}] and C = [\begin{matrix} 1 \\ - 1 \end{matrix}]$

(ii) $A = [\begin{matrix} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{matrix}], B = [\begin{matrix} 1 & - 1 & 1 \\ 0 & 1 & 2 \\ 2 & - 1 & 1 \end{matrix}] and C = [\begin{matrix} 1 & 2 & - 1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{matrix}]$ .

Answer:

$(i) (A B) C = A (B C) \Rightarrow ([\begin{matrix} 1 & 2 & 0 \\ - 1 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 \\ - 1 & 2 \\ 0 & 3 \end{matrix}]) [\begin{matrix} 1 \\ - 1 \end{matrix}] = [\begin{matrix} 1 & 2 & 0 \\ - 1 & 0 & 1 \end{matrix}] ([\begin{matrix} 1 & 0 \\ - 1 & 2 \\ 0 & 3 \end{matrix}] [\begin{matrix} 1 \\ - 1 \end{matrix}]) \Rightarrow ([\begin{matrix} 1 - 2 + 0 & 0 + 4 + 0 \\ - 1 - 0 + 0 & 0 + 0 + 3 \end{matrix}]) [\begin{matrix} 1 \\ - 1 \end{matrix}] = [\begin{matrix} 1 & 2 & 0 \\ - 1 & 0 & 1 \end{matrix}] ([\begin{matrix} 1 - 0 \\ - 1 - 2 \\ 0 - 3 \end{matrix}]) \Rightarrow [\begin{matrix} - 1 & 4 \\ - 1 & 3 \end{matrix}] [\begin{matrix} 1 \\ - 1 \end{matrix}] = [\begin{matrix} 1 & 2 & 0 \\ - 1 & 0 & 1 \end{matrix}] [[\begin{matrix} 1 \\ - 3 \\ - 3 \end{matrix}]] \Rightarrow [\begin{matrix} - 1 - 4 \\ - 1 - 3 \end{matrix}] = [\begin{matrix} 1 - 6 - 0 \\ - 1 - 0 - 3 \end{matrix}] \Rightarrow [\begin{matrix} - 5 \\ - 4 \end{matrix}] = [\begin{matrix} - 5 \\ - 4 \end{matrix}] ∴ LHS = RHS Hence proved .$

$(i i) (A B) C = A (B C) \Rightarrow ([\begin{matrix} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & - 1 & 1 \\ 0 & 1 & 2 \\ 2 & - 1 & 1 \end{matrix}]) [\begin{matrix} 1 & 2 & - 1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{matrix}] ([\begin{matrix} 1 & - 1 & 1 \\ 0 & 1 & 2 \\ 2 & - 1 & 1 \end{matrix}] [\begin{matrix} 1 & 2 & - 1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{matrix}]) \Rightarrow ([\begin{matrix} 4 + 0 + 6 & - 4 + 2 - 3 & 4 + 4 + 3 \\ 1 + 0 + 4 & - 1 + 1 - 2 & 1 + 2 + 2 \\ 3 + 0 + 2 & - 3 + 0 - 1 & 3 + 0 + 1 \end{matrix}]) [\begin{matrix} 1 & 2 & - 1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{matrix}] ([\begin{matrix} 1 - 3 + 0 & 2 - 0 + 0 & - 1 - 1 + 1 \\ 0 + 3 + 0 & 0 + 0 + 0 & 0 + 1 + 2 \\ 2 - 3 + 0 & 4 - 0 + 0 & - 2 - 1 + 1 \end{matrix}]) \Rightarrow [\begin{matrix} 10 & - 5 & 11 \\ 5 & - 2 & 5 \\ 5 & - 4 & 4 \end{matrix}] [\begin{matrix} 1 & 2 & - 1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{matrix}] [\begin{matrix} - 2 & 2 & - 1 \\ 3 & 0 & 3 \\ - 1 & 4 & - 2 \end{matrix}] \Rightarrow [\begin{matrix} 10 - 15 + 0 & 20 - 0 + 0 & - 10 - 5 + 11 \\ 5 - 6 + 0 & 10 - 0 + 0 & - 5 - 2 + 5 \\ 5 - 12 + 0 & 10 - 0 + 0 & - 5 - 4 + 4 \end{matrix}] = [\begin{matrix} - 8 + 6 - 3 & 8 + 0 + 12 & - 4 + 6 - 6 \\ - 2 + 3 - 2 & 2 + 0 + 8 & - 1 + 3 - 4 \\ - 6 + 0 - 1 & 6 + 0 + 4 & - 3 + 0 - 2 \end{matrix}] \Rightarrow [\begin{matrix} - 5 & 20 & - 4 \\ - 1 & 10 & - 2 \\ - 7 & 10 & - 5 \end{matrix}] = [\begin{matrix} - 5 & 20 & - 4 \\ - 1 & 10 & - 2 \\ - 7 & 10 & - 5 \end{matrix}] ∴ LHS = RHS Hence proved .$

Page No 4.42:

Question 17:

For the following matrices verify the distributivity of matrix multiplication over matrix addition i.e. A (B + C) = AB + AC:
(i) $A = [\begin{matrix} 1 & - 1 \\ 0 & 2 \end{matrix}], B = [\begin{matrix} - 1 & 0 \\ 2 & 1 \end{matrix}] and C = [\begin{matrix} 0 & 1 \\ 1 & - 1 \end{matrix}]$
(ii) $A = [\begin{matrix} 2 & - 1 \\ 1 & 1 \\ - 1 & 2 \end{matrix}], B = [\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}] and C = [\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}] .$

Answer:

$(i) A (B + C) = A B + A C \Rightarrow [\begin{matrix} 1 & - 1 \\ 0 & 2 \end{matrix}] ([\begin{matrix} - 1 & 0 \\ 2 & 1 \end{matrix}] + [\begin{matrix} 0 & 1 \\ 1 & - 1 \end{matrix}]) = [\begin{matrix} 1 & - 1 \\ 0 & 2 \end{matrix}] [\begin{matrix} - 1 & 0 \\ 2 & 1 \end{matrix}] + [\begin{matrix} 1 & - 1 \\ 0 & 2 \end{matrix}] [\begin{matrix} 0 & 1 \\ 1 & - 1 \end{matrix}] \Rightarrow [\begin{matrix} 1 & - 1 \\ 0 & 2 \end{matrix}] [\begin{matrix} - 1 + 0 & 0 + 1 \\ 2 + 1 & 1 - 1 \end{matrix}] = [\begin{matrix} - 1 - 2 & 0 - 1 \\ 0 + 4 & 0 + 2 \end{matrix}] + [\begin{matrix} 0 - 1 & 1 + 1 \\ 0 + 2 & 0 - 2 \end{matrix}] \Rightarrow [\begin{matrix} 1 & - 1 \\ 0 & 2 \end{matrix}] [\begin{matrix} - 1 & 1 \\ 3 & 0 \end{matrix}] = [\begin{matrix} - 3 & - 1 \\ 4 & 2 \end{matrix}] + [\begin{matrix} - 1 & 2 \\ 2 & - 2 \end{matrix}] \Rightarrow [\begin{matrix} - 1 - 3 & 1 - 0 \\ 0 + 6 & 0 + 0 \end{matrix}] = [\begin{matrix} - 3 - 1 & - 1 + 2 \\ 4 + 2 & 2 - 2 \end{matrix}] \Rightarrow [\begin{matrix} - 4 & 1 \\ 6 & 0 \end{matrix}] = [\begin{matrix} - 4 & 1 \\ 6 & 0 \end{matrix}] ∴ LHS = RHS Hence proved .$

$(ii) A (B + C) = A B + A C \Rightarrow [\begin{matrix} 2 & - 1 \\ 1 & 1 \\ - 1 & 2 \end{matrix}] ([\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}] + [\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}]) = [\begin{matrix} 2 & - 1 \\ 1 & 1 \\ - 1 & 2 \end{matrix}] [\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}] + [\begin{matrix} 2 & - 1 \\ 1 & 1 \\ - 1 & 2 \end{matrix}] [\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} 2 & - 1 \\ 1 & 1 \\ - 1 & 2 \end{matrix}] [\begin{matrix} 0 + 1 & 1 - 1 \\ 1 + 0 & 1 + 1 \end{matrix}] = [\begin{matrix} 0 - 1 & 2 - 1 \\ 0 + 1 & 1 + 1 \\ 0 + 2 & - 1 + 2 \end{matrix}] + [\begin{matrix} 2 - 0 & - 2 - 1 \\ 1 + 0 & - 1 + 1 \\ - 1 + 0 & 1 + 2 \end{matrix}] \Rightarrow [\begin{matrix} 2 & - 1 \\ 1 & 1 \\ - 1 & 2 \end{matrix}] [\begin{matrix} 1 & 0 \\ 1 & 2 \end{matrix}] = [\begin{matrix} - 1 & 1 \\ 1 & 2 \\ 2 & 1 \end{matrix}] + [\begin{matrix} 2 & - 3 \\ 1 & 0 \\ - 1 & 3 \end{matrix}] \Rightarrow [\begin{matrix} 2 - 1 & 0 - 2 \\ 1 + 1 & 0 + 2 \\ - 1 + 2 & 0 + 4 \end{matrix}] = [\begin{matrix} - 1 + 2 & 1 - 3 \\ 1 + 1 & 2 + 0 \\ 2 - 1 & 1 + 3 \end{matrix}] \Rightarrow [\begin{matrix} 1 & - 2 \\ 2 & 2 \\ 1 & 4 \end{matrix}] = [\begin{matrix} 1 & - 2 \\ 2 & 2 \\ 1 & 4 \end{matrix}] ∴ LHS = RHS Hence proved .$

Page No 4.42:

Question 18:

If $A = [\begin{matrix} 1 & 0 & - 2 \\ 3 & - 1 & 0 \\ - 2 & 1 & 1 \end{matrix}], B = [\begin{matrix} 0 & 5 & - 4 \\ - 2 & 1 & 3 \\ - 1 & 0 & 2 \end{matrix}] and C = [\begin{matrix} 1 & 5 & 2 \\ - 1 & 1 & 0 \\ 0 & - 1 & 1 \end{matrix}]$ , verify that A (B − C) = AB − AC.

Answer:

$Given : A (B - C) = A B - A C \Rightarrow [\begin{matrix} 1 & 0 & - 2 \\ 3 & - 1 & 0 \\ - 2 & 1 & 1 \end{matrix}] ([\begin{matrix} 0 & 5 & - 4 \\ - 2 & 1 & 3 \\ - 1 & 0 & 2 \end{matrix}] - [\begin{matrix} 1 & 5 & 2 \\ - 1 & 1 & 0 \\ 0 & - 1 & 1 \end{matrix}]) = [\begin{matrix} 1 & 0 & - 2 \\ 3 & - 1 & 0 \\ - 2 & 1 & 1 \end{matrix}] [\begin{matrix} 0 & 5 & - 4 \\ - 2 & 1 & 3 \\ - 1 & 0 & 2 \end{matrix}] - [\begin{matrix} 1 & 0 & - 2 \\ 3 & - 1 & 0 \\ - 2 & 1 & 1 \end{matrix}] [\begin{matrix} 1 & 5 & 2 \\ - 1 & 1 & 0 \\ 0 & - 1 & 1 \end{matrix}] \Rightarrow [\begin{matrix} 1 & 0 & - 2 \\ 3 & - 1 & 0 \\ - 2 & 1 & 1 \end{matrix}] [\begin{matrix} 0 - 1 & 5 - 5 & - 4 - 2 \\ - 2 + 1 & 1 - 1 & 3 - 0 \\ - 1 - 0 & 0 + 1 & 2 - 1 \end{matrix}] = [\begin{matrix} 0 - 0 + 2 & 5 + 0 - 0 & - 4 + 0 - 4 \\ 0 + 2 - 0 & 15 - 1 + 0 & - 12 - 3 + 0 \\ 0 - 2 - 1 & - 10 + 1 + 0 & 8 + 3 + 2 \end{matrix}] - [\begin{matrix} 1 - 0 - 0 & 5 + 0 + 2 & 2 + 0 - 2 \\ 3 + 1 + 0 & 15 - 1 - 0 & 6 - 0 + 0 \\ - 2 - 1 + 0 & - 10 + 1 - 1 & - 4 + 0 + 1 \end{matrix}] \Rightarrow [\begin{matrix} 1 & 0 & - 2 \\ 3 & - 1 & 0 \\ - 2 & 1 & 1 \end{matrix}] [\begin{matrix} - 1 & 0 & - 6 \\ - 1 & 0 & 3 \\ - 1 & 1 & 1 \end{matrix}] = [\begin{matrix} 2 & 5 & - 8 \\ 2 & 14 & - 15 \\ - 3 & - 9 & 13 \end{matrix}] - [\begin{matrix} 1 & 7 & 0 \\ 4 & 14 & 6 \\ - 3 & - 10 & - 3 \end{matrix}] \Rightarrow [\begin{matrix} - 1 - 0 + 2 & 0 + 0 - 2 & - 6 + 0 - 2 \\ - 3 + 1 - 0 & 0 - 0 + 0 & - 18 - 3 + 0 \\ 2 - 1 - 1 & 0 + 0 + 1 & 12 + 3 + 1 \end{matrix}] = [\begin{matrix} 2 - 1 & 5 - 7 & - 8 - 0 \\ 2 - 4 & 14 - 14 & - 15 - 6 \\ - 3 + 3 & - 9 + 10 & 13 + 3 \end{matrix}] \Rightarrow [\begin{matrix} 1 & - 2 & - 8 \\ - 2 & 0 & - 21 \\ 0 & 1 & 16 \end{matrix}] = [\begin{matrix} 1 & - 2 & - 8 \\ - 2 & 0 & - 21 \\ 0 & 1 & 16 \end{matrix}] ∴ LHS = RHS Hence proved .$

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Question 19:

Compute the elements a₄₃ and a₂₂ of the matrix:
$A = [\begin{matrix} 0 & 1 & 0 \\ 2 & 0 & 2 \\ 0 & 3 & 2 \\ 4 & 0 & 4 \end{matrix}] [\begin{matrix} 2 & - 1 \\ - 3 & 2 \\ 4 & 3 \end{matrix}] [\begin{matrix} 0 & 1 & - 1 & 2 & - 2 \\ 3 & - 3 & 4 & - 4 & 0 \end{matrix}]$

Answer:

We have,

$Given : A = [\begin{matrix} 0 & 1 & 0 \\ 2 & 0 & 2 \\ 0 & 3 & 2 \\ 4 & 0 & 4 \end{matrix}] [\begin{matrix} 2 & - 1 \\ - 3 & 2 \\ 4 & 3 \end{matrix}] [\begin{matrix} 0 & 1 & - 1 & 2 & - 2 \\ 3 & - 3 & 4 & - 4 & 0 \end{matrix}] \Rightarrow A = [\begin{matrix} 0 & 1 & 0 \\ 2 & 0 & 2 \\ 0 & 3 & 2 \\ 4 & 0 & 4 \end{matrix}] [\begin{matrix} 0 - 3 & 2 + 3 & - 2 - 4 & 4 + 4 & - 4 - 0 \\ 0 + 6 & - 3 - 6 & 3 + 8 & - 6 - 8 & 6 + 0 \\ 0 + 9 & 4 - 9 & - 4 + 12 & 8 - 12 & - 8 + 0 \end{matrix}] \Rightarrow A = [\begin{matrix} 0 & 1 & 0 \\ 2 & 0 & 2 \\ 0 & 3 & 2 \\ 4 & 0 & 4 \end{matrix}] [\begin{matrix} - 3 & 5 & - 6 & 8 & - 4 \\ 6 & - 9 & 11 & - 14 & 6 \\ 9 & - 5 & 8 & - 4 & - 8 \end{matrix}] \Rightarrow A = [\begin{matrix} 0 + 6 + 0 & 0 - 9 - 0 & 0 + 11 + 0 & 0 - 14 - 0 & 0 + 6 - 0 \\ - 6 + 0 + 18 & 10 - 0 - 10 & - 12 + 0 + 16 & 16 - 0 - 8 & - 8 + 0 - 16 \\ 0 + 18 + 18 & 0 - 27 - 10 & 0 + 33 + 16 & 0 - 42 - 8 & 0 + 18 - 16 \\ - 12 + 0 + 36 & 20 - 0 - 20 & - 24 + 0 + 32 & 32 - 0 - 16 & - 16 + 0 - 32 \end{matrix}] \Rightarrow A = [\begin{matrix} 6 & - 9 & 11 & - 14 & 6 \\ 12 & 0 & 4 & 8 & - 24 \\ 36 & - 37 & 49 & - 50 & 2 \\ 24 & 0 & 8 & 16 & - 48 \end{matrix}] ∴ a_{43} = 8 and a_{22} = 0$

Page No 4.43:

Question 20:

If $A = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{matrix}]$ , and I is the identity matrix of order 3, show that A³ = pI + qA +rA².

Answer:

$Given : A = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{matrix}] Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{matrix}] [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 0 + 0 + 0 & 0 + 0 + 0 & 0 + 1 + 0 \\ 0 + 0 + p & 0 + 0 + q & 0 + 0 + r \\ 0 + 0 + r p & p + 0 + r q & 0 + q + r^{2} \end{matrix}] A^{2} = [\begin{matrix} 0 & 0 & 1 \\ p & q & r \\ r p & p + r q & q + r^{2} \end{matrix}] A^{3} = A^{2} A \Rightarrow A^{3} = [\begin{matrix} 0 & 0 & 1 \\ p & q & r \\ r p & p + r q & q + r^{2} \end{matrix}] [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{matrix}] \Rightarrow A^{3} = [\begin{matrix} 0 + 0 + p & 0 + 0 + q & 0 + 0 + r \\ 0 + 0 + r p & p + 0 + r q & 0 + q + r^{2} \\ 0 + 0 + p q + r^{2} p & r p + 0 + q^{2} + r^{2} q & 0 + p + r q + r q + r^{3} \end{matrix}] \Rightarrow A^{3} = [\begin{matrix} p & q & r \\ r p & p + r q & q + r^{2} \\ p q + r^{2} p & r p + q^{2} + r^{2} q & p + 2 r q + r^{3} \end{matrix}] . . . (1) p I + q A + r A^{2} \Rightarrow p I + q A + r A^{2} = p [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] + q [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{matrix}] + r [\begin{matrix} 0 & 0 & 1 \\ p & q & r \\ r p & p + r q & q + r^{2} \end{matrix}] \Rightarrow p I + q A + r A^{2} = [\begin{matrix} p & 0 & 0 \\ 0 & p & 0 \\ 0 & 0 & p \end{matrix}] + [\begin{matrix} 0 & q & 0 \\ 0 & 0 & q \\ p q & q^{2} & q r \end{matrix}] + [\begin{matrix} 0 & 0 & r \\ r p & r q & r^{2} \\ r^{2} p & r p + r^{2} q & r q + r^{3} \end{matrix}] \Rightarrow p I + q A + r A^{2} = [\begin{matrix} p + 0 + 0 & 0 + q + 0 & 0 + 0 + r \\ 0 + 0 + r p & p + 0 + r q & 0 + q + r^{2} \\ 0 + p q + r^{2} p & 0 + q^{2} + r p + r^{2} q & p + q r + q r + r^{3} \end{matrix}] \Rightarrow p I + q A + r A^{2} = [\begin{matrix} p & q & r \\ r p & p + r q & q + r^{2} \\ p q + r^{2} p & q^{2} + r^{2} q + r p & p + 2 q r + r^{3} \end{matrix}] . . . (2) A^{3} = p I + q A + r A^{2} [From eqs . (1) and (2)]$

Page No 4.43:

Question 21:

If w is a complex cube root of unity, show that
$([\begin{matrix} 1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w \end{matrix}] + [\begin{matrix} w & w^{2} & 1 \\ w^{2} & 1 & w \\ w & w^{2} & 1 \end{matrix}]) [\begin{matrix} 1 \\ w \\ w^{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

Answer:

$Here, LHS = ([\begin{matrix} 1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w \end{matrix}] + [\begin{matrix} w & w^{2} & 1 \\ w^{2} & 1 & w \\ w & w^{2} & 1 \end{matrix}]) [\begin{matrix} 1 \\ w \\ w^{2} \end{matrix}] = [\begin{matrix} 1 + w & w + w^{2} & w^{2} + 1 \\ w + w^{2} & w^{2} + 1 & 1 + w \\ w^{2} + w & 1 + w^{2} & w + 1 \end{matrix}] [\begin{matrix} 1 \\ w \\ w^{2} \end{matrix}] = [\begin{matrix} - w^{2} & - 1 & - w \\ - 1 & - w & - w^{2} \\ - 1 & - w & - w^{2} \end{matrix}] [\begin{matrix} 1 \\ w \\ w^{2} \end{matrix}] (∵ 1 + w + w^{2} = 0 and w^{3} = 1) = [\begin{matrix} - w^{2} - w - w^{3} \\ - 1 - w^{2} - w^{4} \\ - 1 - w^{2} - w^{4} \end{matrix}] = [\begin{matrix} - w (1 + w + w^{2}) \\ - 1 - w^{2} - w^{3} w \\ - 1 - w^{2} - w^{3} w \end{matrix}] = [\begin{matrix} - w \times 0 \\ - 1 - w^{2} - w \\ - 1 - w^{2} - w \end{matrix}] (∵ 1 + w + w^{2} = 0 and w^{3} = 1) = [\begin{matrix} 0 \\ - 0 \\ - 0 \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}] ∴ ([\begin{matrix} 1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w \end{matrix}] + [\begin{matrix} w & w^{2} & 1 \\ w^{2} & 1 & w \\ w & w^{2} & 1 \end{matrix}]) [\begin{matrix} 1 \\ w \\ w^{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

Page No 4.43:

Question 22:

If $A = [\begin{matrix} 2 & - 3 & - 5 \\ - 1 & 4 & 5 \\ 1 & - 3 & - 4 \end{matrix}]$ , show that A² = A.

Answer:

$Here, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 2 & - 3 & - 5 \\ - 1 & 4 & 5 \\ 1 & - 3 & - 4 \end{matrix}] [\begin{matrix} 2 & - 3 & - 5 \\ - 1 & 4 & 5 \\ 1 & - 3 & - 4 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 4 + 3 - 5 & - 6 - 12 + 15 & - 10 - 15 + 20 \\ - 2 - 4 + 5 & 3 + 16 - 15 & 5 + 20 - 20 \\ 2 + 3 - 4 & - 3 - 12 + 12 & - 5 - 15 + 16 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 2 & - 3 & - 5 \\ - 1 & 4 & 5 \\ 1 & - 3 & - 4 \end{matrix}] ∴ A^{2} = A$

Page No 4.43:

Question 23:

If $A = [\begin{matrix} 4 & - 1 & - 4 \\ 3 & 0 & - 4 \\ 3 & - 1 & - 3 \end{matrix}]$ , show that A² = I₃.

Answer:

$Here, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 4 & - 1 & - 4 \\ 3 & 0 & - 4 \\ 3 & - 1 & - 3 \end{matrix}] [\begin{matrix} 4 & - 1 & - 4 \\ 3 & 0 & - 4 \\ 3 & - 1 & - 3 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 16 - 3 - 12 & - 4 + 0 + 4 & - 16 + 4 + 12 \\ 12 + 0 - 12 & - 3 + 0 + 4 & - 12 + 0 + 12 \\ 12 - 3 - 9 & - 3 + 0 + 3 & - 12 + 4 + 9 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] ∴ A^{2} = I_{3}$

Page No 4.43:

Question 24:

(i) If [1 1 x] $[\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 1 & 0 \end{matrix}] [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}]$ = 0, find x.
(ii) If $[\begin{matrix} 2 & 3 \\ 5 & 7 \end{matrix}] [\begin{matrix} 1 & - 3 \\ - 2 & 4 \end{matrix}] = [\begin{matrix} - 4 & 6 \\ - 9 & x \end{matrix}]$ , find x.

Answer:

(i)
$Given : [\begin{matrix} 1 & 1 & x \end{matrix}] [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 1 & 0 \end{matrix}] [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = 0 \Rightarrow [\begin{matrix} 1 + 0 + 2 x & 0 + 2 + x & 2 + 1 + 0 \end{matrix}] [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = 0 \Rightarrow [\begin{matrix} 1 + 2 x & 2 + x & 3 \end{matrix}] [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = 0 \Rightarrow [\begin{matrix} 1 + 2 x + 2 + x + 3 \end{matrix}] = 0 \Rightarrow 6 + 3 x = 0 \Rightarrow 3 x = - 6 \Rightarrow x = \frac{- 6}{3} ∴ x = - 2$

(ii)

$Given : [\begin{matrix} 2 & 3 \\ 5 & 7 \end{matrix}] [\begin{matrix} 1 & - 3 \\ - 2 & 4 \end{matrix}] = [\begin{matrix} - 4 & 6 \\ - 9 & x \end{matrix}] \Rightarrow [\begin{matrix} 2 - 6 & - 6 + 12 \\ 5 - 14 & - 15 + 28 \end{matrix}] = [\begin{matrix} - 4 & 6 \\ - 9 & x \end{matrix}] \Rightarrow [\begin{matrix} - 4 & 6 \\ - 9 & 13 \end{matrix}] = [\begin{matrix} - 4 & 6 \\ - 9 & x \end{matrix}] \Rightarrow x = 13 ∴ x = 13$

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Question 25:

If [x 4 1] $[\begin{matrix} 2 & 1 & 2 \\ 1 & 0 & 2 \\ 0 & 2 & - 4 \end{matrix}] [\begin{matrix} x \\ 4 \\ - 1 \end{matrix}]$ = 0, find x.

Answer:

$Given : [\begin{matrix} x & 4 & 1 \end{matrix}] [\begin{matrix} 2 & 1 & 2 \\ 1 & 0 & 2 \\ 0 & 2 & - 4 \end{matrix}] [\begin{matrix} x \\ 4 \\ - 1 \end{matrix}] = 0 \Rightarrow [\begin{matrix} 2 x + 4 + 0 & x + 0 + 2 & 2 x + 8 - 4 \end{matrix}] [\begin{matrix} x \\ 4 \\ - 1 \end{matrix}] = 0 \Rightarrow [\begin{matrix} 2 x + 4 & x + 2 & 2 x + 4 \end{matrix}] [\begin{matrix} x \\ 4 \\ - 1 \end{matrix}] = 0 \Rightarrow [\begin{matrix} 2 x^{2} + 4 x + 4 x + 8 - 2 x - 4 \end{matrix}] = 0 \Rightarrow 2 x^{2} + 6 x + 4 = 0 \Rightarrow x^{2} + 3 x + 2 = 0 \Rightarrow x^{2} + 2 x + x + 2 = 0 \Rightarrow x (x + 2) + 1 (x + 2) = 0 \Rightarrow (x + 2) (x + 1) = 0 \Rightarrow x + 2 = 0 or x + 1 = 0 \Rightarrow x = - 2 or x = - 1 ∴ x = - 2, - 1$

Page No 4.43:

Question 26:

If [1 −1 x] $[\begin{matrix} 0 & 1 & - 1 \\ 2 & 1 & 3 \\ 1 & 1 & 1 \end{matrix}] [\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}]$ = 0, find x.

Answer:

$Given : [\begin{matrix} 1 & - 1 & x \end{matrix}] [\begin{matrix} 0 & 1 & - 1 \\ 2 & 1 & 3 \\ 1 & 1 & 1 \end{matrix}] [\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}] = 0 \Rightarrow [\begin{matrix} 0 - 2 + x & 1 - 1 + x & - 1 - 3 + x \end{matrix}] [\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}] = 0 \Rightarrow [\begin{matrix} - 2 + x & x & - 4 + x \end{matrix}] [\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}] = 0 \Rightarrow [\begin{matrix} 0 + x - 4 + x \end{matrix}] = 0 \Rightarrow 2 x - 4 = 0 \Rightarrow 2 x = 4 \Rightarrow x = \frac{4}{2} ∴ x = 2$

Page No 4.43:

Question 27:

If $A = [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}] and I = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ , then prove that A² − A + 2I = O.

Answer:

$Given : A = [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}] Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}] [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 9 - 8 & - 6 + 4 \\ 12 - 8 & - 8 + 4 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 & - 2 \\ 4 & - 4 \end{matrix}] A^{2} - A + 2 I = [\begin{matrix} 1 & - 2 \\ 4 & - 4 \end{matrix}] - [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}] + 2 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow A^{2} - A + 2 I = [\begin{matrix} 1 - 3 & - 2 + 2 \\ 4 - 4 & - 4 + 2 \end{matrix}] + [\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}] \Rightarrow A^{2} - A + 2 I = [\begin{matrix} - 2 & 0 \\ 0 & - 2 \end{matrix}] + [\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}] \Rightarrow A^{2} - A + 2 I = [\begin{matrix} - 2 + 2 & 0 + 0 \\ 0 + 0 & - 2 + 2 \end{matrix}] \Rightarrow A^{2} - A + 2 I = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \Rightarrow A^{2} - A + 2 I = 0 Hence proved .$

Page No 4.43:

Question 28:

If $A = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] and I = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ , then find λ so that A² = 5A + λI.

Answer:

Page No 4.43:

Question 29:

If $A = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}]$ , show that A² − 5A + 7I₂ = O

Answer:

Page No 4.43:

Question 30:

If $A = [\begin{matrix} 2 & 3 \\ - 1 & 0 \end{matrix}]$ , show that A² − 2A + 3I₂ = O

Answer:

$Given : A = [\begin{matrix} 2 & 3 \\ - 1 & 0 \end{matrix}] Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 2 & 3 \\ - 1 & 0 \end{matrix}] [\begin{matrix} 2 & 3 \\ - 1 & 0 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 4 - 3 & 6 + 0 \\ - 2 + 0 & - 3 + 0 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 & 6 \\ - 2 & - 3 \end{matrix}] A^{2} - 2 A + 3 I_{2} \Rightarrow A^{2} - 2 A + 3 I_{2} = [\begin{matrix} 1 & 6 \\ - 2 & - 3 \end{matrix}] - 2 [\begin{matrix} 2 & 3 \\ - 1 & 0 \end{matrix}] + 3 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow A^{2} - 2 A + 3 I_{2} = [\begin{matrix} 1 & 6 \\ - 2 & - 3 \end{matrix}] - [\begin{matrix} 4 & 6 \\ - 2 & 0 \end{matrix}] + [\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}] \Rightarrow A^{2} - 2 A + 3 I_{2} = [\begin{matrix} 1 - 4 + 3 & 6 - 6 + 0 \\ - 2 + 2 + 0 & - 3 + 0 + 3 \end{matrix}] \Rightarrow A^{2} - 2 A + 3 I_{2} = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = 0 Hence proved .$

Page No 4.43:

Question 31:

Show that the matrix $A = [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}]$ satisfies the equation A³ − 4A² + A = O

Answer:

$We have, A = [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] ∴ A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 4 + 3 & 6 + 6 \\ 2 + 2 & 3 + 4 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 7 & 12 \\ 4 & 7 \end{matrix}] \Rightarrow A^{3} = A^{2} A \Rightarrow A^{3} = [\begin{matrix} 7 & 12 \\ 4 & 7 \end{matrix}] [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] \Rightarrow A^{3} = [\begin{matrix} 14 + 12 & 21 + 24 \\ 8 + 7 & 12 + 14 \end{matrix}] \Rightarrow A^{3} = [\begin{matrix} 26 & 45 \\ 15 & 26 \end{matrix}] Now, A^{3} - 4 A^{2} + A \Rightarrow A^{3} - 4 A^{2} + A = [\begin{matrix} 26 & 45 \\ 15 & 26 \end{matrix}] - 4 [\begin{matrix} 7 & 12 \\ 4 & 7 \end{matrix}] + [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] \Rightarrow A^{3} - 4 A^{2} + A = [\begin{matrix} 26 & 45 \\ 15 & 26 \end{matrix}] - [\begin{matrix} 28 & 48 \\ 16 & 28 \end{matrix}] + [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] \Rightarrow A^{3} - 4 A^{2} + A = [\begin{matrix} 26 - 28 + 2 & 45 - 48 + 3 \\ 15 - 16 + 1 & 26 - 28 + 2 \end{matrix}] \Rightarrow A^{3} - 4 A^{2} + A = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = O Hence proved .$

Page No 4.43:

Question 32:

Show that the matrix $A = [\begin{matrix} 5 & 3 \\ 12 & 7 \end{matrix}]$ is root of the equation A² − 12A − I = O

Answer:

$Given : A = [\begin{matrix} 5 & 3 \\ 12 & 7 \end{matrix}] Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 5 & 3 \\ 12 & 7 \end{matrix}] [\begin{matrix} 5 & 3 \\ 12 & 7 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 25 + 36 & 15 + 21 \\ 60 + 84 & 36 + 49 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 61 & 36 \\ 144 & 85 \end{matrix}] A^{2} - 12 A - I \Rightarrow A^{2} - 12 A - I = [\begin{matrix} 61 & 36 \\ 144 & 85 \end{matrix}] - 12 [\begin{matrix} 5 & 3 \\ 12 & 7 \end{matrix}] - [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow A^{2} - 12 A - I = [\begin{matrix} 61 & 36 \\ 144 & 85 \end{matrix}] - [\begin{matrix} 60 & 36 \\ 144 & 84 \end{matrix}] - [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow A^{2} - 12 A - I = [\begin{matrix} 61 - 60 - 1 & 36 - 36 + 0 \\ 144 - 144 + 0 & 85 - 84 - 1 \end{matrix}] \Rightarrow A^{2} - 12 A - I = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] Since A is satisfying the equation A^{2} - 12 A - I, A is the root of the equation A^{2} - 12 A - I .$

Page No 4.43:

Question 33:

If $A = [\begin{matrix} 1 & - 3 \\ - 4 & 3 \end{matrix}]$ , find A² − 3A − 7I.

Answer:

$Given : A = [\begin{matrix} 1 & - 3 \\ - 4 & 3 \end{matrix}] Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 1 & - 3 \\ - 4 & 3 \end{matrix}] [\begin{matrix} 1 & - 3 \\ - 4 & 3 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 + 12 & - 3 - 6 \\ - 4 - 12 & 12 + 9 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 13 & - 9 \\ - 16 & 21 \end{matrix}] A^{2} - 5 A - 14 I \Rightarrow A^{2} - 3 A - 7 I = [\begin{matrix} 13 & - 9 \\ - 16 & 21 \end{matrix}] - 3 [\begin{matrix} 1 & - 3 \\ - 4 & 3 \end{matrix}] - 7 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow A^{2} - 3 A - 7 I = [\begin{matrix} 13 & - 9 \\ - 16 & 21 \end{matrix}] - [\begin{matrix} 3 & - 9 \\ - 12 & 9 \end{matrix}] - [\begin{matrix} 7 & 0 \\ 0 & 7 \end{matrix}] \Rightarrow A^{2} - 3 A - 7 I = [\begin{matrix} 3 & 0 \\ - 4 & 5 \end{matrix}]$

Page No 4.44:

Question 34:

If $A = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}]$ , show that A² − 5A + 7I = O use this to find A⁴.

Answer:

$Given : A = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 9 - 1 & 3 + 2 \\ - 3 - 2 & - 1 + 4 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 8 & 5 \\ - 5 & 3 \end{matrix}] A^{2} - 5 A + 7 I \Rightarrow A^{2} - 5 A + 7 I = [\begin{matrix} 8 & 5 \\ - 5 & 3 \end{matrix}] - 5 [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] + 7 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow A^{2} - 5 A + 7 I = [\begin{matrix} 8 & 5 \\ - 5 & 3 \end{matrix}] - [\begin{matrix} 15 & 5 \\ - 5 & 10 \end{matrix}] + [\begin{matrix} 7 & 0 \\ 0 & 7 \end{matrix}] \Rightarrow A^{2} - 5 A + 7 I = [\begin{matrix} 8 - 15 + 7 & 5 - 5 + 0 \\ - 5 + 5 + 0 & 3 - 10 + 7 \end{matrix}] \Rightarrow A^{2} - 5 A + 7 I = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = 0 Hence proved . Given : A^{2} - 5 A + 7 I = 0 \Rightarrow A^{2} = 5 A - 7 I . . . (1) \Rightarrow A^{3} = A (5 A - 7 I) (Multilpying by A on both sides) \Rightarrow A^{3} = 5 A^{2} - 7 A I \Rightarrow A^{3} = 5 (5 A - 7 I) - 7 A [From eq . (1)] \Rightarrow A^{3} = 25 A - 35 I - 7 A \Rightarrow A^{3} = 18 A - 35 I \Rightarrow A^{4} = (18 A - 35 I) A (Multilpying by A on both sides) \Rightarrow A^{4} = 18 A^{2} - 35 A \Rightarrow A^{4} = 18 (5 A - 7 I) - 35 A [From eq . (1)] \Rightarrow A^{4} = 90 A - 126 I - 35 A \Rightarrow A^{4} = 55 A - 126 I \Rightarrow A^{4} = 55 [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] - 126 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow A^{4} = [\begin{matrix} 165 & 55 \\ - 55 & 110 \end{matrix}] - [\begin{matrix} 126 & 0 \\ 0 & 126 \end{matrix}] \Rightarrow A^{4} = [\begin{matrix} 165 - 126 & 55 - 0 \\ - 55 - 0 & 110 - 126 \end{matrix}] \Rightarrow A^{4} = [\begin{matrix} 39 & 55 \\ - 55 & - 16 \end{matrix}]$

Page No 4.44:

Question 35:

If $A = [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}]$ , find k such that A² = kA − 2I₂

Answer:

$Given : A = [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}] Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}] [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 9 - 8 & - 6 + 4 \\ 12 - 8 & - 8 + 4 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 & - 2 \\ 4 & - 4 \end{matrix}] A^{2} = k A - 2 I_{2} \Rightarrow [\begin{matrix} 1 & - 2 \\ 4 & - 4 \end{matrix}] = k [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}] - 2 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} 1 & - 2 \\ 4 & - 4 \end{matrix}] = [\begin{matrix} 3 k & - 2 k \\ 4 k & - 2 k \end{matrix}] - [\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}] \Rightarrow [\begin{matrix} 1 & - 2 \\ 4 & - 4 \end{matrix}] = [\begin{matrix} 3 k - 2 & - 2 k - 0 \\ 4 k - 0 & - 2 k - 2 \end{matrix}] The corresponding elements of two equal matrices are equal . ∴ 1 = 3 k - 2 \Rightarrow 1 + 2 = 3 k \Rightarrow 3 = 3 k \Rightarrow k = 1$

Page No 4.44:

Question 36:

If $A = [\begin{matrix} 1 & 0 \\ - 1 & 7 \end{matrix}]$ , find k such that A² − 8A + kI = 0.

Answer:

$Given : A = [\begin{matrix} 1 & 0 \\ - 1 & 7 \end{matrix}] N o w, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 1 & 0 \\ - 1 & 7 \end{matrix}] [\begin{matrix} 1 & 0 \\ - 1 & 7 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 - 0 & 0 + 0 \\ - 1 - 7 & 0 + 49 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 & 0 \\ - 8 & 49 \end{matrix}] A^{2} - 8 A + k I = 0 \Rightarrow [\begin{matrix} 1 & 0 \\ - 8 & 49 \end{matrix}] - 8 [\begin{matrix} 1 & 0 \\ - 1 & 7 \end{matrix}] + k [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = 0 \Rightarrow [\begin{matrix} 1 & 0 \\ - 8 & 49 \end{matrix}] - [\begin{matrix} 8 & 0 \\ - 8 & 56 \end{matrix}] + [\begin{matrix} k & 0 \\ 0 & k \end{matrix}] = 0 \Rightarrow [\begin{matrix} 1 - 8 + k & 0 - 0 + 0 \\ - 8 + 8 + 0 & 49 - 56 + k \end{matrix}] = 0 \Rightarrow [\begin{matrix} - 7 + k & 0 \\ 0 & - 7 + k \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] The corresponding elements of two equal matrices are equal . ∴ - 7 + k = 0 \Rightarrow k = 7$

Page No 4.44:

Question 37:

If $A = [\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix}]$ , f (x) = x² − 2x − 3, show that f (A) = 0

Answer:

$Here, f (x) = x^{2} - 2 x - 3 \Rightarrow f (A) = A^{2} - 2 A - 3 I_{2} Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix}] [\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 + 4 & 2 + 2 \\ 2 + 2 & 4 + 1 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 5 & 4 \\ 4 & 5 \end{matrix}] f (A) = A^{2} - 2 A - 3 I_{2} \Rightarrow f (A) = [\begin{matrix} 5 & 4 \\ 4 & 5 \end{matrix}] - 2 [\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix}] - 3 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow f (A) = [\begin{matrix} 5 & 4 \\ 4 & 5 \end{matrix}] - [\begin{matrix} 2 & 4 \\ 4 & 2 \end{matrix}] - [\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}] \Rightarrow f (A) = [\begin{matrix} 5 - 2 - 3 & 4 - 4 - 0 \\ 4 - 4 - 0 & 5 - 2 - 3 \end{matrix}] \Rightarrow f (A) = [\begin{matrix} 5 - 5 & 0 \\ 0 & 5 - 5 \end{matrix}] \Rightarrow f (A) = 0$

Page No 4.44:

Question 38:

If $A = [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] and I = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}],$ then find λ, μ so that A² = λA + μI

Answer:

$Given : A = [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 4 + 3 & 6 + 6 \\ 2 + 2 & 3 + 4 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 7 & 12 \\ 4 & 7 \end{matrix}] A^{2} = λ A + μ I \Rightarrow [\begin{matrix} 7 & 12 \\ 4 & 7 \end{matrix}] = λ [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] + μ [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} 7 & 12 \\ 4 & 7 \end{matrix}] = [\begin{matrix} 2 λ & 3 λ \\ λ & 2 λ \end{matrix}] + [\begin{matrix} μ & 0 \\ 0 & μ \end{matrix}] \Rightarrow [\begin{matrix} 7 & 12 \\ 4 & 7 \end{matrix}] = [\begin{matrix} 2 λ + μ & 3 λ + 0 \\ λ + 0 & 2 λ + μ \end{matrix}] \Rightarrow [\begin{matrix} 7 & 12 \\ 4 & 7 \end{matrix}] = [\begin{matrix} 2 λ + μ & 3 λ \\ λ & 2 λ + μ \end{matrix}] The corresponding elements of two equal matrices are equal . ∴ 7 = 2 λ + μ . . . (1) 12 = 3 λ \Rightarrow λ = \frac{12}{3} = 4 Putting the value of λ in eq . (1), we get 7 = 2 (4) + μ \Rightarrow 7 - 8 = μ ∴ μ = - 1$

Page No 4.44:

Question 39:

Find the value of x for which the matrix product
$[\begin{matrix} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & - 2 & 1 \end{matrix}] [\begin{matrix} - x & 14 x & 7 x \\ 0 & 1 & 0 \\ x & - 4 x & - 2 x \end{matrix}]$ equal an identity matrix.

Answer:

$Here, [\begin{matrix} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & - 2 & 1 \end{matrix}] [\begin{matrix} - x & 14 x & 7 x \\ 0 & 1 & 0 \\ x & - 4 x & - 2 x \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} - 2 x + 0 + 7 x & 28 x + 0 - 28 x & 14 x + 0 - 14 x \\ 0 + 0 + 0 & 0 + 1 - 0 & 0 + 0 - 0 \\ - x - 0 + x & 14 x - 2 - 4 x & 7 x - 0 - 2 x \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} 5 x & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 10 x - 2 & 5 x \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] The corresponding elements of two equal matrices are equal . ∴ 5 x = 1 \Rightarrow x = \frac{1}{5}$

Page No 4.44:

Question 40:

Solve the matrix equations:
(i) $[x 1] [\begin{matrix} 1 & 0 \\ - 2 & - 3 \end{matrix}] [\begin{matrix} x \\ 5 \end{matrix}] = 0$

(ii) $[1 2 1] [\begin{matrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{matrix}] [\begin{matrix} 0 \\ 2 \\ x \end{matrix}] = 0$

(iii) $[x - 5 - 1] [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] [\begin{matrix} x \\ 4 \\ 1 \end{matrix}] = 0$

(iv) $[\begin{matrix} 2 x & 3 \end{matrix}] [\begin{matrix} 1 & 2 \\ - 3 & 0 \end{matrix}] [\begin{matrix} x \\ 8 \end{matrix}] = 0$

Answer:

$(i) [\begin{matrix} x & 1 \end{matrix}] [\begin{matrix} 1 & 0 \\ - 2 & - 3 \end{matrix}] [\begin{matrix} x \\ 5 \end{matrix}] = 0 \Rightarrow [\begin{matrix} x - 2 & 0 - 3 \end{matrix}] [\begin{matrix} x \\ 5 \end{matrix}] = 0 \Rightarrow [\begin{matrix} x - 2 & - 3 \end{matrix}] [\begin{matrix} x \\ 5 \end{matrix}] = 0 \Rightarrow [\begin{matrix} x^{2} - 2 x - 15 \end{matrix}] = 0 \Rightarrow x^{2} - 2 x - 15 = 0 \Rightarrow x^{2} - 5 x + 3 x - 15 = 0 \Rightarrow x (x - 5) + 3 (x - 5) = 0 \Rightarrow (x - 5) (x + 3) = 0 \Rightarrow x - 5 = 0 or x + 3 = 0 \Rightarrow x = 5 or x = - 3$

$(i i) [\begin{matrix} 1 & 2 & 1 \end{matrix}] [\begin{matrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{matrix}] [\begin{matrix} 0 \\ 2 \\ x \end{matrix}] = 0 \Rightarrow [\begin{matrix} 1 + 4 + 1 & 2 + 0 + 0 & 0 + 2 + 2 \end{matrix}] [\begin{matrix} 0 \\ 2 \\ x \end{matrix}] = 0 \Rightarrow [\begin{matrix} 6 & 2 & 4 \end{matrix}] [\begin{matrix} 0 \\ 2 \\ x \end{matrix}] = 0 \Rightarrow [\begin{matrix} 0 + 4 + 4 x \end{matrix}] = 0 \Rightarrow 4 + 4 x = 0 \Rightarrow 4 x = - 4 ∴ x = \frac{- 4}{4} = - 1$

$(i i i) [\begin{matrix} x & - 5 & - 1 \end{matrix}] [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] [\begin{matrix} x \\ 4 \\ 1 \end{matrix}] = 0 \Rightarrow [\begin{matrix} x - 0 - 2 & 0 - 10 - 0 & 2 x - 5 - 3 \end{matrix}] [\begin{matrix} x \\ 4 \\ 1 \end{matrix}] = 0 \Rightarrow [\begin{matrix} x - 2 & - 10 & 2 x - 8 \end{matrix}] [\begin{matrix} x \\ 4 \\ 1 \end{matrix}] = 0 \Rightarrow [\begin{matrix} x^{2} - 2 x - 40 + 2 x - 8 \end{matrix}] = 0 \Rightarrow x^{2} - 48 = 0 \Rightarrow x^{2} = 48 \Rightarrow x = \pm \sqrt{48}$

$(iv) [\begin{matrix} 2 x & 3 \end{matrix}] [\begin{matrix} 1 & 2 \\ - 3 & 0 \end{matrix}] [\begin{matrix} x \\ 8 \end{matrix}] = 0 \Rightarrow [\begin{matrix} 2 x - 9 & 4 x \end{matrix}] [\begin{matrix} x \\ 8 \end{matrix}] = 0 \Rightarrow [x (2 x - 9) + 32 x] = 0 \Rightarrow [2 x^{2} - 9 x + 32 x] = 0 \Rightarrow [2 x^{2} + 23 x] = 0 \Rightarrow 2 x^{2} + 23 x = 0 \Rightarrow x (2 x + 23) = 0 \Rightarrow x = 0 or x = - \frac{23}{2} ∴ x = 0 or x = - \frac{23}{2}$

Page No 4.44:

Question 41:

If $A = [\begin{matrix} 1 & 2 & 0 \\ 3 & - 4 & 5 \\ 0 & - 1 & 3 \end{matrix}]$ , compute A² − 4A + 3I₃.

Answer:

$Given : A = [\begin{matrix} 1 & 2 & 0 \\ 3 & - 4 & 5 \\ 0 & - 1 & 3 \end{matrix}] Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 1 & 2 & 0 \\ 3 & - 4 & 5 \\ 0 & - 1 & 3 \end{matrix}] [\begin{matrix} 1 & 2 & 0 \\ 3 & - 4 & 5 \\ 0 & - 1 & 3 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 + 6 + 0 & 2 - 8 - 0 & 0 + 10 + 0 \\ 3 - 12 + 0 & 6 + 16 - 5 & 0 - 20 + 15 \\ 0 - 3 + 0 & 0 + 4 - 3 & 0 - 5 + 9 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 7 & - 6 & 10 \\ - 9 & 17 & - 5 \\ - 3 & 1 & 4 \end{matrix}] A^{2} - 4 A + 3 I_{3} \Rightarrow A^{2} - 4 A + 3 I_{3} = [\begin{matrix} 7 & - 6 & 10 \\ - 9 & 17 & - 5 \\ - 3 & 1 & 4 \end{matrix}] - 4 [\begin{matrix} 1 & 2 & 0 \\ 3 & - 4 & 5 \\ 0 & - 1 & 3 \end{matrix}] + 3 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow A^{2} - 4 A + 3 I_{3} = [\begin{matrix} 7 & - 6 & 10 \\ - 9 & 17 & - 5 \\ - 3 & 1 & 4 \end{matrix}] - [\begin{matrix} 4 & 8 & 0 \\ 12 & - 16 & 20 \\ 0 & - 4 & 12 \end{matrix}] + [\begin{matrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{matrix}] \Rightarrow A^{2} - 4 A + 3 I_{3} = [\begin{matrix} 7 - 4 + 3 & - 6 - 8 + 0 & 10 - 0 + 0 \\ - 9 - 12 + 0 & 17 + 16 + 3 & - 5 - 20 + 0 \\ - 3 - 0 + 0 & 1 + 4 + 0 & 4 - 12 + 3 \end{matrix}] \Rightarrow A^{2} - 4 A + 3 I_{3} = [\begin{matrix} 6 & - 14 & 10 \\ - 21 & 36 & - 25 \\ - 3 & 5 & - 5 \end{matrix}]$

Page No 4.44:

Question 42:

If f (x) = x² − 2x, find f (A), where $A = [\begin{matrix} 0 & 1 & 2 \\ 4 & 5 & 0 \\ 0 & 2 & 3 \end{matrix}]$

Answer:

$Given : f (x) = x^{2} - 2 x f (A) = A^{2} - 2 A Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 0 & 1 & 2 \\ 4 & 5 & 0 \\ 0 & 2 & 3 \end{matrix}] [\begin{matrix} 0 & 1 & 2 \\ 4 & 5 & 0 \\ 0 & 2 & 3 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 0 + 4 + 0 & 0 + 5 + 4 & 0 + 0 + 6 \\ 0 + 20 + 0 & 4 + 25 + 0 & 8 + 0 + 0 \\ 0 + 8 + 0 & 0 + 10 + 6 & 0 + 0 + 9 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 4 & 9 & 6 \\ 20 & 29 & 8 \\ 8 & 16 & 9 \end{matrix}] f (A) = A^{2} - 2 A \Rightarrow f (A) = [\begin{matrix} 4 & 9 & 6 \\ 20 & 29 & 8 \\ 8 & 16 & 9 \end{matrix}] - 2 [\begin{matrix} 0 & 1 & 2 \\ 4 & 5 & 0 \\ 0 & 2 & 3 \end{matrix}] \Rightarrow f (A) = [\begin{matrix} 4 & 9 & 6 \\ 20 & 29 & 8 \\ 8 & 16 & 9 \end{matrix}] - [\begin{matrix} 0 & 2 & 4 \\ 8 & 10 & 0 \\ 0 & 4 & 6 \end{matrix}] \Rightarrow f (A) = [\begin{matrix} 4 - 0 & 9 - 2 & 6 - 4 \\ 20 - 8 & 29 - 10 & 8 - 0 \\ 8 - 0 & 16 - 4 & 9 - 6 \end{matrix}] \Rightarrow f (A) = [\begin{matrix} 4 & 7 & 2 \\ 12 & 19 & 8 \\ 8 & 12 & 3 \end{matrix}]$

Page No 4.44:

Question 43:

If f (x) = x³ + 4x² − x, find f (A), where $A = [\begin{matrix} 0 & 1 & 2 \\ 2 & - 3 & 0 \\ 1 & - 1 & 0 \end{matrix}]$

Answer:

$Given : f (x) = x^{3} + 4 x^{2} - x f (A) = A^{3} + 4 A^{2} - A Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 0 & 1 & 2 \\ 2 & - 3 & 0 \\ 1 & - 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 & 2 \\ 2 & - 3 & 0 \\ 1 & - 1 & 0 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 0 + 2 + 2 & 0 - 3 - 2 & 0 + 0 + 0 \\ 0 - 6 + 0 & 2 + 9 - 0 & 4 - 0 + 0 \\ 0 - 2 + 0 & 1 + 3 - 0 & 2 - 0 + 0 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 4 & - 5 & 0 \\ - 6 & 11 & 4 \\ - 2 & 4 & 2 \end{matrix}] A^{3} = A^{2} A \Rightarrow A^{3} = [\begin{matrix} 4 & - 5 & 0 \\ - 6 & 11 & 4 \\ - 2 & 4 & 2 \end{matrix}] [\begin{matrix} 0 & 1 & 2 \\ 2 & - 3 & 0 \\ 1 & - 1 & 0 \end{matrix}] \Rightarrow A^{3} = [\begin{matrix} 0 - 10 + 0 & 4 + 15 - 0 & 8 - 0 + 0 \\ 0 + 22 + 4 & - 6 - 33 - 4 & - 12 + 0 + 0 \\ 0 + 8 + 2 & - 2 - 12 - 2 & - 4 + 0 + 0 \end{matrix}] \Rightarrow A^{3} = [\begin{matrix} - 10 & 19 & 8 \\ 26 & - 43 & - 12 \\ 10 & - 16 & - 4 \end{matrix}] f (A) = A^{3} + 4 A^{2} - A \Rightarrow f (A) = [\begin{matrix} - 10 & 19 & 8 \\ 26 & - 43 & - 12 \\ 10 & - 16 & - 4 \end{matrix}] + 4 [\begin{matrix} 4 & - 5 & 0 \\ - 6 & 11 & 4 \\ - 2 & 4 & 2 \end{matrix}] - [\begin{matrix} 0 & 1 & 2 \\ 2 & - 3 & 0 \\ 1 & - 1 & 0 \end{matrix}] \Rightarrow f (A) = [\begin{matrix} - 10 & 19 & 8 \\ 26 & - 43 & - 12 \\ 10 & - 16 & - 4 \end{matrix}] + [\begin{matrix} 16 & - 20 & 0 \\ - 24 & 44 & 16 \\ - 8 & 16 & 8 \end{matrix}] - [\begin{matrix} 0 & 1 & 2 \\ 2 & - 3 & 0 \\ 1 & - 1 & 0 \end{matrix}] \Rightarrow f (A) = [\begin{matrix} - 10 + 16 - 0 & 19 - 20 - 1 & 8 + 0 - 2 \\ 26 - 24 - 2 & - 43 + 44 + 3 & - 12 + 16 - 0 \\ 10 - 8 - 1 & - 16 + 16 + 1 & - 4 + 8 + 0 \end{matrix}] \Rightarrow f (A) = [\begin{matrix} 6 & - 2 & 6 \\ 0 & 4 & 4 \\ 1 & 1 & 4 \end{matrix}]$

Page No 4.44:

Question 44:

If $A = [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}]$ , then show that A is a root of the polynomial f (x) = x³ − 6x² + 7x + 2.

Answer:

$Given : f (x) = x^{3} - 6 x^{2} + 7 x + 2 f (A) = A^{3} - 6 A^{2} + 7 A + 2 I_{3} Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 + 0 + 4 & 0 + 0 + 0 & 2 + 0 + 6 \\ 0 + 0 + 2 & 0 + 4 + 0 & 0 + 2 + 3 \\ 2 + 0 + 6 & 0 + 0 + 0 & 4 + 0 + 9 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix}] A^{3} = A^{2} A \Rightarrow A^{3} = [\begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix}] [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] \Rightarrow A^{3} = [\begin{matrix} 5 + 0 + 16 & 0 + 0 + 0 & 10 + 0 + 24 \\ 2 + 0 + 10 & 0 + 8 + 0 & 4 + 4 + 15 \\ 8 + 0 + 26 & 0 + 0 + 0 & 16 + 0 + 39 \end{matrix}] \Rightarrow A^{3} = [\begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix}] A^{3} - 6 A^{2} + 7 A + 2 I_{3} \Rightarrow A^{3} - 6 A^{2} + 7 A + 2 I_{3} = [\begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix}] - 6 [\begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix}] + 7 [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] + 2 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow A^{3} - 6 A^{2} + 7 A + 2 I_{3} = [\begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix}] - [\begin{matrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{matrix}] + [\begin{matrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{matrix}] + [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}] \Rightarrow A^{3} - 6 A^{2} + 7 A + 2 I_{3} = [\begin{matrix} 21 - 30 + 7 + 2 & 0 - 0 + 0 + 0 & 34 - 48 + 14 + 0 \\ 12 - 12 + 0 + 0 & 8 - 24 + 14 + 2 & 23 - 30 + 7 + 0 \\ 34 - 48 + 14 + 0 & 0 - 0 + 0 + 0 & 55 - 78 + 21 + 2 \end{matrix}] \Rightarrow A^{3} - 6 A^{2} + 7 A + 2 I_{3} = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] = 0 Since f (A) = 0, A is the root of f (x) = x^{3} - 6 x^{2} + 7 x + 2 .$

Page No 4.44:

Question 45:

If $A = [\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}]$ , then prove that A² − 4A − 5I = O.

Answer:

$Given : A = [\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}] N o w, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}] [\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 + 4 + 4 & 2 + 2 + 4 & 2 + 4 + 2 \\ 2 + 2 + 4 & 4 + 1 + 4 & 4 + 2 + 2 \\ 2 + 4 + 2 & 4 + 2 + 2 & 4 + 4 + 1 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{matrix}] A^{2} - 4 A - 5 I \Rightarrow A^{2} - 4 A - 5 I = [\begin{matrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{matrix}] - 4 [\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}] - 5 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow A^{2} - 4 A - 5 I = [\begin{matrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{matrix}] - [\begin{matrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{matrix}] - [\begin{matrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{matrix}] \Rightarrow A^{2} - 4 A - 5 I = [\begin{matrix} 9 - 4 - 5 & 8 - 8 - 0 & 8 - 8 - 0 \\ 8 - 8 - 0 & 9 - 4 - 5 & 8 - 8 - 0 \\ 8 - 8 - 0 & 8 - 8 - 0 & 9 - 4 - 5 \end{matrix}] \Rightarrow A^{2} - 4 A - 5 I = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] = 0$

Hence proved.

Page No 4.44:

Question 46:

If $A = [\begin{matrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{matrix}]$ , show that A² − 7A + 10I₃ = O

Answer:

$Given : A = [\begin{matrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{matrix}] Here, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{matrix}] [\begin{matrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 9 + 2 + 0 & 6 + 8 + 0 & 0 + 0 + 0 \\ 3 + 4 + 0 & 2 + 16 + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + 0 + 0 & 0 + 0 + 25 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{matrix}] Now, A^{2} - 7 A + 10 I_{3} \Rightarrow A^{2} - 7 A + 10 I_{3} = [\begin{matrix} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{matrix}] - 7 [\begin{matrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{matrix}] + 10 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow A^{2} - 7 A + 10 I_{3} = [\begin{matrix} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{matrix}] - [\begin{matrix} 21 & 14 & 0 \\ 7 & 28 & 0 \\ 0 & 0 & 35 \end{matrix}] + [\begin{matrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{matrix}] \Rightarrow A^{2} - 7 A + 10 I_{3} = [\begin{matrix} 11 - 21 + 10 & 14 - 14 + 0 & 0 - 0 + 0 \\ 7 - 7 + 0 & 18 - 28 + 10 & 0 - 0 + 0 \\ 0 - 0 + 0 & 0 - 0 + 0 & 25 - 35 + 10 \end{matrix}] \Rightarrow A^{2} - 7 A + 10 I_{3} = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] = 0$

Page No 4.44:

Question 47:

Without using the concept of inverse of a matrix, find the matrix $[\begin{matrix} x & y \\ z & u \end{matrix}]$ such that
$[\begin{matrix} 5 & - 7 \\ - 2 & 3 \end{matrix}] [\begin{matrix} x & y \\ z & u \end{matrix}] = [\begin{matrix} - 16 & - 6 \\ 7 & 2 \end{matrix}]$

Answer:

$Given : [\begin{matrix} 5 & - 7 \\ - 2 & 3 \end{matrix}] [\begin{matrix} x & y \\ z & u \end{matrix}] = [\begin{matrix} - 16 & - 6 \\ 7 & 2 \end{matrix}] \Rightarrow [\begin{matrix} 5 x - 7 z & 5 y - 7 u \\ - 2 x + 3 z & - 2 y + 3 u \end{matrix}] = [\begin{matrix} - 16 & - 6 \\ 7 & 2 \end{matrix}] The corresponding elements of two equal matrices are equal . ∴ 5 x - 7 z = - 16 . . . (1) 5 y - 7 u = - 6 . . . (2) - 2 y + 3 u = 2 \Rightarrow 3 u = 2 + 2 y \Rightarrow u = \frac{2 + 2 y}{3} . . . (3) - 2 x + 3 z = 7 \Rightarrow 3 z = 7 + 2 x \Rightarrow z = \frac{7 + 2 x}{3} . . . (4) Putting the value of z in eq . (1), we get 5 x - 7 (\frac{7 + 2 x}{3}) = - 16 \Rightarrow 5 x - \frac{49 + 14 x}{3} = - 16 \Rightarrow \frac{15 x - 49 - 14 x}{3} = - 16 \Rightarrow x - 49 = - 48 \Rightarrow x = - 48 + 49 ∴ x = 1 Putting the value of x in eq . (4), we get z = \frac{7 + 2 (1)}{3} \Rightarrow z = \frac{9}{3} = 3 Putting the value of u in eq . (2), we get 5 y - 7 (\frac{2 + 2 y}{3}) = - 6 \Rightarrow 5 y - \frac{14 + 14 y}{3} = - 6 \Rightarrow \frac{15 y - 14 - 14 y}{3} = - 6 \Rightarrow y - 14 = - 18 \Rightarrow y = - 18 + 14 ∴ y = - 4 Putting the value of y in eq . (3), we get u = \frac{2 + 2 (- 4)}{3} \Rightarrow u = - 2 ∴ [\begin{matrix} x & y \\ z & u \end{matrix}] = [\begin{matrix} 1 & - 4 \\ 3 & - 2 \end{matrix}]$

Page No 4.45:

Question 48:

Find the matrix A such that
(i) $[\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] A = [\begin{matrix} 3 & 3 & 5 \\ 1 & 0 & 1 \end{matrix}]$

(ii) $A [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix}] = [\begin{matrix} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{matrix}]$

(iii) $[\begin{matrix} 4 \\ 1 \\ 3 \end{matrix}] A = [\begin{matrix} - 4 & 8 & 4 \\ - 1 & 2 & 1 \\ - 3 & 6 & 3 \end{matrix}]$

(iv) $[\begin{matrix} 2 & 1 & 3 \end{matrix}] [\begin{matrix} - 1 & 0 & - 1 \\ - 1 & 1 & 0 \\ 0 & 1 & 1 \end{matrix}] [\begin{matrix} 1 \\ 0 \\ - 1 \end{matrix}] = A$

(v) $[\begin{matrix} 2 & - 1 \\ 1 & 0 \\ - 3 & 4 \end{matrix}]$ A $= [\begin{matrix} - 1 & - 8 & - 10 \\ 1 & - 2 & - 5 \\ 9 & 22 & 15 \end{matrix}]$

(vi) A $= [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix}] = [\begin{matrix} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \\ 11 & 10 & 9 \end{matrix}]$

Answer:

$(i) Let A = [\begin{matrix} x & y & z \\ a & b & c \end{matrix}] \Rightarrow [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} x & y & z \\ a & b & c \end{matrix}] = [\begin{matrix} 3 & 3 & 5 \\ 1 & 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} x + a & y + b & z + c \\ 0 + a & 0 + b & 0 + c \end{matrix}] = [\begin{matrix} 3 & 3 & 5 \\ 1 & 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} x + a & y + b & z + c \\ a & b & c \end{matrix}] = [\begin{matrix} 3 & 3 & 5 \\ 1 & 0 & 1 \end{matrix}] The corresponding elements of two equal matrices are equal . \Rightarrow x + a = 3 . . . (1) y + b = 3 . . . (2) z + c = 5 . . . (3) \Rightarrow a = 1, b = 0 and c = 1 Putting the value of a in eq . (1), we get x + 1 = 3 \Rightarrow x = 3 - 1 ∴ x = 2 Putting the value of b in eq . (2), we get y + b = 3 \Rightarrow y + 0 = 3 ∴ y = 3 Putting the value of c in eq . (3), we get z + 1 = 5 \Rightarrow z = 5 - 1 ∴ z = 4 ∴ A = [\begin{matrix} 2 & 3 & 4 \\ 1 & 0 & 1 \end{matrix}]$

$(ii) Let A = [\begin{matrix} w & x \\ y & z \end{matrix}] \Rightarrow A [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix}] = [\begin{matrix} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{matrix}] \Rightarrow [\begin{matrix} w & x \\ y & z \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix}] = [\begin{matrix} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{matrix}] \Rightarrow [\begin{matrix} w + 4 x & 2 w + 5 x & 3 w + 6 x \\ y + 4 z & 2 y + 5 z & 3 y + 6 z \end{matrix}] = [\begin{matrix} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{matrix}] The correspnding elements of two equal matrices are equal . ∴ 3 w + 6 x = - 9 . . . (1) y + 4 z = 2 y = 2 - 4 z . . . (2) w + 4 x = - 7 \Rightarrow w = - 7 - 4 x . . . (3) 2 y + 5 z = 4 . . . (4) Putting the value of w in eq . (1), we get 3 (- 7 - 4 x) + 6 x = - 9 \Rightarrow - 21 - 12 x + 6 x = - 9 \Rightarrow - 6 x = 12 \Rightarrow x = - 2 Putting the value of x in eq . (3), we get w = - 7 - 4 (- 2) \Rightarrow w = - 7 + 8 \Rightarrow w = 1 Putting the value of y in eq . (4), we get 2 (2 - 4 z) + 5 z = 4 \Rightarrow 4 - 8 z + 5 z = 4 \Rightarrow 4 - 3 z = 4 \Rightarrow - 3 z = 0 \Rightarrow z = 0 Putting the value of z in eq . (2), we get y = 2 - 4 (0) \Rightarrow y = 2 ∴ A = [\begin{matrix} 1 & - 2 \\ 2 & 0 \end{matrix}]$

$(iii) Let A = [\begin{matrix} x & y & z \end{matrix}] \Rightarrow [\begin{matrix} 4 \\ 1 \\ 3 \end{matrix}] [\begin{matrix} x & y & z \end{matrix}] = [\begin{matrix} - 4 & 8 & 4 \\ - 1 & 2 & 1 \\ - 3 & 6 & 3 \end{matrix}] \Rightarrow [\begin{matrix} 4 x & 4 y & 4 z \\ x & y & z \\ 3 x & 3 y & 3 z \end{matrix}] = [\begin{matrix} - 4 & 8 & 4 \\ - 1 & 2 & 1 \\ - 3 & 6 & 3 \end{matrix}] The corresponding elements of two equal matrices are equal . \Rightarrow 4 x = - 4 . . . (1) 4 y = 8 . . . (2) 4 z = 4 . . . (3) \Rightarrow x = - 1, y = 2 and z = 1 ∴ A = [\begin{matrix} - 1 & 2 & 1 \end{matrix}]$

$(iv) Let A = [\begin{matrix} x \end{matrix}] \Rightarrow [\begin{matrix} 2 & 1 & 3 \end{matrix}] [\begin{matrix} - 1 & 0 & - 1 \\ - 1 & 1 & 0 \\ 0 & 1 & 1 \end{matrix}] [\begin{matrix} 1 \\ 0 \\ - 1 \end{matrix}] = A \Rightarrow [\begin{matrix} 2 & 1 & 3 \end{matrix}] [\begin{matrix} - 1 & 0 & - 1 \\ - 1 & 1 & 0 \\ 0 & 1 & 1 \end{matrix}] [\begin{matrix} 1 \\ 0 \\ - 1 \end{matrix}] = [\begin{matrix} x \end{matrix}] \Rightarrow [\begin{matrix} - 2 - 1 + 0 & 0 + 1 + 3 & - 2 + 0 + 3 \end{matrix}] [\begin{matrix} 1 \\ 0 \\ - 1 \end{matrix}] = [\begin{matrix} x \end{matrix}] \Rightarrow [\begin{matrix} - 3 & 4 & 1 \end{matrix}] [\begin{matrix} 1 \\ 0 \\ - 1 \end{matrix}] = [\begin{matrix} x \end{matrix}] \Rightarrow [\begin{matrix} - 3 + 0 - 1 \end{matrix}] = [\begin{matrix} x \end{matrix}] \Rightarrow [\begin{matrix} - 4 \end{matrix}] = [\begin{matrix} x \end{matrix}] The corresponding elements of two equal matrices are equal . ∴ x = - 4 ∴ A = [- 4]$

$(v) [\begin{matrix} 2 & - 1 \\ 1 & 0 \\ - 3 & 4 \end{matrix}] A = [\begin{matrix} - 1 & - 8 & - 10 \\ 1 & - 2 & - 5 \\ 9 & 22 & 15 \end{matrix}] Let A = [\begin{matrix} x & y & z \\ a & b & c \end{matrix}] \Rightarrow [\begin{matrix} 2 & - 1 \\ 1 & 0 \\ - 3 & 4 \end{matrix}] [\begin{matrix} x & y & z \\ a & b & c \end{matrix}] = [\begin{matrix} - 1 & - 8 & - 10 \\ 1 & - 2 & - 5 \\ 9 & 22 & 15 \end{matrix}] \Rightarrow [\begin{matrix} 2 x - a & 2 y - b & 2 z - c \\ x & y & z \\ - 3 x + 4 a & - 3 y + 4 b & - 3 z + 4 c \end{matrix}] = [\begin{matrix} - 1 & - 8 & - 10 \\ 1 & - 2 & - 5 \\ 9 & 22 & 15 \end{matrix}] By comparing the elements of second row, we get x = 1, y = - 2, z = - 5 By comparing the elements of first row, we get 2 x - a = - 1 \Rightarrow 2 - a = - 1 \Rightarrow a = 3 Also, 2 y - b = - 8 \Rightarrow - 4 - b = - 8 \Rightarrow b = 4 And, 2 z - c = - 10 \Rightarrow - 10 - c = - 10 \Rightarrow c = 0 ∴ A = [\begin{matrix} 1 & - 2 & - 5 \\ 3 & 4 & 0 \end{matrix}]$

$(vi) A [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix}] = [\begin{matrix} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \\ 11 & 10 & 9 \end{matrix}] Let A = [\begin{matrix} x & a \\ y & b \\ z & c \end{matrix}] \Rightarrow [\begin{matrix} x & a \\ y & b \\ z & c \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix}] = [\begin{matrix} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \\ 11 & 10 & 9 \end{matrix}] \Rightarrow [\begin{matrix} x + 4 a & 2 x + 5 a & 3 x + 6 a \\ y + 4 b & 2 y + 5 b & 3 y + 6 b \\ z + 4 c & 2 z + 5 c & 3 z + 6 c \end{matrix}] = [\begin{matrix} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \\ 11 & 10 & 9 \end{matrix}] By comparing the corresponding elements, we get x + 4 a = - 7 and 2 x + 5 a = - 8 \Rightarrow a = - 2 and x = 1 Also, y + 4 b = 2 and 2 y + 5 b = 4 \Rightarrow b = 0 and y = 2 And, z + 4 c = 11 and 2 z + 5 c = 10 \Rightarrow c = 4 and z = - 5 ∴ A = [\begin{matrix} 1 & - 2 \\ 2 & 0 \\ - 5 & 4 \end{matrix}]$

Page No 4.45:

Question 49:

Find a 2 × 2 matrix A such that
$A [\begin{matrix} 1 & - 2 \\ 1 & 4 \end{matrix}] = 6 I_{2}$

Answer:

Let A = $[\begin{matrix} w & x \\ y & z \end{matrix}]$

Now,

$[\begin{matrix} w & x \\ y & z \end{matrix}] [\begin{matrix} 1 & - 2 \\ 1 & 4 \end{matrix}] = 6 I_{2} \Rightarrow [\begin{matrix} w + x & - 2 w + 4 x \\ y + z & - 2 y + 4 z \end{matrix}] = 6 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} w + x & - 2 w + 4 x \\ y + z & - 2 y + 4 z \end{matrix}] = [\begin{matrix} 6 & 0 \\ 0 & 6 \end{matrix}] The corresponding elements of two equal matrices are equal . ∴ w + x = 6 \Rightarrow w = 6 - x . . . (1) - 2 w + 4 x = 0 . . . (2) Putting the value of w in e q . (2), we get - 2 (6 - x) + 4 x = 0 \Rightarrow - 12 + 2 x + 4 x = 0 \Rightarrow - 12 + 6 x = 0 \Rightarrow 6 x = 12 \Rightarrow x = 2 Putting the value of x in eq . (1), we get w = 6 - 2 \Rightarrow w = 4 Now, y + z = 0 \Rightarrow y = - z . . . (3) - 2 y + 4 z = 6 . . . (4) Putting the value of y in eq . (4), we get - 2 (- z) + 4 z = 6 \Rightarrow 2 z + 4 z = 6 \Rightarrow 6 z = 6 \Rightarrow z = 1 Putting the value of z in eq . (3), we get y = - 1 ∴ A = [\begin{matrix} 4 & 2 \\ - 1 & 1 \end{matrix}]$

Page No 4.45:

Question 50:

If $A = [\begin{matrix} 0 & 0 \\ 4 & 0 \end{matrix}]$ , find A¹⁶.

Answer:

$Given : A = [\begin{matrix} 0 & 0 \\ 4 & 0 \end{matrix}] Here, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 0 & 0 \\ 4 & 0 \end{matrix}] [\begin{matrix} 0 & 0 \\ 4 & 0 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] A^{4} = A^{2} A^{2} \Rightarrow A^{4} = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \Rightarrow A^{4} = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] A^{8} = A^{4} A^{4} \Rightarrow A^{8} = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \Rightarrow A^{8} = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] A^{16} = A^{8} A^{8} \Rightarrow A^{16} = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] ∴ A^{16} = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] Thus, A^{16} is a null matrix .$

Page No 4.45:

Question 51:

If $A = [\begin{matrix} 0 & - x \\ x & 0 \end{matrix}], B = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}]$ and x² = −1, then show that (A + B)² = A² + B².

Answer:

Given: $A = [\begin{matrix} 0 & - x \\ x & 0 \end{matrix}], B = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}]$ and x² = −1

To show: (A + B)² = A² + B²

LHS:
$A + B = [\begin{matrix} 0 & - x \\ x & 0 \end{matrix}] + [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] = [\begin{matrix} 0 + 0 & - x + 1 \\ x + 1 & 0 + 0 \end{matrix}] = [\begin{matrix} 0 & - x + 1 \\ x + 1 & 0 \end{matrix}] {(A + B)}^{2} = [\begin{matrix} 0 & - x + 1 \\ x + 1 & 0 \end{matrix}] [\begin{matrix} 0 & - x + 1 \\ x + 1 & 0 \end{matrix}] = [\begin{matrix} 0 + (1 - x) (1 + x) & 0 + 0 \\ 0 + 0 & (x + 1) (1 - x) + 0 \end{matrix}] = [\begin{matrix} 1 - x^{2} & 0 \\ 0 & 1 - x^{2} \end{matrix}] . . . (1)$

RHS:
$A = [\begin{matrix} 0 & - x \\ x & 0 \end{matrix}] A^{2} = [\begin{matrix} 0 & - x \\ x & 0 \end{matrix}] [\begin{matrix} 0 & - x \\ x & 0 \end{matrix}] = [\begin{matrix} 0 - x^{2} & 0 + 0 \\ 0 + 0 & - x^{2} + 0 \end{matrix}] = [\begin{matrix} - x^{2} & 0 \\ 0 & - x^{2} \end{matrix}] . . . (2) B = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] B^{2} = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] = [\begin{matrix} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] . . . (3) Adding (2) and (3), we get A^{2} + B^{2} = [\begin{matrix} - x^{2} & 0 \\ 0 & - x^{2} \end{matrix}] + [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 - x^{2} & 0 \\ 0 & 1 - x^{2} \end{matrix}] . . . (4)$

Comparing (1) and (4), we get

(A + B)² = A² + B²

Page No 4.45:

Question 52:

If $A = [\begin{matrix} 1 & 0 & - 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}]$ , then verify that A² + A = A(A + I), where I is the identity matrix.

Answer:

To verify: A² + A = A(A + I),

Given: $A = [\begin{matrix} 1 & 0 & - 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}]$
$A^{2} = [\begin{matrix} 1 & 0 & - 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}] [\begin{matrix} 1 & 0 & - 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}] = [\begin{matrix} 1 + 0 + 0 & 0 + 0 - 3 & - 3 + 0 - 3 \\ 2 + 2 + 0 & 0 + 1 + 3 & - 6 + 3 + 3 \\ 0 + 2 + 0 & 0 + 1 + 1 & 0 + 3 + 1 \end{matrix}] = [\begin{matrix} 1 & - 3 & - 6 \\ 4 & 4 & 0 \\ 2 & 2 & 4 \end{matrix}]$

LHS:
$A^{2} + A = [\begin{matrix} 1 & - 3 & - 6 \\ 4 & 4 & 0 \\ 2 & 2 & 4 \end{matrix}] + [\begin{matrix} 1 & 0 & - 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}] = [\begin{matrix} 1 + 1 & - 3 + 0 & - 6 - 3 \\ 4 + 2 & 4 + 1 & 0 + 3 \\ 2 + 0 & 2 + 1 & 4 + 1 \end{matrix}] = [\begin{matrix} 2 & - 3 & - 9 \\ 6 & 5 & 3 \\ 2 & 3 & 5 \end{matrix}]$

RHS:
$A + I = [\begin{matrix} 1 & 0 & - 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}] + [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 + 1 & 0 + 0 & - 3 + 0 \\ 2 + 0 & 1 + 1 & 3 + 0 \\ 0 + 0 & 1 + 0 & 1 + 1 \end{matrix}] = [\begin{matrix} 2 & 0 & - 3 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{matrix}] A (A + I) = [\begin{matrix} 1 & 0 & - 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}] [\begin{matrix} 2 & 0 & - 3 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{matrix}] = [\begin{matrix} 2 + 0 + 0 & 0 + 0 - 3 & - 3 + 0 - 6 \\ 4 + 2 + 0 & 0 + 2 + 3 & - 6 + 3 + 6 \\ 0 + 2 + 0 & 0 + 2 + 1 & 0 + 3 + 2 \end{matrix}] = [\begin{matrix} 2 & - 3 & - 9 \\ 6 & 5 & 3 \\ 2 & 3 & 5 \end{matrix}]$

Therefore, LHS = RHS.

Hence, A² + A = A(A + I) is verified.

Page No 4.45:

Question 53:

If $A = [\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}]$ , then find A² − 5A − 14I. Hence, obtain A³.

Answer:

Given: $A = [\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}]$

$A^{2} = [\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}] [\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}] = [\begin{matrix} 9 + 20 & - 15 - 10 \\ - 12 - 8 & 20 + 4 \end{matrix}] = [\begin{matrix} 29 & - 25 \\ - 20 & 24 \end{matrix}]$

$A^{2} - 5 A - 14 I = [\begin{matrix} 29 & - 25 \\ - 20 & 24 \end{matrix}] - 5 [\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}] - 14 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 29 - 15 - 14 & - 25 + 25 - 0 \\ - 20 + 20 - 0 & 24 - 10 - 14 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

Therefore, A² − 5A − 14I = 0 ...(1)

Premultiplying the (1) by A, we get

A(A² − 5A − 14I) = A.0
⇒ A³ − 5A² − 14A = 0
⇒ A³ = 5A² + 14A
$∴ A^{3} = 5 [\begin{matrix} 29 & - 25 \\ - 20 & 24 \end{matrix}] + 14 [\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}] = [\begin{matrix} 145 + 42 & - 125 - 70 \\ - 100 - 56 & 120 + 28 \end{matrix}] = [\begin{matrix} 187 & - 195 \\ - 156 & 148 \end{matrix}]$

Page No 4.45:

Question 54:

(i) If $P (x) = [\begin{matrix} \cos x & \sin x \\ - \sin x & \cos x \end{matrix}]$ , then show that P(x) P(y) = P(x + y) = P(y) P(x).

(ii) If $P = [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}] and Q = [\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix}], prove that P Q = [\begin{matrix} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{matrix}] = Q P$

Answer:

(i) Given: $P (x) = [\begin{matrix} \cos x & \sin x \\ - \sin x & \cos x \end{matrix}]$

then, $P (y) = [\begin{matrix} \cos y & \sin y \\ - \sin y & \cos y \end{matrix}]$

Now,
$P (x) P (y) = [\begin{matrix} \cos x & \sin x \\ - \sin x & \cos x \end{matrix}] [\begin{matrix} \cos y & \sin y \\ - \sin y & \cos y \end{matrix}] = [\begin{matrix} \cos x \cos y - \sin x \sin y & \cos x \sin y + \sin x \cos y \\ - \sin x \cos y - \cos x \sin y & - \sin x \sin y + \cos x \cos y \end{matrix}] = [\begin{matrix} \cos (x + y) & \sin (x + y) \\ - \sin (x + y) & \cos (x + y) \end{matrix}] . . . (1)$

Also, $P (x + y) = [\begin{matrix} \cos (x + y) & \sin (x + y) \\ - \sin (x + y) & \cos (x + y) \end{matrix}] . . . (2)$

Now,
$P (y) P (x) = [\begin{matrix} \cos y & \sin y \\ - \sin y & \cos y \end{matrix}] [\begin{matrix} \cos x & \sin x \\ - \sin x & \cos x \end{matrix}] = [\begin{matrix} \cos y \cos x - \sin y \sin x & \cos y \sin x + \sin y \cos x \\ - \sin y \cos x - \cos y \sin x & - \sin y \sin x + \cos y \cos x \end{matrix}] = [\begin{matrix} \cos (x + y) & \sin (x + y) \\ - \sin (x + y) & \cos (x + y) \end{matrix}] . . . (3)$

From (1), (2) and (3), we get

P(x) P(y) = P(x + y) = P(y) P(x)

(ii) Given: $P = [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}] and Q = [\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix}]$

Now,
$P Q = [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}] [\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix}] = [\begin{matrix} x a + 0 + 0 & 0 + 0 + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + y b + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + 0 + 0 & 0 + 0 + z c \end{matrix}] = [\begin{matrix} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{matrix}] . . . (4)$

Also,
$Q P = [\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix}] [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}] = [\begin{matrix} a x + 0 + 0 & 0 + 0 + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + b y + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + 0 + 0 & 0 + 0 + c z \end{matrix}] = [\begin{matrix} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{matrix}] . . . (5)$

From (4) and (5), we get
$P Q = [\begin{matrix} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{matrix}] = Q P$

Page No 4.45:

Question 55:

If $A = [\begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{matrix}]$ , find A² − 5A + 4I and hence find a matrix X such that A² − 5A + 4I + X = 0.

Answer:

Given: $A = [\begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{matrix}]$

$A^{2} = [\begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{matrix}] [\begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{matrix}] = [\begin{matrix} 4 + 0 + 1 & 0 + 0 - 1 & 2 + 0 + 0 \\ 4 + 2 + 3 & 0 + 1 - 3 & 2 + 3 + 0 \\ 2 - 2 + 0 & 0 - 1 - 0 & 1 - 3 + 0 \end{matrix}] = [\begin{matrix} 5 & - 1 & 2 \\ 9 & - 2 & 5 \\ 0 & - 1 & - 2 \end{matrix}]$

Now,
$A^{2} - 5 A + 4 I = [\begin{matrix} 5 & - 1 & 2 \\ 9 & - 2 & 5 \\ 0 & - 1 & - 2 \end{matrix}] - 5 [\begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{matrix}] + 4 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 5 - 10 + 4 & - 1 - 0 + 0 & 2 - 5 + 0 \\ 9 - 10 + 0 & - 2 - 5 + 4 & 5 - 15 + 0 \\ 0 - 5 + 0 & - 1 + 5 + 0 & - 2 - 0 + 4 \end{matrix}] = [\begin{matrix} - 1 & - 1 & - 3 \\ - 1 & - 3 & - 10 \\ - 5 & 4 & 2 \end{matrix}]$

Now, A² − 5A + 4I + X = 0
⇒ X = −(A² − 5A + 4I)
$∴ X = - [\begin{matrix} - 1 & - 1 & - 3 \\ - 1 & - 3 & - 10 \\ - 5 & 4 & 2 \end{matrix}] = [\begin{matrix} 1 & 1 & 3 \\ 1 & 3 & 10 \\ 5 & - 4 & - 2 \end{matrix}]$

Page No 4.45:

Question 56:

If $A = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}]$ , prove that $A^{n} = [\begin{matrix} 1 & n \\ 0 & 1 \end{matrix}]$ for all positive integers n.

Answer:

We shall prove the result by the principle of mathematical induction on n.

Step 1: If n = 1, by definition of integral powers of matrix, we have
$A^{1} = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] = A$
So, the result is true for n = 1.

Step 2: Let the result be true for n = m. Then,
$A^{m} = [\begin{matrix} 1 & m \\ 0 & 1 \end{matrix}]$ ...(1)

Now, we shall show that the result is true for $n = m + 1$ .
Here,
$A^{m + 1} = [\begin{matrix} 1 & m + 1 \\ 0 & 1 \end{matrix}]$

By definition of integral power of matrix, we have
$A^{m + 1} = A^{m} A = [\begin{matrix} 1 & m \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] [From eq . (1)] = [\begin{matrix} 1 + 0 & 1 + m \\ 0 + 0 & 0 + 1 \end{matrix}] = [\begin{matrix} 1 & 1 + m \\ 0 & 1 \end{matrix}]$

This shows that when the result is true for n = m, it is also true for n = m + 1.

Hence, by the principle of mathematical induction, the result is valid for any positive integer n.

Page No 4.45:

Question 57:

If $A = [\begin{matrix} a & b \\ 0 & 1 \end{matrix}]$ , prove that $A^{n} = [\begin{matrix} a^{n} & b (a^{n} - 1) / a - 1 \\ 0 & 1 \end{matrix}]$ for every positive integer n.

Answer:

We shall prove the result by the principle of mathematical induction on n.

Step 1: If n = 1, by definition of integral power of a matrix, we have
$A^{1} = [\begin{matrix} a^{1} & b (a^{1} - 1) / a - 1 \\ 0 & 1 \end{matrix}] = [\begin{matrix} a & b \\ 0 & 1 \end{matrix}] = A$

So, the result is true for n = 1.

Step 2: Let the result be true for n = m. Then,
$A^{m} = [\begin{matrix} a^{m} & b (a^{m} - 1) / a - 1 \\ 0 & 1 \end{matrix}]$ ...(1)

Now, we shall show that the result is true for $n = m + 1$ .
Here,
$A^{m + 1} = [\begin{matrix} a^{m + 1} & b (a^{m + 1} - 1) / a - 1 \\ 0 & 1 \end{matrix}]$

By definition of integral power of matrix, we have
$A^{m + 1} = A^{m} A \Rightarrow A^{m + 1} = [\begin{matrix} a^{m} & b (a^{m} - 1) / a - 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} a & b \\ 0 & 1 \end{matrix}] [From eq . (1)] \Rightarrow A^{m + 1} = [\begin{matrix} a^{m} a + 0 & \{a^{m} b + b (a^{m} - 1)\} / a - 1 \\ 0 + 0 & 0 + 1 \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} a^{m + 1} & (a^{m + 1} b - a^{m} b + a^{m} b - b) / a - 1 \\ 0 & 1 \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} a^{m + 1} & b (a^{m + 1} - 1) / a - 1 \\ 0 & 1 \end{matrix}]$

This shows that when the result is true for n = m, it is also true for n = m +1.

Hence, by the principle of mathematical induction, the result is valid for any positive integer n.

Page No 4.45:

Question 58:

If $A = [\begin{matrix} cosθ & i sinθ \\ i sinθ & cosθ \end{matrix}]$ , then prove by principle of mathematical induction that

$A^{n} = [\begin{matrix} \cos n θ & i sinθ \\ i \sin n θ & \cos n θ \end{matrix}]$ for all n ∈ N.

Answer:

We shall prove the result by the principle of mathematical induction on n.

Step 1: If n = 1, by definition of integral power of a matrix, we have

$A^{1} = [\begin{matrix} \cos 1 θ & i \sin 1 θ \\ i \sin 1 θ & \cos 1 θ \end{matrix}] = [\begin{matrix} \cos θ & i \sin θ \\ i \sin θ & \cos θ \end{matrix}] = A$

Thus, the result is true for n=1.

Step 2: Let the result be true for n = m. Then,

$A^{m} = [\begin{matrix} \cos m θ & i \sin m θ \\ i \sin m θ & \cos m θ \end{matrix}]$

Now we shall show that the result is true for $n = m + 1$ .
Here,
$A^{m + 1} = [\begin{matrix} \cos (m + 1) θ & i \sin (m + 1) θ \\ i \sin (m + 1) θ & \cos (m + 1) θ \end{matrix}]$ ...(1)

By definition of integral power of matrix, we have
$A^{m + 1} = A^{m} A \Rightarrow A^{m + 1} = [\begin{matrix} \cos m θ & i \sin m θ \\ i \sin m θ & \cos m θ \end{matrix}] [\begin{matrix} \cos θ & i \sin θ \\ i \sin θ & \cos θ \end{matrix}] [From eq . (1)] \Rightarrow A^{m + 1} = [\begin{matrix} \cos m θ . \cos θ + i \sin m θ . i \sin θ & \cos m θ . i \sin θ + i \sin m θ . \cos θ \\ i \sin m θ . \cos θ + \cos m θ . i \sin θ & i \sin m θ . i \sin θ + \cos m θ . \cos θ \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} \cos m θ . \cos θ - \sin m θ . \sin θ & i (\cos m θ . \sin θ + \sin m θ . \cos θ) \\ i (\sin m θ . \cos θ + \cos m θ \sin θ) & - \sin m θ . \sin θ + \cos m θ . \cos θ \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} \cos m θ . \cos θ - \sin m θ . \sin θ & i (\cos m θ . \sin θ + \sin m θ . \cos θ) \\ i (\sin m θ . \cos θ + \cos m θ \sin θ) & \cos m θ . \cos θ - \sin m θ . \sin θ \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} \cos (m θ + θ) & i \sin (m θ + θ) \\ i \sin (m θ + θ) & \cos (m θ + θ) \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} \cos (m + 1) θ & i \sin (m + 1) θ \\ i \sin (m + 1) θ & \cos (m + 1) θ \end{matrix}]$

This shows that when the result is true for n = m, it is true for $n = m + 1$ .
Hence, by the principle of mathematical induction, the result is valid for all n $\in N$ .

Disclaimer: n is missing before $θ$ in a₁₂ in Aⁿ.

Page No 4.46:

Question 59:

If $A = [\begin{matrix} \cos α + \sin α & \sqrt{2} \sin α \\ - \sqrt{2} \sin α & \cos α - \sin α \end{matrix}]$ , prove that

$A^{n} = [\begin{matrix} \cos n α + \sin n α & \sqrt{2} \sin n α \\ - \sqrt{2} \sin n α & \cos n α - \sin n α \end{matrix}]$ for all n ∈ N.

Answer:

We shall prove the result by the principle of mathematical induction on n.

Step 1: If n = 1, by definition of integral power of a matrix, we have

$A^{1} = [\begin{matrix} \cos 1 α + \sin 1 α & \sqrt{2} \sin 1 α \\ - \sqrt{2} \sin 1 α & \cos 1 α - \sin 1 α \end{matrix}] = [\begin{matrix} \cos α + \sin α & \sqrt{2} \sin α \\ - \sqrt{2} \sin α & \cos α - \sin α \end{matrix}] = A$

So, the result is true for n = 1.

Step 2: Let the result be true for n = m. Then,
$A^{m} = [\begin{matrix} \cos m α + \sin m α & \sqrt{2} \sin m α \\ - \sqrt{2} \sin m α & \cos m α - \sin m α \end{matrix}]$ ...(1)

Now we shall show that the result is true for $n = m + 1$ .
Here,
$A^{m + 1} = [\begin{matrix} \cos (m + 1) α + \sin (m + 1) α & \sqrt{2} \sin (m + 1) α \\ - \sqrt{2} \sin (m + 1) α & \cos (m + 1) α - \sin (m + 1) α \end{matrix}]$

By definition of integral power of matrix, we have
$A^{m + 1} = A^{m} . A \Rightarrow A^{m + 1} = [\begin{matrix} \cos m α + \sin m α & \sqrt{2} \sin m α \\ - \sqrt{2} \sin m α & \cos m α - \sin m α \end{matrix}] [\begin{matrix} \cos α + \sin α & \sqrt{2} \sin α \\ - \sqrt{2} \sin α & \cos α - \sin α \end{matrix}] [From eq . (1)] \Rightarrow A^{m + 1} = [\begin{matrix} (\cos m α + \sin m α) (\cos α + \sin α) - \sqrt{2} \sin m α (\sqrt{2} \sin α) & (\cos m α + \sin m α) (\sqrt{2} \sin α) + \sqrt{2} \sin m α (\cos α - \sin α) \\ - \sqrt{2} \sin m α (\cos α + \sin α) - (\cos m α - \sin m α) (\sqrt{2} \sin α) & - \sqrt{2} \sin m α (\sqrt{2} \sin α) + (\cos m α - \sin m α) (\cos α - \sin α) \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} \cos m α \cos α + \sin m α \cos α + \cos m α \sin α + \sin m α \sin α - 2 \sin m α \sin α & \sqrt{2} \sin α \cos m α + \sqrt{2} \sin α \sin m α + \sqrt{2} \sin m α \cos α - \sqrt{2} \sin m a \sin α \\ - \sqrt{2} \sin m a \cos α - \sqrt{2} \sin m a \sin α - \sqrt{2} \sin α \cos m α + \sqrt{2} \sin α \sin m α & - 2 \sin α \sin m α + \cos m α \cos α - \sin m α \cos α - \cos m α \sin α + \sin m α \sin α \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} \cos (m α - α) + \sin (m α + α) - \cos (m α - α) + \cos (m α + α) & \sqrt{2} \sin (m α + α) \\ - \sqrt{2} \sin (m α + α) & \cos (m α + α) - \sin (m α + α) \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} \cos (m α + α) + \sin (m α + α) & \sqrt{2} \sin (m α + a) \\ - \sqrt{2} \sin (m α + α) & \cos (m α + α) - \sin (m α + α) \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} \cos (m + 1) α + \sin (m + 1) α & \sqrt{2} \sin (m + 1) α \\ - \sqrt{2} \sin (m + 1) α & \cos (m + 1) α - \sin (m + 1) α \end{matrix}]$

This show that when the result is true for n = m, it is also true for n = m +1.

Hence, by the principle of mathematical induction, the result is valid for all n $\in N$ .

Page No 4.46:

Question 60:

Let $A = [\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}]$ . Use the principle of mathematical induction to show that

$A^{n} = [\begin{matrix} 1 & n & n (n + 1) / 2 \\ 0 & 1 & n \\ 0 & 0 & 1 \end{matrix}]$ for every positive integer n.

Answer:

We shall prove the result by the principle of mathematical induction on n.

Step 1: If n = 1, by definition of integral power of a matrix, we have

$A^{1} = [\begin{matrix} 1 & 1 & 1 (1 + 1) / 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}] = A$

Thus, the result is true for n = 1.

Step 2: Let the result be true for n = m. Then,

$A^{m} = [\begin{matrix} 1 & m & m (m + 1) / 2 \\ 0 & 1 & m \\ 0 & 0 & 1 \end{matrix}]$ ...(1)

Now, we shall show that the result is true for $n = m + 1$ .
Here,

$A^{m + 1} = [\begin{matrix} 1 & m + 1 & m + 1 (m + 1 + 1) / 2 \\ 0 & 1 & m + 1 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} 1 & m + 1 & (m + 1) (m + 2) / 2 \\ 0 & 1 & m + 1 \\ 0 & 0 & 1 \end{matrix}]$

By definition of integral power of matrix, we have
$A^{m + 1} = A^{m} A \Rightarrow A^{m + 1} = [\begin{matrix} 1 & m & m (m + 1) / 2 \\ 0 & 1 & m \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}] [From eq . (1)] \Rightarrow A^{m + 1} = [\begin{matrix} 1 + 0 + 0 & 1 + m + 0 & 1 + m + m (m + 1) / 2 \\ 0 + 0 + 0 & 0 + 1 + 0 & 0 + 1 + m \\ 0 + 0 + 0 & 0 + 0 + 0 & 0 + 0 + 1 \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} 1 & 1 + m & (2 + 2 m + m^{2} + m) / 2 \\ 0 & 1 & 1 + m \\ 0 & 0 & 1 \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} 1 & 1 + m & (m^{2} + 3 m + 2) / 2 \\ 0 & 1 & 1 + m \\ 0 & 0 & 1 \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} 1 & 1 + m & (m + 1) (m + 2) / 2 \\ 0 & 1 & 1 + m \\ 0 & 0 & 1 \end{matrix}]$

This shows that when the result is true for n = m, it is also true for n = m + 1.

Hence, by the principle of mathematical induction, the result is valid for any positive integer n.

Page No 4.46:

Question 61:

If B, C are n rowed square matrices and if A = B + C, BC = CB, C² = O, then show that for every n ∈ N, Aⁿ⁺¹ = Bⁿ (B + (n + 1) C).

Answer:

Let $P (n)$ be the statement given by $P (n) : A^{n + 1} = B^{n} (B + (n + 1) C)$ .

For n = 1, we have
$P (1) : A^{2} = B (B + 2 C) Here, LHS = A^{2} = (B + C) (B + C) = B (B + C) + C (B + C) = B^{2} + B C + C B + C^{2} = B^{2} + 2 B C [∵ B C = C B and C^{2} = O] = B (B + 2 C) = RHS$

Hence, the statement is true for n = 1.

If the statement is true for n = k, then
$P (k) : A^{k + 1} = B^{k} (B + (k + 1) C)$ ...(1)

For $P (k + 1)$ to be true, we must have
$P (k + 1) : A^{k + 2} = B^{k + 1} (B + (k + 2) C)$

Now,
$A^{k + 2} = A^{k + 1} A = [B^{k} (B + (k + 1) C)] (B + C) [From eqn . (1)] = [B^{k + 1} + (k + 1) B^{k} C] (B + C) = B^{k + 1} (B + C) + (k + 1) B^{k} C (B + C) = B^{k + 2} + B^{k + 1} C + (k + 1) B^{k} C B + (k + 1) B^{k} C^{2} = B^{k + 2} + B^{k + 1} C + (k + 1) B^{k} B C [∵ B C = C B a n d C^{2} = 0] = B^{k + 2} + B^{k + 1} C + (k + 1) B^{k + 1} C = B^{k + 2} + (k + 2) B^{k + 1} C = B^{k + 1} [B + (k + 2) C]$

So the statement is true for n = k+1.
Hence, by the principle of mathematical induction, $P (n)$ is true for all $n \in N$ .

Page No 4.46:

Question 62:

If A = diag (a, b, c), show that Aⁿ = diag (aⁿ, bⁿ, cⁿ) for all positive integer n.

Answer:

We shall prove the result by the principle of mathematical induction on n.

Step 1: If n = 1, by definition of integral power of a matrix, we have

$A^{1} = [\begin{matrix} a^{1} & 0 & 0 \\ 0 & b^{1} & 0 \\ 0 & 0 & c^{1} \end{matrix}] = [\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix}] = A$

So, the result is true for n = 1.

Step 2: Let the result be true for n = m. Then,

$A^{m} = [\begin{matrix} a^{m} & 0 & 0 \\ 0 & b^{m} & 0 \\ 0 & 0 & c^{m} \end{matrix}]$ ...(1)

Now, we shall check if the result is true for $n = m + 1$ .
Here,
$A^{m + 1} = [\begin{matrix} a^{m + 1} & 0 & 0 \\ 0 & b^{m + 1} & 0 \\ 0 & 0 & c^{m + 1} \end{matrix}]$

By definition of integral power of matrix, we have

$A^{m + 1} = A^{m} A \Rightarrow A^{m + 1} = [\begin{matrix} a^{m} & 0 & 0 \\ 0 & b^{m} & 0 \\ 0 & 0 & c^{m} \end{matrix}] [\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix}] [From eq . (1)] \Rightarrow A^{m + 1} = [\begin{matrix} a a^{m} + 0 + 0 & 0 + 0 + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + b b^{m} + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + 0 + 0 & 0 + 0 + c c^{m} \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} a^{m + 1} & 0 & 0 \\ 0 & b^{m + 1} & 0 \\ 0 & 0 & c^{m + 1} \end{matrix}]$

This shows that when the result is true for n = m, it is also true for $n = m + 1$ .
Hence, by the principle of mathematical induction, the result is valid for any positive integer n.

Page No 4.46:

Question 63:

If A is a square matrix, using mathematical induction prove that (A^T)ⁿ = (Aⁿ)^T for all n ∈ ℕ.

Answer:

Let the given statement P(n), be given as
P(n): (A^T)ⁿ = (Aⁿ)^T for all n ∈ ℕ.

We observe that
P(1): (A^T)¹ = A^T = (A¹)^T
Thus, P(n) is true for n = 1.

Assume that P(n) is true for n = k ∈ ℕ.
i.e., P(k): (A^T)^k = (A^k)^T

To prove that P(k + 1) is true, we have
(A^T)^k^{+ 1} = (A^T)^k.(A^T)¹
= (A^k)^T.(A¹)^T
= (A^k^{+ 1})^T
Thus, P(k + 1) is true, whenever P(k) is true.

Hence, by the Principle of mathematical induction, P(n) is true for all n ∈ ℕ.

Page No 4.46:

Question 64:

A matrix X has a + b rows and a + 2 columns while the matrix Y has b + 1 rows and a + 3 columns. Both matrices XY and YX exist. Find a and b. Can you say XY and YX are of the same type? Are they equal.

Answer:

$Here, [\begin{matrix} X \end{matrix}]_{(a + b) \times (a + 2)} {[\begin{matrix} Y \end{matrix}]}_{(b + 1) \times (a + 3)} Since X Y exists, the number of columns in X is equal to the number of rows in Y . \Rightarrow a + 2 = b + 1 . . . (1) Similarly, \sin c e Y X exists, the number of columns in Y is equal to the number of rows in X . \Rightarrow a + b = a + 3 \Rightarrow b = 3 Putting the value of b in (1), we get a + 2 = 3 + 1 \Rightarrow a = 2$

Since the order of the matrices XY and YX is not same, XY and YX are not of the same type and they are unequal.

Page No 4.46:

Question 65:

Give examples of matrices
(i) A and B such that AB ≠ BA
(ii) A and B such that AB = O but A ≠ 0, B ≠ 0.
(iii) A and B such that AB = O but BA ≠ O.
(iv) A, B and C such that AB = AC but B ≠ C, A ≠ 0.

Answer:

$(i) Let A = [\begin{matrix} 1 & - 2 \\ 3 & 2 \end{matrix}] and B = [\begin{matrix} 2 & 3 \\ - 1 & 2 \end{matrix}] A B = [\begin{matrix} 1 & - 2 \\ 3 & 2 \end{matrix}] [\begin{matrix} 2 & 3 \\ - 1 & 2 \end{matrix}] \Rightarrow A B = [\begin{matrix} 2 + 2 & 3 - 4 \\ 6 - 2 & 9 + 4 \end{matrix}] \Rightarrow A B = [\begin{matrix} 4 & - 1 \\ 4 & 13 \end{matrix}] Now, B A = [\begin{matrix} 2 & 3 \\ - 1 & 2 \end{matrix}] [\begin{matrix} 1 & - 2 \\ 3 & 2 \end{matrix}] \Rightarrow B A = [\begin{matrix} 2 + 9 & - 4 + 6 \\ - 1 + 6 & 2 + 4 \end{matrix}] \Rightarrow B A = [\begin{matrix} 11 & 2 \\ 5 & 6 \end{matrix}]$

Thus, AB ≠ BA.

$(ii) Let A = [\begin{matrix} 0 & 2 \\ 0 & 0 \end{matrix}] and B = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] ∴ A B = [\begin{matrix} 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = O$

Thus, AB = O while A ≠ 0 and B ≠ 0.

$(iii) Let A = [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] and B = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] ∴ A B = O and B A = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] \Rightarrow B A = [\begin{matrix} 0 + 0 & 1 + 0 \\ 0 + 0 & 0 + 0 \end{matrix}] \Rightarrow B A = [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]$

Thus, AB = O but BA ≠ O.

$(i v) Let A = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}], B = [\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}] and C = [\begin{matrix} 0 & 0 \\ 0 & 2 \end{matrix}] ∴ A B = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow A B = [\begin{matrix} 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] and A C = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix} 0 & 0 \\ 0 & 2 \end{matrix}] \Rightarrow A C = [\begin{matrix} 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

Thus,
AB = AC
But B ≠ C and A ≠ 0.

Page No 4.46:

Question 66:

Let A and B be square matrices of the same order. Does (A + B)² = A² + 2AB + B² hold? If not, why?

Answer:

$LHS = {(A + B)}^{2} = (A + B) (A + B) = A (A + B) + B (A + B) = A^{2} + A B + B A + B^{2}$

We know that a matrix does not have commutative property. So,
AB ≠ BA
Thus,
${(A + B)}^{2}$ ≠ $A^{2} + 2 A B + B^{2}$

Page No 4.46:

Question 67:

If A and B are square matrices of the same order, explain, why in general
(i) (A + B)² ≠ A² + 2AB + B²
(ii) (A − B)² ≠ A² − 2AB + B²
(iii) (A + B) (A − B) ≠ A² − B².

Answer:

$(i) LHS = {(A + B)}^{2} = (A + B) (A + B) = A (A + B) + B (A + B) = A^{2} + A B + B A + B^{2}$

We know that a matrix does not have commutative property. So,
AB ≠ BA
Thus,
${(A + B)}^{2}$ ≠ $A^{2} + 2 A B + B^{2}$

$(i i) LHS = {(A - B)}^{2} = (A - B) (A - B) = A (A - B) - B (A - B) = A^{2} - A B - B A + B^{2}$

We know that a matrix does not have commutative property. So,
AB ≠ BA
Thus,
${(A - B)}^{2}$ ≠ $A^{2} - 2 A B + B^{2}$

$(i i i) LHS = (A + B) (A - B) = A (A - B) + B (A - B) = A^{2} - A B + B A - B^{2}$

We know that a matrix does not have commutative property. So,
AB ≠ BA
Thus,
$(A + B) (A - B)$ ≠ $A^{2} - B^{2}$

Page No 4.46:

Question 68:

Let A and B be square matrices of the order 3 × 3. Is (AB)² = A² B²? Give reasons.

Answer:

Yes, (AB)² = A² B² if AB = BA.

If AB = BA, then
(AB)² = (AB)(AB)
= A(BA)B (associative law)
= A(AB)B
= A² B²

Page No 4.46:

Question 69:

If A and B are square matrices of the same order such that AB = BA, then show that (A + B)² = A² + 2AB + B².

Answer:

(A + B)² = (A + B)(A + B)
= A² + AB + BA + B²
= A² + 2AB + B² (∵ AB = BA)

Hence, (A + B)² = A² + 2AB + B².

Page No 4.46:

Question 70:

Let $A = [\begin{matrix} 1 & 1 & 1 \\ 3 & 3 & 3 \end{matrix}], B = [\begin{matrix} 3 & 1 \\ 5 & 2 \\ - 2 & 4 \end{matrix}] and C = [\begin{matrix} 4 & 2 \\ - 3 & 5 \\ 5 & 0 \end{matrix}] .$
Verify that AB = AC though B ≠ C, A ≠ O.

Answer:

$Here, A = [\begin{matrix} 1 & 1 & 1 \\ 3 & 3 & 3 \end{matrix}], B = [\begin{matrix} 3 & 1 \\ 5 & 2 \\ - 2 & 4 \end{matrix}] and C = [\begin{matrix} 4 & 2 \\ - 3 & 5 \\ 5 & 0 \end{matrix}] Now, A B = [\begin{matrix} 1 & 1 & 1 \\ 3 & 3 & 3 \end{matrix}] [\begin{matrix} 3 & 1 \\ 5 & 2 \\ - 2 & 4 \end{matrix}] \Rightarrow A B = [\begin{matrix} 3 + 5 - 2 & 1 + 2 + 4 \\ 9 + 15 - 6 & 3 + 6 + 12 \end{matrix}] \Rightarrow A B = [\begin{matrix} 6 & 7 \\ 18 & 21 \end{matrix}] A C = [\begin{matrix} 1 & 1 & 1 \\ 3 & 3 & 3 \end{matrix}] [\begin{matrix} 4 & 2 \\ - 3 & 5 \\ 5 & 0 \end{matrix}] \Rightarrow A C = [\begin{matrix} 4 - 3 + 5 & 2 + 5 + 0 \\ 12 - 9 + 15 & 6 + 15 + 0 \end{matrix}] \Rightarrow A C = [\begin{matrix} 6 & 7 \\ 18 & 21 \end{matrix}]$

So, AB = AC though B ≠ C , A ≠ O.

Page No 4.46:

Question 71:

Three shopkeepers A, B and C go to a store to buy stationary. A purchases 12 dozen notebooks, 5 dozen pens and 6 dozen pencils. B purchases 10 dozen notebooks, 6 dozen pens and 7 dozen pencils. C purchases 11 dozen notebooks, 13 dozen pens and 8 dozen pencils. A notebook costs 40 paise, a pen costs Rs. 1.25 and a pencil costs 35 paise. Use matrix multiplication to calculate each individual's bill.

Answer:

Shopkeepers	Notebooks In dozen	Pens In dozen	Pencils In dozen
A	12	5	6
B	10	6	7
C	11	3	8

Here,
Cost of notebooks per dozen =

(12 \times 40) paise

= Rs 4.80
Cost of pens per dozen =

Rs . (12 \times 1.25)

= Rs 15
Cost ofpPencils per dozen =

(12 \times 35) paise

= Rs 4.20

∴ [\begin{matrix} 12 & 5 & 6 \\ 10 & 6 & 7 \\ 11 & 13 & 8 \end{matrix}] [\begin{matrix} 4.80 \\ 15 \\ 4.20 \end{matrix}] = [\begin{matrix} 12 \times 4.80 + 5 \times 15 + 6 \times 4.20 \\ 10 \times 4.80 + 6 \times 15 + 7 \times 4.20 \\ 11 \times 4.80 + 13 \times 15 + 8 \times 4.20 \end{matrix}] = [\begin{matrix} 57.60 + 75 + 25.20 \\ 48 + 90 + 29.40 \\ 52.80 + 195 + 33.60 \end{matrix}] = [\begin{matrix} 157.80 \\ 167.40 \\ 281.40 \end{matrix}]

Thus, the bills of A, B and C are Rs 157.80, Rs 167.40 and Rs 281.40, respectively.

Page No 4.46:

Question 72:

The cooperative stores of a particular school has 10 dozen physics books, 8 dozen chemistry books and 5 dozen mathematics books. Their selling prices are Rs. 8.30, Rs. 3.45 and Rs. 4.50 each respectively. Find the total amount the store will receive from selling all the items.

Answer:

Stock of various types of books in the store is given by

$Physics Chemistry Mathematics X = [\begin{matrix} 120 & 96 & 60 \end{matrix}]$

Selling price of various types of books in the store is given by

$Y = [\begin{matrix} 8.30 \\ 3.45 \\ 4.50 \end{matrix}] \begin{matrix} Physics \\ Chemistry \\ Mathematics \end{matrix}$

Total amount received by the store from selling all the items is given by

$X Y = [\begin{matrix} 120 & 96 & 60 \end{matrix}] [\begin{matrix} 8.30 \\ 3.45 \\ 4.50 \end{matrix}] = [\begin{matrix} (120) (8.30) + (96) (3.45) + (60) (4.50) \end{matrix}] = [\begin{matrix} 996 + 331.20 + 270 \end{matrix}] = [\begin{matrix} 1597.20 \end{matrix}]$

Required amount = Rs 1597.20

Page No 4.47:

Question 73:

In a legislative assembly election, a political group hired a public relations firm to promote its candidates in three ways: telephone, house calls and letters. The cost per contact (in paise) is given matrix A as
$A = \begin{matrix} Cost per contact \\ \begin{matrix} [\begin{matrix} 40 \\ 100 \\ 50 \end{matrix}] \end{matrix} \end{matrix} \begin{matrix} \begin{matrix} Telephone \\ House call \\ Letter \end{matrix} \end{matrix}$
The number of contacts of each type made in two cities X and Y is given in matrix B as
$\begin{matrix} \begin{matrix} Telephone & House call & Letter \end{matrix} \\ B = [\begin{matrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{matrix}] \end{matrix} \begin{matrix} \to X \\ \to Y \end{matrix}$
Find the total amount spent by the group in the two cities X and Y.

Answer:

The cost per contact $(i n p a i s e)$ is given by

$A = [\begin{matrix} 40 \\ 100 \\ 50 \end{matrix}] \begin{matrix} Telephone \\ Housecall \\ Letter \end{matrix}$

The number of contacts of each type made in the two cities X and Y is given by

$Telephone Housecall Letter B = [\begin{matrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{matrix}] \begin{matrix} X \\ Y \end{matrix}$

Total amount spent by the group in the two cities X and Y is given by

$B A = [\begin{matrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{matrix}] [\begin{matrix} 40 \\ 100 \\ 50 \end{matrix}] = [\begin{matrix} 40000 + 50000 + 250000 \\ 120000 + 100000 + 500000 \end{matrix}] = [\begin{matrix} 340000 \\ 720000 \end{matrix}] \begin{matrix} X \\ Y \end{matrix}$

Thus,
Amount spent on X = Rs 3400
Amount spent on Y = Rs 7200

Page No 4.47:

Question 74:

A trust fund has Rs 30000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30000 among the two types of bonds. If the trust fund must obtain an annual total interest of
(i) Rs 1800 (ii) Rs 2000

Answer:

If Rs x are invested in the first type of bond and Rs $(30000 - x)$ are invested in the second type of bond, then the matrix $A = [\begin{matrix} x & 30000 - x \end{matrix}]$ represents investment and the matrix $B = [\begin{matrix} \frac{5}{100} \\ \frac{7}{100} \end{matrix}]$ represents rate of interest.

$(i) [\begin{matrix} x & 30000 - x \end{matrix}] [\begin{matrix} \frac{5}{100} \\ \frac{7}{100} \end{matrix}] = [\begin{matrix} 1800 \end{matrix}] \Rightarrow [\begin{matrix} \frac{5 x}{100} + \end{matrix} \frac{7 (30000 - x)}{100}] = [\begin{matrix} 1800 \end{matrix}] \Rightarrow \frac{5 x + 210000 - 7 x}{100} = 1800 \Rightarrow 210000 - 2 x = 180000 \Rightarrow 2 x = 30000 \Rightarrow x = 15000$

Thus,
Amount invested in the first bond = Rs 15000

Amount invested in the second bond = Rs $(30000 - 15000)$
= Rs 15000

$(ii) [\begin{matrix} x & 30000 - x \end{matrix}] [\begin{matrix} \frac{5}{100} \\ \frac{7}{100} \end{matrix}] = [\begin{matrix} 2000 \end{matrix}] \Rightarrow [\begin{matrix} \frac{5 x}{100} + \end{matrix} \frac{7 (30000 - x)}{100}] = [\begin{matrix} 2000 \end{matrix}] \Rightarrow \frac{5 x + 210000 - 7 x}{100} = 2000 \Rightarrow 210000 - 2 x = 200000 \Rightarrow 2 x = 10000 \Rightarrow x = 5000$

Thus,
Amount invested in the first bond = Rs 5000

Amount invested in the second bond = Rs $(30000 - 5000)$
= Rs 25000

Page No 4.47:

Question 75:

To promote making of toilets for women, an organisation tried to generate awarness through (i) house calls, (ii) letters, and (iii) announcements. The cost for each mode per attempt is given below:
(i) ₹50 (ii) ₹20   (iii) ₹40

The number of attempts made in three villages X, Y and Z are given below:
(i) (ii) (iii)
X   400 300 100
Y   300 250 75
Z   500 400 150

Find the total cost incurred by the organisation for three villages separately, using matrices.

Answer:

According to the question,

Let A be the matrix showing number of attempts made in three villages X, Y and Z.
$A = [\begin{matrix} 400 & 300 & 100 \\ 300 & 250 & 75 \\ 500 & 400 & 150 \end{matrix}]$

And, B be a matrix showing the cost for each mode per attempt.
$B = [\begin{matrix} 50 \\ 20 \\ 40 \end{matrix}]$

Now, the total cost per village will be shown by AB.
$A B = [\begin{matrix} 400 & 300 & 100 \\ 300 & 250 & 75 \\ 500 & 400 & 150 \end{matrix}] [\begin{matrix} 50 \\ 20 \\ 40 \end{matrix}] = [\begin{matrix} 20000 + 6000 + 4000 \\ 15000 + 5000 + 3000 \\ 25000 + 8000 + 6000 \end{matrix}] = [\begin{matrix} 30000 \\ 23000 \\ 39000 \end{matrix}]$

Hence, the total cost incurred by the organisation for three villages separately is
X: ₹30,000
Y: ₹23,000
Z: ₹39,000

Page No 4.47:

Question 76:

There are 2 families A and B. There are 4 men, 6 women and 2 children in family A, and 2 men, 2 women and 4 children in family B. The recommend daily amount of calories is 2400 for men, 1900 for women, 1800 for children and 45 grams of proteins for men, 55 grams for women and 33 grams for children. Represent the above information using matrix. Using matrix multiplication, calculate the total requirement of calories and proteins for each of the two families. What awareness can you create among people about the planned diet from this question?

Answer:

According to the question,

Let X be the matrix showing number of family members in family A and B.
$X = [\begin{matrix} 4 & 6 & 2 \\ 2 & 2 & 4 \end{matrix}] \begin{matrix} Family A \\ Family B \end{matrix}$

And, Y be a matrix showing the recommend daily amount of calories.
$Y = [\begin{matrix} 2400 \\ 1900 \\ 1800 \end{matrix}]$

And, Z be a matrix showing the recommend daily amount of proteins.
$Z = [\begin{matrix} 45 \\ 55 \\ 33 \end{matrix}]$

Now, the total requirement of calories of the two families will be shown by XY.
$X Y = [\begin{matrix} 4 & 6 & 2 \\ 2 & 2 & 4 \end{matrix}] [\begin{matrix} 2400 \\ 1900 \\ 1800 \end{matrix}] = [\begin{matrix} 9600 + 11400 + 3600 \\ 4800 + 3800 + 7200 \end{matrix}] = [\begin{matrix} 24600 \\ 15800 \end{matrix}]$

Also, the total requirement of proteins of the two families will be shown by XZ.
$X Z = [\begin{matrix} 4 & 6 & 2 \\ 2 & 2 & 4 \end{matrix}] [\begin{matrix} 45 \\ 55 \\ 33 \end{matrix}] = [\begin{matrix} 180 + 330 + 66 \\ 90 + 110 + 132 \end{matrix}] = [\begin{matrix} 576 \\ 332 \end{matrix}]$

Hence, the total requirement of calories and proteins for each of the two families is shown as:

$\begin{matrix} Calories & Proteins \end{matrix} \begin{matrix} Family A : \\ Family B : \end{matrix} \begin{matrix} 24600 \\ 15800 \end{matrix} \begin{matrix} 576 \\ 332 \end{matrix}$

Page No 4.47:

Question 77:

In a parliament election, a political party hired a public relations firm to promote its candidates in three ways − telephone, house calls and letters. The cost per contact (in paisa) is given in matrix A as

$A = [\begin{matrix} 140 \\ 200 \\ 150 \end{matrix}] \begin{matrix} Telephone \\ House calls \\ Letters \end{matrix}$

The number of contacts of each type made in two cities X and Y is given in the matrix B as

$\begin{matrix} Telephone & House calls & Letters \end{matrix} B = [\begin{matrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{matrix}] \begin{matrix} City X \\ City Y \end{matrix}$

Find the total amount spent by the party in the two cities.

What should one consider before casting his/her vote − party's promotional activity of their social activities?

Answer:

According to the question,

Let A be the matrix showing the cost per contact (in paisa).
$A = [\begin{matrix} 140 \\ 200 \\ 150 \end{matrix}] \begin{matrix} Telephone \\ House calls \\ Letters \end{matrix}$

And, B be a matrix showing the number of contacts of each type made in two cities X and Y.
$\begin{matrix} Telephone & House calls & Letters \end{matrix} B = [\begin{matrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{matrix}] \begin{matrix} City X \\ City Y \end{matrix}$

Now, the total amount spent by the party in the two cities will be shown by BA.
$B A = [\begin{matrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{matrix}] [\begin{matrix} 140 \\ 200 \\ 150 \end{matrix}] = [\begin{matrix} 140000 + 100000 + 750000 \\ 420000 + 200000 + 1500000 \end{matrix}] = [\begin{matrix} 990000 \\ 2120000 \end{matrix}]$

Hence, the total amount spent by the party in the two cities is
X: ₹9900
Y: ₹21200

One should consider social activities of a party before casting his/her vote.

Page No 4.48:

Question 78:

The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves ₹ 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value?

Answer:

Page No 4.48:

Question 79:

A trust invested some money in two type of bonds. The first bond pays 10% interest and second bond pays 12% interest. The trust received ₹ 2800 as interest. However, if trust had interchanged money in bonds, they would have got ₹ 100 less as interes. Using matrix method, find the amount invested by the trust.

Answer:

Let Rs x be invested in the first bond and Rs y be invested in the second bond.
Let A be the investment matrix and B be the interest per rupee matrix. Then,

$A = [\begin{matrix} x & y \end{matrix}] and B = [\begin{matrix} \frac{10}{100} \\ \frac{12}{100} \end{matrix}] Total annual interest = AB = [\begin{matrix} x & y \end{matrix}] [\begin{matrix} \frac{10}{100} \\ \frac{12}{100} \end{matrix}] = \frac{10 x}{100} + \frac{12 y}{100} ∴ \frac{10 x}{100} + \frac{12 y}{100} = 2800 \Rightarrow 10 x + 12 y = 280000 . . . . . (1)$
If the rates of interest had been interchanged, then the total interest earned is Rs 100 less than the previous interest.
$∴ \frac{12 x}{100} + \frac{10 y}{100} = 2700 \Rightarrow 12 x + 10 y = 270000 . . . . . (2)$
The system of equations (1) and (2) can be expressed as
PX = Q, where $P = [\begin{matrix} 10 & 12 \\ 12 & 10 \end{matrix}], X = [\begin{matrix} x \\ y \end{matrix}], Q = [\begin{matrix} 280000 \\ 270000 \end{matrix}]$
$|P| = |\begin{matrix} 10 & 12 \\ 12 & 10 \end{matrix}| = 100 - 144 = - 44 \neq 0$
Thus, P is invertible.
$∴ X = P^{- 1} Q \Rightarrow X = \frac{a d j P}{|P|} Q \Rightarrow [\begin{matrix} x \\ y \end{matrix}] = \frac{1}{(- 44)} {[\begin{matrix} 10 & - 12 \\ - 12 & 10 \end{matrix}]}^{T} [\begin{matrix} 280000 \\ 270000 \end{matrix}] \Rightarrow [\begin{matrix} x \\ y \end{matrix}] = \frac{1}{(- 44)} [\begin{matrix} 10 & - 12 \\ - 12 & 10 \end{matrix}] [\begin{matrix} 280000 \\ 270000 \end{matrix}] \Rightarrow [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} \frac{2800000 - 3240000}{- 44} \\ \frac{- 3360000 + 2700000}{- 44} \end{matrix}] = [\begin{matrix} 10000 \\ 15000 \end{matrix}] \Rightarrow x = 10000 and y = 15000$

Therefore, Rs 10,000 be invested in the first bond and Rs 15,000 be invested in the second bond.

Page No 4.54:

Question 1:

Let $A = [\begin{matrix} 2 & - 3 \\ - 7 & 5 \end{matrix}] and B = [\begin{matrix} 1 & 0 \\ 2 & - 4 \end{matrix}]$ , verify that
(i) (2A)^T = 2A^T
(ii) (A + B)^T = A^T + B^T
(iii) (A − B)^T = A^T − B^T
(iv) (AB)^T = B^T A^T

Answer:

$Given : A = [\begin{matrix} 2 & - 3 \\ - 7 & 5 \end{matrix}] A^{T} = [\begin{matrix} 2 & - 7 \\ - 3 & 5 \end{matrix}] B = [\begin{matrix} 1 & 0 \\ 2 & - 4 \end{matrix}] B^{T} = [\begin{matrix} 1 & 2 \\ 0 & - 4 \end{matrix}] (i) {(2 A)}^{T} = 2 A^{T} \Rightarrow {(2 [\begin{matrix} 2 & - 3 \\ - 7 & 5 \end{matrix}])}^{T} = 2 [\begin{matrix} 2 & - 7 \\ - 3 & 5 \end{matrix}] \Rightarrow {[\begin{matrix} 4 & - 6 \\ - 14 & 10 \end{matrix}]}^{T} = [\begin{matrix} 4 & - 14 \\ - 6 & 10 \end{matrix}] \Rightarrow [\begin{matrix} 4 & - 14 \\ - 6 & 10 \end{matrix}] = [\begin{matrix} 4 & - 14 \\ - 6 & 10 \end{matrix}] ∴ LHS = RHS (i i) {(A + B)}^{T} = A^{T} + B^{T} \Rightarrow {([\begin{matrix} 2 & - 3 \\ - 7 & 5 \end{matrix}] + [\begin{matrix} 1 & 0 \\ 2 & - 4 \end{matrix}])}^{T} = [\begin{matrix} 2 & - 7 \\ - 3 & 5 \end{matrix}] + [\begin{matrix} 1 & 2 \\ 0 & - 4 \end{matrix}] \Rightarrow {([\begin{matrix} 2 + 1 & - 3 + 0 \\ - 7 + 2 & 5 - 4 \end{matrix}])}^{T} = [\begin{matrix} 2 + 1 & - 7 + 2 \\ - 3 + 0 & 5 - 4 \end{matrix}] \Rightarrow {([\begin{matrix} 3 & - 3 \\ - 5 & 1 \end{matrix}])}^{T} = [\begin{matrix} 3 & - 5 \\ - 3 & 1 \end{matrix}] \Rightarrow [\begin{matrix} 3 & - 5 \\ - 3 & 1 \end{matrix}] = [\begin{matrix} 3 & - 5 \\ - 3 & 1 \end{matrix}] ∴ LHS = RHS (i i i) {(A - B)}^{T} = A^{T} - B^{T} \Rightarrow {([\begin{matrix} 2 & - 3 \\ - 7 & 5 \end{matrix}] - [\begin{matrix} 1 & 0 \\ 2 & - 4 \end{matrix}])}^{T} = [\begin{matrix} 2 & - 7 \\ - 3 & 5 \end{matrix}] - [\begin{matrix} 1 & 2 \\ 0 & - 4 \end{matrix}] \Rightarrow {([\begin{matrix} 2 - 1 & - 3 - 0 \\ - 7 - 2 & 5 + 4 \end{matrix}])}^{T} = [\begin{matrix} 2 - 1 & - 7 - 2 \\ - 3 - 0 & 5 + 4 \end{matrix}] \Rightarrow {([\begin{matrix} 1 & - 3 \\ - 9 & 9 \end{matrix}])}^{T} = [\begin{matrix} 1 & - 9 \\ - 3 & 9 \end{matrix}] \Rightarrow [\begin{matrix} 1 & - 9 \\ - 3 & 9 \end{matrix}] = [\begin{matrix} 1 & - 9 \\ - 3 & 9 \end{matrix}] ∴ LHS = RHS (i v) {(A B)}^{T} = B^{T} A^{T} \Rightarrow {([\begin{matrix} 2 & - 3 \\ - 7 & 5 \end{matrix}] [\begin{matrix} 1 & 0 \\ 2 & - 4 \end{matrix}])}^{T} = [\begin{matrix} 1 & 2 \\ 0 & - 4 \end{matrix}] [\begin{matrix} 2 & - 7 \\ - 3 & 5 \end{matrix}] \Rightarrow {([\begin{matrix} 2 - 6 & 0 + 12 \\ - 7 + 10 & 0 - 20 \end{matrix}])}^{T} = [\begin{matrix} 2 - 6 & - 7 + 10 \\ 0 + 12 & 0 - 20 \end{matrix}] \Rightarrow {([\begin{matrix} - 4 & 12 \\ 3 & - 20 \end{matrix}])}^{T} = [\begin{matrix} - 4 & 3 \\ 12 & - 20 \end{matrix}] \Rightarrow [\begin{matrix} - 4 & 3 \\ 12 & - 20 \end{matrix}] = [\begin{matrix} - 4 & 3 \\ 12 & - 20 \end{matrix}] ∴ LHS = RHS$

Page No 4.54:

Question 2:

If $A = [\begin{matrix} 3 \\ 5 \\ 2 \end{matrix}]$ and B = [1 0 4], verify that (AB)^T = B^T A^T

Answer:

$Given : A = [\begin{matrix} 3 \\ 5 \\ 2 \end{matrix}] A^{T} = [\begin{matrix} 3 & 5 & 2 \end{matrix}] B = [\begin{matrix} 1 & 0 & 4 \end{matrix}] B^{T} = [\begin{matrix} 1 \\ 0 \\ 4 \end{matrix}] Now, A B = [\begin{matrix} 3 \\ 5 \\ 2 \end{matrix}] [\begin{matrix} 1 & 0 & 4 \end{matrix}] \Rightarrow A B = [\begin{matrix} 3 & 0 & 12 \\ 5 & 0 & 20 \\ 2 & 0 & 8 \end{matrix}] \Rightarrow {(A B)}^{T} = [\begin{matrix} 3 & 5 & 2 \\ 0 & 0 & 0 \\ 12 & 20 & 8 \end{matrix}] . . . (1) B^{T} A^{T} = [\begin{matrix} 1 \\ 0 \\ 4 \end{matrix}] [\begin{matrix} 3 & 5 & 2 \end{matrix}] \Rightarrow B^{T} A^{T} = [\begin{matrix} 3 & 5 & 2 \\ 0 & 0 & 0 \\ 12 & 20 & 8 \end{matrix}] . . . (2) \Rightarrow {(A B)}^{T} = B^{T} A^{T} [From eqs . (1) and (2)]$

Page No 4.54:

Question 3:

Let $A = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 1 & 3 \\ 1 & 2 & 1 \end{matrix}] and B = [\begin{matrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}] .$ Find A^T, B^T and verify that
(i) (A + B)^T = A^T + B^T
(ii) (AB)^T = B^T A^T
(iii) (2A)^T = 2A^T.

Answer:

$Given : A = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 1 & 3 \\ 1 & 2 & 1 \end{matrix}] and B = [\begin{matrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}] A^{T} = [\begin{matrix} 1 & 2 & 1 \\ - 1 & 1 & 2 \\ 0 & 3 & 1 \end{matrix}] and B^{T} = [\begin{matrix} 1 & 2 & 0 \\ 2 & 1 & 1 \\ 3 & 3 & 1 \end{matrix}] (i) A + B = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 1 & 3 \\ 1 & 2 & 1 \end{matrix}] + [\begin{matrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}] \Rightarrow A + B = [\begin{matrix} 1 + 1 & - 1 + 2 & 0 + 3 \\ 2 + 2 & 1 + 1 & 3 + 3 \\ 1 + 0 & 2 + 1 & 1 + 1 \end{matrix}] \Rightarrow A + B = [\begin{matrix} 2 & 1 & 3 \\ 4 & 2 & 6 \\ 1 & 3 & 2 \end{matrix}] \Rightarrow {(A + B)}^{T} = [\begin{matrix} 2 & 4 & 1 \\ 1 & 2 & 3 \\ 3 & 6 & 2 \end{matrix}] . . . (1) Now, A^{T} + B^{T} = [\begin{matrix} 1 & 2 & 1 \\ - 1 & 1 & 2 \\ 0 & 3 & 1 \end{matrix}] + [\begin{matrix} 1 & 2 & 0 \\ 2 & 1 & 1 \\ 3 & 3 & 1 \end{matrix}] \Rightarrow A^{T} + B^{T} = [\begin{matrix} 1 + 1 & 2 + 2 & 1 + 0 \\ - 1 + 2 & 1 + 1 & 2 + 1 \\ 0 + 3 & 3 + 3 & 1 + 1 \end{matrix}] \Rightarrow A^{T} + B^{T} = [\begin{matrix} 2 & 4 & 1 \\ 1 & 2 & 3 \\ 3 & 6 & 2 \end{matrix}] . . . (2) \Rightarrow {(A + B)}^{T} = A^{T} + B^{T} [From eqs . (1) and (2)] (i i) A B = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 1 & 3 \\ 1 & 2 & 1 \end{matrix}] [\begin{matrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{matrix}] \Rightarrow A B = [\begin{matrix} 1 - 2 + 0 & 2 - 1 + 0 & 3 - 3 + 0 \\ 2 + 2 + 0 & 4 + 1 + 3 & 6 + 3 + 3 \\ 1 + 4 + 0 & 2 + 2 + 1 & 3 + 6 + 1 \end{matrix}] \Rightarrow A B = [\begin{matrix} - 1 & 1 & 0 \\ 4 & 8 & 12 \\ 5 & 5 & 10 \end{matrix}] \Rightarrow {(A B)}^{T} = [\begin{matrix} - 1 & 4 & 5 \\ 1 & 8 & 5 \\ 0 & 12 & 10 \end{matrix}] . . . (1) Now, B^{T} A^{T} = [\begin{matrix} 1 & 2 & 0 \\ 2 & 1 & 1 \\ 3 & 3 & 1 \end{matrix}] [\begin{matrix} 1 & 2 & 1 \\ - 1 & 1 & 2 \\ 0 & 3 & 1 \end{matrix}] \Rightarrow B^{T} A^{T} = [\begin{matrix} 1 - 2 + 0 & 2 + 2 + 0 & 1 + 4 + 0 \\ 2 - 1 + 0 & 4 + 1 + 3 & 2 + 2 + 1 \\ 3 - 3 + 0 & 6 + 3 + 3 & 3 + 6 + 1 \end{matrix}] \Rightarrow B^{T} A^{T} = [\begin{matrix} - 1 & 4 & 5 \\ 1 & 8 & 5 \\ 0 & 12 & 10 \end{matrix}] . . . (2) \Rightarrow {(A B)}^{T} = B^{T} A^{T} [From eqs . (1) and (2)] (iii) 2 A = 2 [\begin{matrix} 1 & - 1 & 0 \\ 2 & 1 & 3 \\ 1 & 2 & 1 \end{matrix}] \Rightarrow 2 A = [\begin{matrix} 2 & - 2 & 0 \\ 4 & 2 & 6 \\ 2 & 4 & 2 \end{matrix}] \Rightarrow {(2 A)}^{T} = [\begin{matrix} 2 & 4 & 2 \\ - 2 & 2 & 4 \\ 0 & 6 & 2 \end{matrix}] . . . (1) Now, 2 A^{T} = 2 [\begin{matrix} 1 & 2 & 1 \\ - 1 & 1 & 2 \\ 0 & 3 & 1 \end{matrix}] \Rightarrow 2 A^{T} = [\begin{matrix} 2 & 4 & 2 \\ - 2 & 2 & 4 \\ 0 & 6 & 2 \end{matrix}] . . . (2) \Rightarrow {(2 A)}^{T} = 2 A^{T} [From eqs . (1) and (2)]$

Page No 4.55:

Question 4:

If $A = [\begin{matrix} - 2 \\ 4 \\ 5 \end{matrix}]$ , B = [1 3 −6], verify that (AB)^T = B^T A^T

Answer:

$Given : A = [\begin{matrix} - 2 \\ 4 \\ 5 \end{matrix}] A^{T} = [\begin{matrix} - 2 & 4 & 5 \end{matrix}] B = [\begin{matrix} 1 & 3 & - 6 \end{matrix}] B^{T} = [\begin{matrix} 1 \\ 3 \\ - 6 \end{matrix}] Now, A B = [\begin{matrix} - 2 \\ 4 \\ 5 \end{matrix}] [\begin{matrix} 1 & 3 & - 6 \end{matrix}] \Rightarrow A B = [\begin{matrix} - 2 & - 6 & 12 \\ 4 & 12 & - 24 \\ 5 & 15 & - 30 \end{matrix}] \Rightarrow {(A B)}^{T} = [\begin{matrix} - 2 & 4 & 5 \\ - 6 & 12 & 15 \\ 12 & - 24 & - 30 \end{matrix}] . . . (1) B^{T} A^{T} = [\begin{matrix} 1 \\ 3 \\ - 6 \end{matrix}] [\begin{matrix} - 2 & 4 & 5 \end{matrix}] \Rightarrow B^{T} A^{T} = [\begin{matrix} - 2 & 4 & 5 \\ - 6 & 12 & 15 \\ 12 & - 24 & - 30 \end{matrix}] . . . (2) ∴ {(A B)}^{T} = B^{T} A^{T} [From eqs . (1) and (2)]$

Page No 4.55:

Question 5:

If $A = [\begin{matrix} 2 & 4 & - 1 \\ - 1 & 0 & 2 \end{matrix}], B = [\begin{matrix} 3 & 4 \\ - 1 & 2 \\ 2 & 1 \end{matrix}]$ , find (AB)^T

Answer:

$Here, A B = [\begin{matrix} 2 & 4 & - 1 \\ - 1 & 0 & 2 \end{matrix}] [\begin{matrix} 3 & 4 \\ - 1 & 2 \\ 2 & 1 \end{matrix}] \Rightarrow A B = [\begin{matrix} 6 - 4 - 2 & 8 + 8 - 1 \\ - 3 - 0 + 4 & - 4 + 0 + 2 \end{matrix}] \Rightarrow A B = [\begin{matrix} 0 & 15 \\ 1 & - 2 \end{matrix}] \Rightarrow {(A B)}^{T} = [\begin{matrix} 0 & 1 \\ 15 & - 2 \end{matrix}]$

Page No 4.55:

Question 6:

(i) For two matrices A and B, $A = [\begin{matrix} 2 & 1 & 3 \\ 4 & 1 & 0 \end{matrix}], B = [\begin{matrix} 1 & - 1 \\ 0 & 2 \\ 5 & 0 \end{matrix}]$ verify that
(AB)^T = B^T A^T.

(ii) For the matrices A and B, verify that (AB)^T = B^T A^T, where
$A = [\begin{matrix} 1 & 3 \\ 2 & 4 \end{matrix}], B = [\begin{matrix} 1 & 4 \\ 2 & 5 \end{matrix}]$

Answer:

$(i) Given : A = [\begin{matrix} 2 & 1 & 3 \\ 4 & 1 & 0 \end{matrix}] A^{T} = [\begin{matrix} 2 & 4 \\ 1 & 1 \\ 3 & 0 \end{matrix}] B = [\begin{matrix} 1 & - 1 \\ 0 & 2 \\ 5 & 0 \end{matrix}] B^{T} = [\begin{matrix} 1 & 0 & 5 \\ - 1 & 2 & 0 \end{matrix}] Now, A B = [\begin{matrix} 2 & 1 & 3 \\ 4 & 1 & 0 \end{matrix}] [\begin{matrix} 1 & - 1 \\ 0 & 2 \\ 5 & 0 \end{matrix}] \Rightarrow A B = [\begin{matrix} 2 + 0 + 15 & - 2 + 2 + 0 \\ 4 + 0 + 0 & - 4 + 2 + 0 \end{matrix}] \Rightarrow A B = [\begin{matrix} 17 & 0 \\ 4 & - 2 \end{matrix}] \Rightarrow {(A B)}^{T} = [\begin{matrix} 17 & 4 \\ 0 & - 2 \end{matrix}] . . . (1) B^{T} A^{T} = [\begin{matrix} 1 & 0 & 5 \\ - 1 & 2 & 0 \end{matrix}] [\begin{matrix} 2 & 4 \\ 1 & 1 \\ 3 & 0 \end{matrix}] \Rightarrow B^{T} A^{T} = [\begin{matrix} 2 + 0 + 15 & 4 + 0 + 0 \\ - 2 + 2 + 0 & - 4 + 2 + 0 \end{matrix}] \Rightarrow B^{T} A^{T} = [\begin{matrix} 17 & 4 \\ 0 & - 2 \end{matrix}] . . . (2) ∴ {(A B)}^{T} = B^{T} A^{T} [From eqs . (1) and (2)]$

$(i i) Given : A = [\begin{matrix} 1 & 3 \\ 2 & 4 \end{matrix}] A^{T} = [\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}] B = [\begin{matrix} 1 & 4 \\ 2 & 5 \end{matrix}] B^{T} = [\begin{matrix} 1 & 2 \\ 4 & 5 \end{matrix}] Now, A B = [\begin{matrix} 1 & 3 \\ 2 & 4 \end{matrix}] [\begin{matrix} 1 & 4 \\ 2 & 5 \end{matrix}] \Rightarrow A B = [\begin{matrix} 1 + 6 & 4 + 15 \\ 2 + 8 & 8 + 20 \end{matrix}] \Rightarrow A B = [\begin{matrix} 7 & 19 \\ 10 & 28 \end{matrix}] \Rightarrow {(A B)}^{T} = [\begin{matrix} 7 & 10 \\ 19 & 28 \end{matrix}] . . . (1) A l s o, B^{T} A^{T} = [\begin{matrix} 1 & 2 \\ 4 & 5 \end{matrix}] [\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}] \Rightarrow B^{T} A^{T} = [\begin{matrix} 1 + 6 & 2 + 8 \\ 4 + 15 & 8 + 20 \end{matrix}] \Rightarrow B^{T} A^{T} = [\begin{matrix} 7 & 10 \\ 19 & 28 \end{matrix}] . . . (2) ∴ {(A B)}^{T} = B^{T} A^{T} [From eqs . (1) and (2)]$

Page No 4.55:

Question 7:

If $A^{T} = [\begin{matrix} 3 & 4 \\ - 1 & 2 \\ 0 & 1 \end{matrix}] and B = [\begin{matrix} - 1 & 2 & 1 \\ 1 & 2 & 3 \end{matrix}]$ , find A^T − B^T.

Answer:

Given: $A^{T} = [\begin{matrix} 3 & 4 \\ - 1 & 2 \\ 0 & 1 \end{matrix}] and B = [\begin{matrix} - 1 & 2 & 1 \\ 1 & 2 & 3 \end{matrix}]$

$B^{T} = [\begin{matrix} - 1 & 1 \\ 2 & 2 \\ 1 & 3 \end{matrix}]$

Now,
$A^{T} - B^{T} = [\begin{matrix} 3 & 4 \\ - 1 & 2 \\ 0 & 1 \end{matrix}] - [\begin{matrix} - 1 & 1 \\ 2 & 2 \\ 1 & 3 \end{matrix}] = [\begin{matrix} 3 + 1 & 4 - 1 \\ - 1 - 2 & 2 - 2 \\ 0 - 1 & 1 - 3 \end{matrix}] = [\begin{matrix} 4 & 3 \\ - 3 & 0 \\ - 1 & - 2 \end{matrix}]$

Therefore, $A^{T} - B^{T} = [\begin{matrix} 4 & 3 \\ - 3 & 0 \\ - 1 & - 2 \end{matrix}]$ .

Page No 4.55:

Question 8:

If $A = [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}]$ , then verify that A^T A = I₂.

Answer:

$Given : A = [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}] A^{T} = [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}] Now, A^{T} A = I_{2} Consider : LHS = A^{T} A = [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}] [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}] = [\begin{matrix} \cos^{2} α + \sin^{2} α & \cos α \sin α - \sin α \cos α \\ \sin α \cos α - \cos α \sin α & \sin^{2} α + \cos^{2} α \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = RHS$

Hence proved.

Page No 4.55:

Question 9:

If $A = [\begin{matrix} \sin α & \cos α \\ - \cos α & \sin α \end{matrix}]$ , verify that A^T A = I₂.

Answer:

$Given : A = [\begin{matrix} \sin α & \cos α \\ - \cos α & \sin α \end{matrix}] A^{T} = [\begin{matrix} \sin α & - \cos α \\ \cos α & \sin α \end{matrix}] Now, A^{T} A = [\begin{matrix} \sin α & - \cos α \\ \cos α & \sin α \end{matrix}] [\begin{matrix} \sin α & \cos α \\ - \cos α & \sin α \end{matrix}] \Rightarrow A^{T} A = [\begin{matrix} (\sin α) (\sin α) + (- \cos α) (- \cos α) & (\sin α) (\cos α) + (- \cos α) (\sin α) \\ (\cos α) (\sin α) + (\sin α) (- \cos α) & (\cos α) (\cos α) + (\sin α) (\sin α) \end{matrix}] \Rightarrow A^{T} A = [\begin{matrix} \sin^{2} α + \cos^{2} α & \sin α \cos α - \sin α \cos α \\ \sin α \cos α - \sin α \cos α & \cos^{2} α + \sin^{2} α \end{matrix}] \Rightarrow A^{T} A = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I$

Page No 4.55:

Question 10:

If l_i, m_i, n_i, i = 1, 2, 3 denote the direction cosines of three mutually perpendicular vectors in space, prove that AA^T = I, where $A = [\begin{matrix} l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \\ l_{3} & m_{3} & n_{3} \end{matrix}]$ .

Answer:

Given,
$(l_{1}, m_{1}, n_{1}), (l_{2}, m_{2}, n_{2}), (l_{3}, m_{3}, n_{3})$ are the direction cosines of three mutually perpendicular vectors in space.

$\begin{matrix} l_{1}^{2} + m_{1}^{2} + n_{1}^{2} = 1 \\ l_{2}^{2} + m_{2}^{2} + n_{2}^{2} = 1 \\ l_{3}^{2} + m_{3}^{2} + n_{3}^{2} = 1 \end{matrix}\} . . . . . (i) \begin{matrix} l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0 \\ l_{2} l_{3} + m_{2} m_{3} + n_{2} n_{3} = 0 \\ l_{3} l_{1} + m_{3} m_{1} + n_{3} n_{1} = 0 \end{matrix}\} . . . . . (ii)$
Let $A = [\begin{matrix} l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \\ l_{3} & m_{3} & n_{3} \end{matrix}]$
$\Rightarrow A^{T} = [\begin{matrix} l_{1} & l_{2} & l_{3} \\ m_{1} & m_{2} & m_{3} \\ n_{1} & n_{2} & n_{3} \end{matrix}]$
$A A^{T} = [\begin{matrix} l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \\ l_{3} & m_{3} & n_{3} \end{matrix}] [\begin{matrix} l_{1} & l_{2} & l_{3} \\ m_{1} & m_{2} & m_{3} \\ n_{1} & n_{2} & n_{3} \end{matrix}] \Rightarrow A A^{T} = [\begin{matrix} {l_{1}}^{2} + {m_{1}}^{2} + {n_{1}}^{2} & l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} & l_{3} l_{1} + m_{3} m_{1} + n_{3} n_{1} \\ l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} & {l_{2}}^{2} + {m_{2}}^{2} + {n_{2}}^{2} & l_{2} l_{3} + m_{2} m_{3} + n_{2} n_{3} \\ l_{3} l_{1} + m_{3} m_{1} + n_{3} n_{1} & l_{2} l_{3} + m_{2} m_{3} + n_{2} n_{3} & {l_{3}}^{2} + {m_{3}}^{2} + {n_{3}}^{2} \end{matrix}]$
From (i) and (ii), we get
$A A^{T} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = I$
Hence proved.

Page No 4.60:

Question 1:

If $A = [\begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix}]$ , prove that A − A^T is a skew-symmetric matrix.

Answer:

$Given : A = [\begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix}] A^{T} = [\begin{matrix} 2 & 4 \\ 3 & 5 \end{matrix}] Now, (A - A^{T}) = [\begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix}] - [\begin{matrix} 2 & 4 \\ 3 & 5 \end{matrix}] \Rightarrow (A - A^{T}) = [\begin{matrix} 2 - 2 & 3 - 4 \\ 4 - 3 & 5 - 5 \end{matrix}] \Rightarrow (A - A^{T}) = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}] . . . (1) {(A - A^{T})}^{T} = {[\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}]}^{T} \Rightarrow {(A - A^{T})}^{T} = [\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}] \Rightarrow {(A - A^{T})}^{T} = - [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}] \Rightarrow (A - A^{T}) = - {(A - A^{T})}^{T} [Using eq . (1)] Thus, (A - A^{T}) is a skew - symmetric matrix .$

Page No 4.6:

Question 1:

If a matrix has 8 elements, what are the possible orders it can have? What if it has 5 elements?

Answer:

We know that if a matrix is of order $m \times n$ , then it has mn elements.

The possible orders of a matrix with 8 elements are given below:
1 $\times$ 8, 2 $\times$ 4, 4 $\times$ 2, 8 $\times$ 1

Thus, there are 4 possible orders of the matrix.

The possible orders of a matrix with 5 elements are given below:
1 $\times$ 5, 5 $\times$ 1

Thus, there are 2 possible orders of the matrix.

Page No 4.6:

Question 2:

If A = [a_ij] = $[\begin{matrix} 2 & 3 & - 5 \\ 1 & 4 & 9 \\ 0 & 7 & - 2 \end{matrix}]$ and B = [b_ij] = $[\begin{matrix} 2 & - 1 \\ - 3 & 4 \\ 1 & 2 \end{matrix}]$
then find (i) a₂₂ + b₂₁ (ii) a₁₁ b₁₁ + a₂₂ b₂₂

Answer:

$(i)$

$a_{22} + b_{21}$

$Here, a_{22} = 4 and b_{21} = - 3 \Rightarrow a_{22} + b_{21} = 4 - 3 = 1$

$(i i)$

$a_{11} b_{11} + a_{22} b_{22}$

$Here, a_{11} = 2, b_{11} = 2, a_{22} = 4 and b_{22} = 4 \Rightarrow a_{11} b_{11} + a_{22} b_{22} = 2 \times 2 + 4 \times 4 \Rightarrow a_{11} b_{11} + a_{22} b_{22} = 4 + 16 \Rightarrow a_{11} b_{11} + a_{22} b_{22} = 20$

Page No 4.6:

Question 3:

Let A be a matrix of order 3 × 4. If R₁ denotes the first row of A and C₂ denotes its second column, then determine the orders of matrices R₁ and C₂

Answer:

The order of $R_{1}$ is $1 \times 4$ and the order of $C_{2} is 3 \times 1$ .

Page No 4.61:

Question 2:

If $A = [\begin{matrix} 3 & - 4 \\ 1 & - 1 \end{matrix}]$ , show that A − A^T is a skewsymmetric matrix.

Answer:

$Given : A = [\begin{matrix} 3 & - 4 \\ 1 & - 1 \end{matrix}] A^{T} = [\begin{matrix} 3 & 1 \\ - 4 & - 1 \end{matrix}] Now, A - A^{T} = [\begin{matrix} 3 & - 4 \\ 1 & - 1 \end{matrix}] - [\begin{matrix} 3 & 1 \\ - 4 & - 1 \end{matrix}] \Rightarrow A - A^{T} = [\begin{matrix} 3 - 3 & - 4 - 1 \\ 1 + 4 & - 1 + 1 \end{matrix}] \Rightarrow A - A^{T} = [\begin{matrix} 0 & - 5 \\ 5 & 0 \end{matrix}] . . . (1) {(A - A^{T})}^{T} = {[\begin{matrix} 0 & - 5 \\ 5 & 0 \end{matrix}]}^{T} \Rightarrow {(A - A^{T})}^{T} = [\begin{matrix} 0 & 5 \\ - 5 & 0 \end{matrix}] \Rightarrow {(A - A^{T})}^{T} = - [\begin{matrix} 0 & - 5 \\ 5 & 0 \end{matrix}] \Rightarrow {(A - A^{T})}^{T} = - (A - A^{T}) [From eq . (1)] Thus, (A - A^{T}) is a skew - symmetric matrix .$

Page No 4.61:

Question 3:

If the matrix $A = [\begin{matrix} 5 & 2 & x \\ y & z & - 3 \\ 4 & t & - 7 \end{matrix}]$ is a symmetric matrix, find x, y, z and t.

Answer:

$Given : A = [\begin{matrix} 5 & 2 & x \\ y & z & - 3 \\ 4 & t & - 7 \end{matrix}] \Rightarrow A^{T} = [\begin{matrix} 5 & y & 4 \\ 2 & z & t \\ x & - 3 & - 7 \end{matrix}] Since A is a symmetric matrix, A^{T} = A . \Rightarrow [\begin{matrix} 5 & y & 4 \\ 2 & z & t \\ x & - 3 & - 7 \end{matrix}] = [\begin{matrix} 5 & 2 & x \\ y & z & - 3 \\ 4 & t & - 7 \end{matrix}] The corresponding elements of two equal matrices are equal . ∴ x = 4 y = 2 z = z t = - 3 Hence, x = 4, y = 2, t = - 3 and z can have any value .$

Page No 4.61:

Question 4:

Let $A = [\begin{matrix} 3 & 2 & 7 \\ 1 & 4 & 3 \\ - 2 & 5 & 8 \end{matrix}] .$ Find matrices X and Y such that X + Y = A, where X is a symmetric and Y is a skew-symmetric matrix.

Answer:

$Given : A = [\begin{matrix} 3 & 2 & 7 \\ 1 & 4 & 3 \\ - 2 & 5 & 8 \end{matrix}] \Rightarrow A^{T} = [\begin{matrix} 3 & 1 & - 2 \\ 2 & 4 & 5 \\ 7 & 3 & 8 \end{matrix}] Let X = \frac{1}{2} (A + A^{T}) = \frac{1}{2} ([\begin{matrix} 3 & 2 & 7 \\ 1 & 4 & 3 \\ - 2 & 5 & 8 \end{matrix}] + [\begin{matrix} 3 & 1 & - 2 \\ 2 & 4 & 5 \\ 7 & 3 & 8 \end{matrix}]) = [\begin{matrix} 3 & \frac{3}{2} & \frac{5}{2} \\ \frac{3}{2} & 4 & 4 \\ \frac{5}{2} & 4 & 8 \end{matrix}] Let Y = \frac{1}{2} (A - A^{T}) = \frac{1}{2} ([\begin{matrix} 3 & 2 & 7 \\ 1 & 4 & 3 \\ - 2 & 5 & 8 \end{matrix}] - [\begin{matrix} 3 & 1 & - 2 \\ 2 & 4 & 5 \\ 7 & 3 & 8 \end{matrix}]) = [\begin{matrix} 0 & \frac{1}{2} & \frac{9}{2} \\ \frac{- 1}{2} & 0 & - 1 \\ \frac{- 9}{2} & 1 & 0 \end{matrix}] X^{T} = {[\begin{matrix} 3 & \frac{3}{2} & \frac{5}{2} \\ \frac{3}{2} & 4 & 4 \\ \frac{5}{2} & 4 & 8 \end{matrix}]}^{T} = {[\begin{matrix} 3 & \frac{3}{2} & \frac{5}{2} \\ \frac{3}{2} & 4 & 4 \\ \frac{5}{2} & 4 & 8 \end{matrix}]}^{T} = X Y^{T} = {[\begin{matrix} 0 & \frac{1}{2} & \frac{9}{2} \\ \frac{- 1}{2} & 0 & - 1 \\ \frac{- 9}{2} & 1 & 0 \end{matrix}]}^{T} = [\begin{matrix} 0 & \frac{- 1}{2} & \frac{- 9}{2} \\ \frac{1}{2} & 0 & 1 \\ \frac{9}{2} & - 1 & 0 \end{matrix}] = - [\begin{matrix} 0 & \frac{1}{2} & \frac{9}{2} \\ \frac{- 1}{2} & 0 & - 1 \\ \frac{- 9}{2} & 1 & 0 \end{matrix}] = Y Thus, X is a symmetric matrix and Y is skew - symmetric matrix . Now, X + Y = [\begin{matrix} 3 & \frac{3}{2} & \frac{5}{2} \\ \frac{3}{2} & 4 & 4 \\ \frac{5}{2} & 4 & 8 \end{matrix}] + [\begin{matrix} 0 & \frac{1}{2} & \frac{9}{2} \\ \frac{- 1}{2} & 0 & - 1 \\ \frac{- 9}{2} & 1 & 0 \end{matrix}] = [\begin{matrix} 3 & 2 & 7 \\ 1 & 4 & 3 \\ - 2 & 5 & 8 \end{matrix}] = A ∴ X = [\begin{matrix} 3 & \frac{3}{2} & \frac{5}{2} \\ \frac{3}{2} & 4 & 4 \\ \frac{5}{2} & 4 & 8 \end{matrix}] and Y = [\begin{matrix} 0 & \frac{1}{2} & \frac{9}{2} \\ \frac{- 1}{2} & 0 & - 1 \\ \frac{- 9}{2} & 1 & 0 \end{matrix}]$

Page No 4.61:

Question 5:

Express the matrix $A = [\begin{matrix} 4 & 2 & - 1 \\ 3 & 5 & 7 \\ 1 & - 2 & 1 \end{matrix}]$ as the sum of a symmetric and a skew-symmetric matrix.

Answer:

$Given : A = [\begin{matrix} 4 & 2 & - 1 \\ 3 & 5 & 7 \\ 1 & - 2 & 1 \end{matrix}] A^{T} = [\begin{matrix} 4 & 3 & 1 \\ 2 & 5 & - 2 \\ - 1 & 7 & 1 \end{matrix}] Let X = \frac{1}{2} (A + A^{T}) = \frac{1}{2} ([\begin{matrix} 4 & 2 & - 1 \\ 3 & 5 & 7 \\ 1 & - 2 & 1 \end{matrix}] + [\begin{matrix} 4 & 3 & 1 \\ 2 & 5 & - 2 \\ - 1 & 7 & 1 \end{matrix}]) = [\begin{matrix} 4 & \frac{5}{2} & 0 \\ \frac{5}{2} & 5 & \frac{5}{2} \\ 0 & \frac{5}{2} & 1 \end{matrix}] X^{T} = {[\begin{matrix} 4 & \frac{5}{2} & 0 \\ \frac{5}{2} & 5 & \frac{5}{2} \\ 0 & \frac{5}{2} & 1 \end{matrix}]}^{T} = [\begin{matrix} 4 & \frac{5}{2} & 0 \\ \frac{5}{2} & 5 & \frac{5}{2} \\ 0 & \frac{5}{2} & 1 \end{matrix}] = X Let Y = \frac{1}{2} (A - A^{T}) = \frac{1}{2} ([\begin{matrix} 4 & 2 & - 1 \\ 3 & 5 & 7 \\ 1 & - 2 & 1 \end{matrix}] - [\begin{matrix} 4 & 3 & 1 \\ 2 & 5 & - 2 \\ - 1 & 7 & 1 \end{matrix}]) = [\begin{matrix} 0 & \frac{- 1}{2} & - 1 \\ \frac{1}{2} & 0 & \frac{9}{2} \\ 1 & \frac{- 9}{2} & 0 \end{matrix}] Y^{T} = {[\begin{matrix} 0 & \frac{- 1}{2} & - 1 \\ \frac{1}{2} & 0 & \frac{9}{2} \\ 1 & \frac{- 9}{2} & 0 \end{matrix}]}^{T} = [\begin{matrix} 0 & \frac{1}{2} & 1 \\ \frac{- 1}{2} & 0 & \frac{- 9}{2} \\ - 1 & \frac{9}{2} & 0 \end{matrix}] = - [\begin{matrix} 0 & \frac{- 1}{2} & - 1 \\ \frac{1}{2} & 0 & \frac{9}{2} \\ 1 & \frac{- 9}{2} & 0 \end{matrix}] = - Y Thus, X is a symmetric matrix and Y is a skew - symmetric matrix . Now, X + Y = [\begin{matrix} 4 & \frac{5}{2} & 0 \\ \frac{5}{2} & 5 & \frac{5}{2} \\ 0 & \frac{5}{2} & 1 \end{matrix}] + [\begin{matrix} 0 & \frac{- 1}{2} & - 1 \\ \frac{1}{2} & 0 & \frac{9}{2} \\ 1 & \frac{- 9}{2} & 0 \end{matrix}] = [\begin{matrix} 4 & 2 & - 1 \\ 3 & 5 & 7 \\ 1 & - 2 & 1 \end{matrix}] = A$

Page No 4.61:

Question 6:

Define a symmetric matrix. Prove that for $A = [\begin{matrix} 2 & 4 \\ 5 & 6 \end{matrix}]$ , A + A^T is a symmetric matrix where A^T is the transpose of A.

Answer:

$A square matrix A is called a symmetric ma t rix, i f A^{T} = A . Given : A = [\begin{matrix} 2 & 4 \\ 5 & 6 \end{matrix}] A^{T} = [\begin{matrix} 2 & 5 \\ 4 & 6 \end{matrix}] Now, A + A^{T} = [\begin{matrix} 2 & 4 \\ 5 & 6 \end{matrix}] + [\begin{matrix} 2 & 5 \\ 4 & 6 \end{matrix}] \Rightarrow A + A^{T} = [\begin{matrix} 4 & 9 \\ 9 & 12 \end{matrix}] . . . (1) {(A + A^{T})}^{T} = {[\begin{matrix} 4 & 9 \\ 9 & 12 \end{matrix}]}^{T} = [\begin{matrix} 4 & 9 \\ 9 & 12 \end{matrix}] = A + A^{T} [From eq . (1)] ∴ {(A + A^{T})}^{T} = (A + A^{T}) Thus, (A + A^{T}) is a symmetric matrix .$

Page No 4.61:

Question 7:

Express the matrix $A = [\begin{matrix} 3 & - 4 \\ 1 & - 1 \end{matrix}]$ as the sum of a symmetric and a skew-symmetric matrix.

Answer:

$Given : A = [\begin{matrix} 3 & - 4 \\ 1 & - 1 \end{matrix}] A^{T} = [\begin{matrix} 3 & 1 \\ - 4 & - 1 \end{matrix}] Let X = \frac{1}{2} (A + A^{T}) = \frac{1}{2} ([\begin{matrix} 3 & - 4 \\ 1 & - 1 \end{matrix}] + [\begin{matrix} 3 & 1 \\ - 4 & - 1 \end{matrix}]) = [\begin{matrix} 3 & \frac{- 3}{2} \\ \frac{- 3}{2} & - 1 \end{matrix}] X^{T} = {[\begin{matrix} 3 & \frac{- 3}{2} \\ \frac{- 3}{2} & - 1 \end{matrix}]}^{T} = [\begin{matrix} 3 & \frac{- 3}{2} \\ \frac{- 3}{2} & - 1 \end{matrix}] = X Let Y = \frac{1}{2} (A - A^{T}) = \frac{1}{2} ([\begin{matrix} 3 & - 4 \\ 1 & - 1 \end{matrix}] - [\begin{matrix} 3 & 1 \\ - 4 & - 1 \end{matrix}]) = [\begin{matrix} 0 & \frac{- 5}{2} \\ \frac{5}{2} & 0 \end{matrix}] Y^{T} = {[\begin{matrix} 0 & \frac{- 5}{2} \\ \frac{5}{2} & 0 \end{matrix}]}^{T} = [\begin{matrix} 0 & \frac{5}{2} \\ \frac{- 5}{2} & 0 \end{matrix}] = - [\begin{matrix} 0 & \frac{- 5}{2} \\ \frac{5}{2} & 0 \end{matrix}] = Y ∴ X is a symmetric matrix and Y is a skew - symmetric matrix . X + Y = [\begin{matrix} 3 & \frac{- 3}{2} \\ \frac{- 3}{2} & - 1 \end{matrix}] + [\begin{matrix} 0 & \frac{- 5}{2} \\ \frac{5}{2} & 0 \end{matrix}] = [\begin{matrix} 3 & - 4 \\ 1 & - 1 \end{matrix}] = A$

Page No 4.61:

Question 8:

Express the following matrix as the sum of a symmetric and skew-symmetric matrix and verify your result: $[\begin{matrix} 3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2 \end{matrix}]$ .

Answer:

$Here, A = [\begin{matrix} 3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2 \end{matrix}] \Rightarrow A^{T} = [\begin{matrix} 3 & 3 & - 1 \\ - 2 & - 2 & 1 \\ - 4 & - 5 & 2 \end{matrix}] Let X = \frac{1}{2} (A + A^{T}) = \frac{1}{2} ([\begin{matrix} 3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2 \end{matrix}] + [\begin{matrix} 3 & 3 & - 1 \\ - 2 & - 2 & 1 \\ - 4 & - 5 & 2 \end{matrix}]) = [\begin{matrix} 3 & \frac{1}{2} & \frac{- 5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ \frac{- 5}{2} & - 2 & 2 \end{matrix}] X^{T} = {[\begin{matrix} 3 & \frac{1}{2} & \frac{- 5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ \frac{- 5}{2} & - 2 & 2 \end{matrix}]}^{T} = [\begin{matrix} 3 & \frac{1}{2} & \frac{- 5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ \frac{- 5}{2} & - 2 & 2 \end{matrix}] = X Let Y = \frac{1}{2} (A - A^{T}) = \frac{1}{2} ([\begin{matrix} 3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2 \end{matrix}] - [\begin{matrix} 3 & 3 & - 1 \\ - 2 & - 2 & 1 \\ - 4 & - 5 & 2 \end{matrix}]) = [\begin{matrix} 0 & \frac{- 5}{2} & \frac{- 3}{2} \\ \frac{5}{2} & 0 & - 3 \\ \frac{3}{2} & 3 & 0 \end{matrix}] Y^{T} = {[\begin{matrix} 0 & \frac{- 5}{2} & \frac{- 3}{2} \\ \frac{5}{2} & 0 & - 3 \\ \frac{3}{2} & 3 & 0 \end{matrix}]}^{T} = [\begin{matrix} 0 & \frac{5}{2} & \frac{3}{2} \\ \frac{- 5}{2} & 0 & 3 \\ \frac{- 3}{2} & - 3 & 0 \end{matrix}] = - [\begin{matrix} 0 & \frac{- 5}{2} & \frac{- 3}{2} \\ \frac{5}{2} & 0 & - 3 \\ \frac{3}{2} & 3 & 0 \end{matrix}] = - Y \Rightarrow X is a symmetric matrix and Y is a skew - symmetric matrix . Now, X + Y = [\begin{matrix} 3 & \frac{1}{2} & \frac{- 5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ \frac{- 5}{2} & - 2 & 2 \end{matrix}] + [\begin{matrix} 0 & \frac{- 5}{2} & \frac{- 3}{2} \\ \frac{5}{2} & 0 & - 3 \\ \frac{3}{2} & 3 & 0 \end{matrix}] = [\begin{matrix} 3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2 \end{matrix}] = A$

Page No 4.61:

Question 9:

For the matrix $A = [\begin{matrix} 2 & 3 \\ 5 & 7 \end{matrix}]$ , find A + A^T and verify that it is a symmetric matrix.

Answer:

The given matrix is

$A = [\begin{matrix} 2 & 3 \\ 5 & 7 \end{matrix}]$ .....(1)

$∴ A^{T} = [\begin{matrix} 2 & 5 \\ 3 & 7 \end{matrix}]$ ......(2)

Adding (1) and (2), we get

$A + A^{T} = [\begin{matrix} 2 & 3 \\ 5 & 7 \end{matrix}] + [\begin{matrix} 2 & 5 \\ 3 & 7 \end{matrix}] \Rightarrow A + A^{T} = [\begin{matrix} 2 + 2 & 3 + 5 \\ 5 + 3 & 7 + 7 \end{matrix}] = [\begin{matrix} 4 & 8 \\ 8 & 14 \end{matrix}]$

A matrix X is said to be symmetric matrix if $X^{T} = X$ .

Now,

${(A + A^{T})}^{T} = {[\begin{matrix} 4 & 8 \\ 8 & 14 \end{matrix}]}^{T} = [\begin{matrix} 4 & 8 \\ 8 & 14 \end{matrix}] ∴ {(A + A^{T})}^{T} = A + A^{T}$

Thus, the matrix $A + A^{T}$ is symmetric matrix.

Page No 4.62:

Question 1:

If $A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & - 1 \end{matrix}]$ , then A² is equal to
(a) a null matrix
(b) a unit matrix
(c) −A
(d) A

Answer:

$(b)$ a unit matrix

$A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & - 1 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & - 1 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 + 0 + 0 & 0 + 0 + 0 & 0 + 0 - 0 \\ 0 + 0 + 0 & 0 + 1 + 0 & 0 + 0 - 0 \\ a + 0 - a & 0 + b - b & 0 + 0 + 1 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

Page No 4.62:

Question 2:

If $A = [\begin{matrix} i & 0 \\ 0 & i \end{matrix}]$ , n ∈ N, then A⁴ⁿ equals

(a) $[\begin{matrix} 0 & i \\ i & 0 \end{matrix}]$

(b) $[\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

(c) $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

(d) $[\begin{matrix} 0 & i \\ i & 0 \end{matrix}]$

Answer:

(c) $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$Here, A = [\begin{matrix} i & 0 \\ 0 & i \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} i & 0 \\ 0 & i \end{matrix}] [\begin{matrix} i & 0 \\ 0 & i \end{matrix}] = [\begin{matrix} i^{2} & 0 \\ 0 & i^{2} \end{matrix}] \Rightarrow A^{3} = A^{2} . A = [\begin{matrix} i^{2} & 0 \\ 0 & i^{2} \end{matrix}] [\begin{matrix} i & 0 \\ 0 & i \end{matrix}] = [\begin{matrix} - i & 0 \\ 0 & - i \end{matrix}] \Rightarrow A^{4} = A^{3} . A = [\begin{matrix} - i & 0 \\ 0 & - i \end{matrix}] [\begin{matrix} i & 0 \\ 0 & i \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

So, $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ is repeated on multiple of 4 and 4n is a multiple of 4.

Thus,
$A^{4 n} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

Page No 4.62:

Question 3:

If A and B are two matrices such that AB = A and BA = B, then B² is equal to
(a) B
(b) A
(c) 1
(d) 0

Answer:

(a) B

$Here, A B = A . . . (1) B A = B . . . (2) \Rightarrow B A B = B B [Multiplying both sides by B] \Rightarrow B A = B^{2} [From eq . (1)] \Rightarrow B = B^{2} [From eq . (2)]$

Page No 4.62:

Question 4:

If AB = A and BA = B, where A and B are square matrices, then
(a) B² = B and A² = A
(b) B² ≠ B and A² = A
(c) A² ≠ A, B² = B
(d) A² ≠ A, B² ≠ B

Answer:

(a) B² = B and A² = A

$Here, A B = A . . . (1) B A = B . . . (2) \Rightarrow A B A = A A [Multiplying both sides by A] B A B = B B [Multiplying both sides by A] \Rightarrow A B = A^{2} [From eq . (2)] B A = B^{2} [From eq . (1)] \Rightarrow A = A^{2} [From eq . (1)] B = B^{2} [From eq . (2)]$

Page No 4.62:

Question 5:

If A and B are two matrices such that AB = B and BA = A, A² + B² is equal to
(a) 2 AB
(b) 2 BA
(c) A + B
(d) AB

Answer:

(c) A + B

$Given : A B = B and B A = A A^{2} + B^{2} = A A + B B \Rightarrow A^{2} + B^{2} = B A B A + A B A B [∵ A B = B and B A = A] \Rightarrow A^{2} + B^{2} = B B A + A A B [∵ A B = B and B A = A] \Rightarrow A^{2} + B^{2} = B A + A B [∵ A B = B and B A = A] \Rightarrow A^{2} + B^{2} = A + B [∵ A B = B and B A = A]$

Page No 4.62:

Question 6:

If ${[\begin{matrix} \cos \frac{2 π}{7} & - \sin \frac{2 π}{7} \\ \sin \frac{2 π}{7} & \cos \frac{2 π}{7} \end{matrix}]}^{k} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ , then the least positive integral value of k is
(a) 3
(b) 4
(c) 6
(d) 7

Answer:

(d) 7

$Here, A = [\begin{matrix} \cos \frac{2 π}{7} & - \sin \frac{2 π}{7} \\ \sin \frac{2 π}{7} & \cos \frac{2 π}{7} \end{matrix}] \Rightarrow A^{2} = A \times A \Rightarrow A^{2} = [\begin{matrix} \cos \frac{2 π}{7} & - \sin \frac{2 π}{7} \\ \sin \frac{2 π}{7} & \cos \frac{2 π}{7} \end{matrix}] [\begin{matrix} \cos \frac{2 π}{7} & - \sin \frac{2 π}{7} \\ \sin \frac{2 π}{7} & \cos \frac{2 π}{7} \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} \cos^{2} \frac{2 π}{7} - \sin^{2} \frac{2 π}{7} & (- 2 \cos \frac{2 π}{7} \sin \frac{2 π}{7}) \\ 2 \cos \frac{2 π}{7} \sin \frac{2 π}{7} & \cos^{2} \frac{2 π}{7} - \sin^{2} \frac{2 π}{7} \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} \cos \frac{4 π}{7} & - \sin \frac{4 π}{7} \\ \sin \frac{4 π}{7} & \cos \frac{4 π}{7} \end{matrix}] [\begin{matrix} ∵ \cos^{2} θ - \sin^{2} θ = \cos 2 θ \\ 2 \sin θ \cos θ = \sin 2 θ \end{matrix}] \Rightarrow A^{3} = A^{2} \times A \Rightarrow A^{3} = [\begin{matrix} \cos \frac{4 π}{7} & - \sin \frac{4 π}{7} \\ \sin \frac{4 π}{7} & \cos \frac{4 π}{7} \end{matrix}] [\begin{matrix} \cos \frac{2 π}{7} & - \sin \frac{2 π}{7} \\ \sin \frac{2 π}{7} & \cos \frac{2 π}{7} \end{matrix}] \Rightarrow A^{3} = [\begin{matrix} (\cos \frac{4 π}{7} \cos \frac{2 π}{7} - \sin \frac{4 π}{7} \sin \frac{2 π}{7}) & (- \cos \frac{4 π}{7} \sin \frac{2 π}{7} - \sin \frac{4 π}{7} \cos \frac{2 π}{7}) \\ (\sin \frac{4 π}{7} \cos \frac{2 π}{7} + \cos \frac{4 π}{7} \sin \frac{2 π}{7}) & (- \sin \frac{2 π}{7} \sin \frac{4 π}{7} + \cos \frac{4 π}{7} \cos \frac{2 π}{7}) \end{matrix}] \Rightarrow A^{3} = [\begin{matrix} \cos \frac{6 π}{7} & - \sin \frac{6 π}{7} \\ \sin \frac{6 π}{7} & \cos \frac{6 π}{7} \end{matrix}] [\begin{matrix} ∵ \cos (A + B) = \cos A \cos B - \sin A \sin B \\ \sin (A + B) = \sin A \cos B + \cos A \sin B \end{matrix}]$

Now we check if the pattern is same for k = 6.
Here,
$A^{6} = A^{3} . A^{3} \Rightarrow A^{6} = [\begin{matrix} \cos \frac{6 π}{7} & - \sin \frac{6 π}{7} \\ \sin \frac{6 π}{7} & \cos \frac{6 π}{7} \end{matrix}] [\begin{matrix} \cos \frac{6 π}{7} & - \sin \frac{6 π}{7} \\ \sin \frac{6 π}{7} & \cos \frac{6 π}{7} \end{matrix}] \Rightarrow A^{6} = [\begin{matrix} \cos \frac{12 π}{7} & - \sin \frac{12 π}{7} \\ \sin \frac{12 π}{7} & \cos \frac{12 π}{7} \end{matrix}]$

Now, we check if the pattern is same for k = 7.
Here,
$A^{7} = A^{6} \times A \Rightarrow A^{7} = [\begin{matrix} \cos \frac{12 π}{7} & - \sin \frac{12 π}{7} \\ \sin \frac{12 π}{7} & \cos \frac{12 π}{7} \end{matrix}] [\begin{matrix} \cos \frac{2 π}{7} & - \sin \frac{2 π}{7} \\ \sin \frac{2 π}{7} & \cos \frac{2 π}{7} \end{matrix}] \Rightarrow A^{7} = [\begin{matrix} \cos \frac{14 π}{7} & - \sin \frac{14 π}{7} \\ \sin \frac{14 π}{7} & \cos \frac{14 π}{7} \end{matrix}] \Rightarrow A^{7} = [\begin{matrix} \cos 2 π & - \sin 2 π \\ \sin 2 π & \cos 2 π \end{matrix}] [∵ \frac{14 π}{7} = 2 π] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

So, the least positive integral value of k is 7.

Page No 4.62:

Question 7:

If the matrix AB is zero, then
(a) It is not necessary that either A = O or, B = O
(b) A = O or B = O
(c) A = O and B = O
(d) all the above statements are wrong

Answer:

(a) It is not necessary that either A = O or, B = O

$Let A = [\begin{matrix} 0 & 2 \\ 0 & 0 \end{matrix}] and B = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] ∴ A B = [\begin{matrix} 0 & 2 \\ 0 & 0 \end{matrix}] [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

Page No 4.62:

Question 8:

Let A = $[\begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix}]$ , then Aⁿ is equal to

(a) $[\begin{matrix} a^{n} & 0 & 0 \\ 0 & a^{n} & 0 \\ 0 & 0 & a \end{matrix}]$

(b) $[\begin{matrix} a^{n} & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix}]$

(c) $[\begin{matrix} a^{n} & 0 & 0 \\ 0 & a^{n} & 0 \\ 0 & 0 & a^{n} \end{matrix}]$

(d) $[\begin{matrix} n a & 0 & 0 \\ 0 & n a & 0 \\ 0 & 0 & n a \end{matrix}]$

Answer:

(c) $[\begin{matrix} a^{n} & 0 & 0 \\ 0 & a^{n} & 0 \\ 0 & 0 & a^{n} \end{matrix}]$

$Here, A = [\begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix}] [\begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix}] = [\begin{matrix} a^{2} & 0 & 0 \\ 0 & a^{2} & 0 \\ 0 & 0 & a^{2} \end{matrix}] \Rightarrow A^{3} = [\begin{matrix} a^{2} & 0 & 0 \\ 0 & a^{2} & 0 \\ 0 & 0 & a^{2} \end{matrix}] [\begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix}] = [\begin{matrix} a^{3} & 0 & 0 \\ 0 & a^{3} & 0 \\ 0 & 0 & a^{3} \end{matrix}] This pattern is applicable on all natural numbers . ∴ A^{n} = [\begin{matrix} a^{n} & 0 & 0 \\ 0 & a^{n} & 0 \\ 0 & 0 & a^{n} \end{matrix}]$

Page No 4.62:

Question 9:

If A, B are square matrices of order 3, A is non-singular and AB = O, then B is a
(a) null matrix
(b) singular matrix
(c) unit-matrix
(d) non-singular matrix

Answer:

$(a)$ null matrix

Since A is non-singular matrix and the determinant of a non-singular matrix is non-zero, B should be a null matrix.

Page No 4.62:

Question 10:

If $A = [\begin{matrix} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{matrix}] and B = [\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c^{1} & c_{2} & c_{3} \end{matrix}]$ , then AB is equal to
(a) B
(b) nB
(c) Bⁿ
(d) A + B

Answer:

(b) nB

$Here, A = [\begin{matrix} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{matrix}] and B = [\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}] ∴ A B = [\begin{matrix} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{matrix}] [\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}] \Rightarrow A B = [\begin{matrix} n a_{1} & n a_{2} & n a_{3} \\ n b_{1} & n b_{2} & n b_{3} \\ n c_{1} & n c_{2} & n c_{3} \end{matrix}] \Rightarrow A B = n [\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}] \Rightarrow A B = n B$

Page No 4.62:

Question 11:

If $A = [\begin{matrix} 1 & a \\ 0 & 1 \end{matrix}]$ , then Aⁿ (where n ∈ N) equals

(a) $[\begin{matrix} 1 & n a \\ 0 & 1 \end{matrix}]$

(b) $[\begin{matrix} 1 & n^{2} a \\ 0 & 1 \end{matrix}]$

(c) $[\begin{matrix} 1 & n a \\ 0 & 0 \end{matrix}]$

(d) $[\begin{matrix} n & n a \\ 0 & n \end{matrix}]$

Answer:

(a) $[\begin{matrix} 1 & n a \\ 0 & 1 \end{matrix}]$

$Here, A = [\begin{matrix} 1 & a \\ 0 & 1 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 & a \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & a \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 + 0 & a + a \\ 0 + 0 & 0 + 1 \end{matrix}] = [\begin{matrix} 1 & 2 a \\ 0 & 1 \end{matrix}] A^{3} = A^{2} \times A = [\begin{matrix} 1 & 2 a \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & a \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 + 0 & a + 2 a \\ 0 + 0 & 0 + 1 \end{matrix}] = [\begin{matrix} 1 & 3 a \\ 0 & 1 \end{matrix}]$

This pattern is applicable for all natural numbers.

$∴ A^{n} = [\begin{matrix} 1 & n a \\ 0 & 1 \end{matrix}]$

Page No 4.63:

Question 12:

If $A = [\begin{matrix} 1 & 2 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] a n d B = [\begin{matrix} 1 & - 2 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$ and AB = I₃, then x + y equals
(a) 0
(b) −1
(c) 2
(d) none of these

Answer:

(a) 0

$Given : A B = I_{3} \Rightarrow [\begin{matrix} 1 & 2 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & - 2 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} 1 & 0 & y + x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] The corresponding elements of two equal matrices are equal . ∴ y + x = 0$

Page No 4.63:

Question 13:

If $A = [\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}], B = [\begin{matrix} a & 1 \\ b & - 1 \end{matrix}]$ and (A + B)² = A² + B², values of a and b are
(a) a = 4, b = 1
(b) a = 1, b = 4
(c) a = 0, b = 4
(d) a = 2, b = 4

Answer:

(b) a = 1, b = 4

$Here, {(A + B)}^{2} = A^{2} + B^{2} \Rightarrow A^{2} + A B + B A + B^{2} = A^{2} + B^{2} \Rightarrow A B + B A = O \Rightarrow A B = - B A \Rightarrow [\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}] [\begin{matrix} a & 1 \\ b & - 1 \end{matrix}] = - [\begin{matrix} a & 1 \\ b & - 1 \end{matrix}] [\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}] \Rightarrow [\begin{matrix} a - b & 2 \\ 2 a - b & 3 \end{matrix}] = - [\begin{matrix} a + 2 & - a - 1 \\ b - 2 & - b + 1 \end{matrix}] \Rightarrow [\begin{matrix} a - b & 2 \\ 2 a - b & 3 \end{matrix}] = [\begin{matrix} - a - 2 & a + 1 \\ b + 2 & b - 1 \end{matrix}] The corresponding elements of two equal matrices are equal . \Rightarrow a + 1 = 2 and b - 1 = 3 ∴ a = 1 and b = 4$

Page No 4.63:

Question 14:

If $A = [\begin{matrix} α & β \\ γ & - α \end{matrix}]$ is such that A² = I, then
(a) 1 + α² + βγ = 0
(b) 1 − α² + βγ = 0
(c) 1 − α² − βγ = 0
(d) 1 + α² − βγ = 0

Answer:

(c) 1 − α² − βγ = 0

$Here, A^{2} = I \Rightarrow [\begin{matrix} α & β \\ γ & - α \end{matrix}] [\begin{matrix} α & β \\ γ & - α \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} α^{2} + β γ & α β - β α \\ λ α - α γ & γ β + α^{2} \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} α^{2} + β γ & 0 \\ 0 & γ β + α^{2} \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] The corresponding elements of two equal matrices are equal . \Rightarrow α^{2} + β γ = 1 \Rightarrow 1 - α^{2} - β γ = 0$

Page No 4.63:

Question 15:

If S = [S_ij] is a scalar matrix such that s_ij = k and A is a square matrix of the same order, then AS = SA = ?
(a) A^k
(b) k + A
(c) kA
(d) kS

Answer:

(c) kA

Here,

$S = [S_{i j}] \Rightarrow S = [\begin{matrix} k & 0 \\ 0 & k \end{matrix}] [∵ S_{i j} = k] Let A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}] [∵ A is square matrix] Now, A S = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}] [\begin{matrix} k & 0 \\ 0 & k \end{matrix}] = [\begin{matrix} k a_{11} & k a_{12} \\ k a_{21} & k a_{22} \end{matrix}] = k [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}] = k A S A = [\begin{matrix} k & 0 \\ 0 & k \end{matrix}] [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}] = [\begin{matrix} k a_{11} & k a_{12} \\ k a_{21} & k a_{22} \end{matrix}] = k [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}] = k A ∴ A S = S A = k A$

Page No 4.63:

Question 16:

If A is a square matrix such that A² = A, then (I + A)³ − 7A is equal to
(a) A
(b) I − A
(c) I
(d) 3A

Answer:

(c) I

$Here, A^{2} = A . . . (1) A^{3} = A^{2} A = A^{2} [From eq . (1)] = A ∴ A^{3} = A . . . (2) We know that {(I + A)}^{3} = I^{3} + 3 {(I)}^{2} A + 3 (I) A^{2} + A^{3} \Rightarrow {(I + A)}^{3} = I + 3 A + 3 A + A [From eqs . (1) and (2)] \Rightarrow {(I + A)}^{3} = I + 7 A \Rightarrow {(I + A)}^{3} - 7 A = I$

Page No 4.63:

Question 17:

If a matrix A is both symmetric and skew-symmetric, then
(a) A is a diagonal matrix
(b) A is a zero matrix
(c) A is a scalar matrix
(d) A is a square matrix

Answer:

(b) A is a zero matrix

Let $A = [a_{i j}]$ be a matrix which is both symmetric and skew-symmetric.

If $A = [a_{i j}]$ is a symmetric matrix, then
$a_{i j} = a_{j i}$ for all i, j                   ...(1)

If $A = [a_{i j}]$ is a skew-symmetric matrix, then
$a_{i j} = - a_{j i}$ for all i, j
$\Rightarrow a_{j i} = - a_{i j}$ for all i,j            ...(2)

From eqs. (1) and (2), we have

$a_{i j} = - a_{i j} \Rightarrow a_{i j} + a_{i j} = 0 \Rightarrow 2 a_{i j} = 0 \Rightarrow a_{i j} = 0 ∴ A = [a_{i j}] is a zero matrix or null matrix .$

Page No 4.63:

Question 18:

The matrix $[\begin{matrix} 0 & 5 & - 7 \\ - 5 & 0 & 11 \\ 7 & - 11 & 0 \end{matrix}]$ is
(a) a skew-symmetric matrix
(b) a symmetric matrix
(c) a diagonal matrix
(d) an uppertriangular matrix

Answer:

(a) a skew-symmetric matrix

Here,

A = $[\begin{matrix} 0 & 5 & - 7 \\ - 5 & 0 & 11 \\ 7 & - 11 & 0 \end{matrix}]$

$\Rightarrow$ A^T = $[\begin{matrix} 0 & - 5 & 7 \\ 5 & 0 & - 11 \\ - 7 & 11 & 0 \end{matrix}]$

$\Rightarrow A^{T} = - [\begin{matrix} 0 & 5 & - 7 \\ - 5 & 0 & 11 \\ 7 & - 11 & 0 \end{matrix}] \Rightarrow A^{T} = - A$

Thus, A is a skew-symmetric matrix.

Page No 4.63:

Question 19:

If A is a square matrix, then AA is a
(a) skew-symmetric matrix
(b) symmetric matrix
(c) diagonal matrix
(d) none of these

Answer:

(d) none of these

Given: A is a square matrix.

$Let A = [\begin{matrix} 1 & 2 \\ 1 & 0 \end{matrix}] \Rightarrow A A = [\begin{matrix} 1 & 2 \\ 1 & 0 \end{matrix}] [\begin{matrix} 1 & 2 \\ 1 & 0 \end{matrix}] = [\begin{matrix} 3 & 2 \\ 1 & 2 \end{matrix}]$

Page No 4.63:

Question 20:

If A and B are symmetric matrices, then ABA is
(a) symmetric matrix
(b) skew-symmetric matrix
(c) diagonal matrix
(d) scalar matrix

Answer:

(a) symmetric matrix

$Since A and B are symmetric matri c e s, we get A = A' and B = B' (A B A)' = (B A)' (A)' = A' B' A' = A B A [∵ A = A' and B = B'] Since (A B A)' = A B A, A B A is a symmetric matrix .$

Page No 4.63:

Question 21:

If $A = [\begin{matrix} 5 & x \\ y & 0 \end{matrix}]$ and A = A^T, then
(a) x = 0, y = 5
(b) x + y = 5
(c) x = y
(d) none of these

Answer:

(c) x = y

$Here, A = [\begin{matrix} 5 & x \\ y & 0 \end{matrix}] A^{T} = [\begin{matrix} 5 & y \\ x & 0 \end{matrix}] Now, A = A^{T} The corresponding elements of two equal matrices are equal . ∴ [\begin{matrix} 5 & x \\ y & 0 \end{matrix}] = [\begin{matrix} 5 & y \\ x & 0 \end{matrix}] \Rightarrow x = y$

Page No 4.63:

Question 22:

If A is 3 × 4 matrix and B is a matrix such that A'B and BA' are both defined. Then, B is of the type
(a) 3 × 4
(b) 3 × 3
(c) 4 × 4
(d) 4 × 3

Answer:

(a) 3 × 4

The order of A is 3 $\times$ 4. So, the order of A' is 4 $\times$ 3.

Now, both $A' B and B A'$ are defined. So, the number of columns in A' should be equal to the number of rows in B for A'B.
Also, the number of columns in B should be equal to number of rows in A' for BA'.

Hence, the order of matrix B is 3 $\times$ 4.

Page No 4.63:

Question 23:

If A = [a_ij] is a square matrix of even order such that a_ij = i² − j², then
(a) A is a skew-symmetric matrix and | A | = 0
(b) A is symmetric matrix and | A | is a square
(c) A is symmetric matrix and | A | = 0
(d) none of these.

Answer:

(d) none of these

$Given : A is a square matrix of even order . Let A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}] \Rightarrow A = [\begin{matrix} 0 & - 3 \\ 3 & 0 \end{matrix}] [∵ a_{i j} = i^{2} - j^{2}] So, it is a skew - symmetric matrix as a_{i j} = - a_{j i} . Now, |A| = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}] = [\begin{matrix} a_{11} a_{22} - a_{21} a_{12} \end{matrix}] = [\begin{matrix} 0 - (- 9) \end{matrix}] = 9$

Page No 4.64:

Question 24:

If $A = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]$ , then A^T + A = I₂, if
(a) θ = n π, n ∈ Z
(b) θ = (2n + 1) $\frac{π}{2}$ , n ∈ Z
(c) θ = 2n π + $\frac{π}{3}$ , n ∈ Z
(d) none of these

Answer:

(c) θ = 2nπ + $\frac{π}{3}$ , n ∈ Z

$Here, A = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}] \Rightarrow A^{T} = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] Now, A^{T} + A = I_{2} \Rightarrow [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] + [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} 2 \cos θ & 0 \\ 0 & 2 \cos θ \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow 2 \cos θ = 1 \Rightarrow \cos θ = \frac{1}{2} \Rightarrow \cos θ = \cos \frac{π}{3} \Rightarrow θ = 2 n π \pm \frac{π}{3} (n \in Z)$

Page No 4.64:

Question 25:

If $A = [\begin{matrix} 2 & 0 & - 3 \\ 4 & 3 & 1 \\ - 5 & 7 & 2 \end{matrix}]$ is expressed as the sum of a symmetric and skew-symmetric matrix, then the symmetric matrix is

(a) $[\begin{matrix} 2 & 2 & - 4 \\ 2 & 3 & 4 \\ - 4 & 4 & 2 \end{matrix}]$

(b) $[\begin{matrix} 2 & 4 & - 5 \\ 0 & 3 & 7 \\ - 3 & 1 & 2 \end{matrix}]$

(c) $[\begin{matrix} 4 & 4 & - 8 \\ 4 & 6 & 8 \\ - 8 & 8 & 4 \end{matrix}]$

(d) $[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

Answer:

(a) $[\begin{matrix} 2 & 2 & - 4 \\ 2 & 3 & 4 \\ - 4 & 4 & 2 \end{matrix}]$

$Here, A = [\begin{matrix} 2 & 0 & - 3 \\ 4 & 3 & 1 \\ - 5 & 7 & 2 \end{matrix}] \Rightarrow A^{T} = [\begin{matrix} 2 & 4 & - 5 \\ 0 & 3 & 7 \\ - 3 & 1 & 2 \end{matrix}] Now, A + A^{T} = [\begin{matrix} 2 & 0 & - 3 \\ 4 & 3 & 1 \\ - 5 & 7 & 2 \end{matrix}] + [\begin{matrix} 2 & 4 & - 5 \\ 0 & 3 & 7 \\ - 3 & 1 & 2 \end{matrix}] \Rightarrow A + A^{T} = [\begin{matrix} 2 + 2 & 0 + 4 & - 3 - 5 \\ 4 + 0 & 3 + 3 & 1 + 7 \\ - 5 - 3 & 7 + 1 & 2 + 2 \end{matrix}] \Rightarrow A + A^{T} = [\begin{matrix} 4 & 4 & - 8 \\ 4 & 6 & 8 \\ - 8 & 8 & 4 \end{matrix}] A - A^{T} = [\begin{matrix} 2 & 0 & - 3 \\ 4 & 3 & 1 \\ - 5 & 7 & 2 \end{matrix}] - [\begin{matrix} 2 & 4 & - 5 \\ 0 & 3 & 7 \\ - 3 & 1 & 2 \end{matrix}] \Rightarrow A - A^{T} = [\begin{matrix} 2 - 2 & 0 - 4 & - 3 + 5 \\ 4 - 0 & 3 - 3 & 1 - 7 \\ - 5 + 3 & 7 - 1 & 2 - 2 \end{matrix}] \Rightarrow A - A^{T} = [\begin{matrix} 0 & - 4 & 2 \\ 4 & 0 & - 6 \\ - 2 & 6 & 0 \end{matrix}] Let P = \frac{1}{2} (A + A^{T}) = \frac{1}{2} [\begin{matrix} 4 & 4 & - 8 \\ 4 & 6 & 8 \\ - 8 & 8 & 4 \end{matrix}] = [\begin{matrix} 2 & 2 & - 4 \\ 2 & 3 & 4 \\ - 4 & 4 & 2 \end{matrix}] Q = \frac{1}{2} (A - A^{T}) = \frac{1}{2} [\begin{matrix} 0 & - 4 & 2 \\ 4 & 0 & - 6 \\ - 2 & 6 & 0 \end{matrix}] = [\begin{matrix} 0 & - 2 & 1 \\ 2 & 0 & - 3 \\ - 1 & 3 & 0 \end{matrix}] Now, P^{T} = {[\begin{matrix} 2 & 2 & - 4 \\ 2 & 3 & 4 \\ - 4 & 4 & 2 \end{matrix}]}^{T} = [\begin{matrix} 2 & 2 & - 4 \\ 2 & 3 & 4 \\ - 4 & 4 & 2 \end{matrix}] = P Q^{T} = {[\begin{matrix} 0 & - 2 & 1 \\ 2 & 0 & - 3 \\ - 1 & 3 & 0 \end{matrix}]}^{T} = [\begin{matrix} 0 & 2 & - 1 \\ - 2 & 0 & 3 \\ 1 & - 3 & 0 \end{matrix}] = - [\begin{matrix} 0 & - 2 & 1 \\ 2 & 0 & - 3 \\ - 1 & 3 & 0 \end{matrix}] = - Q Thus, P is symmetric and Q is skew - symmetric . P + Q = [\begin{matrix} 2 & 2 & - 4 \\ 2 & 3 & 4 \\ - 4 & 4 & 2 \end{matrix}] + [\begin{matrix} 0 & - 2 & 1 \\ 2 & 0 & - 3 \\ - 1 & 3 & 0 \end{matrix}] = [\begin{matrix} 2 + 0 & 2 - 2 & - 4 + 1 \\ 2 + 2 & 3 + 0 & 4 - 3 \\ - 4 - 1 & 4 + 3 & 2 + 0 \end{matrix}] = [\begin{matrix} 2 & 0 & - 3 \\ 4 & 3 & 1 \\ - 5 & 7 & 2 \end{matrix}] = A Thus, we have expressed A is the sum of a symmetric and a skew - symmetric matrix . Hence, the symmetric matrix is [\begin{matrix} 2 & 2 & - 4 \\ 2 & 3 & 4 \\ - 4 & 4 & 2 \end{matrix}] .$

Page No 4.64:

Question 26:

Out of the given matrices, choose that matrix which is a scalar matrix:

(a) $[\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

(b) $[\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$

(c) $[\begin{matrix} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{matrix}]$

(d) $[\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

Answer:

$(a) [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

A diagonal matrix in which all the diagonal elements are equal is called the scalar matrix.

Page No 4.64:

Question 27:

The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is
(a) 27
(b) 18
(c) 81
(d) 512

Answer:

(d) 512

There are 9 elements in a 3 $\times$ 3 matrix and one element can be filled in two ways, either with 0 or 1.

Thus,
Total possible matrices = $2^{9}$ = 512

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Question 28:

Which of the given values of x and y make the following pairs of matrices equal?
$[\begin{matrix} 3 x + 7 & 5 \\ y + 1 & 2 - 3 x \end{matrix}], [\begin{matrix} 0 & y - 2 \\ 8 & 4 \end{matrix}]$
(a) x = $- \frac{1}{3}$ , y = 7
(b) y = 7, x = $- \frac{2}{3}$
(c) x = $- \frac{1}{3}$ , 4 = $- \frac{2}{5}$
(d) Not possible to find

Answer:

(d) Not possible to find

$Here, [\begin{matrix} 3 x + 7 & 5 \\ y + 1 & 2 - 3 x \end{matrix}] = [\begin{matrix} 0 & y - 2 \\ 8 & 4 \end{matrix}] We know that for two equal matrice s the corresponding elements are equal . ∴ 3 x + 7 = 0, 5 = y - 2, y + 1 = 8 and 2 - 3 x = 4 \Rightarrow 3 x = - 7, 5 + 2 = y, y = 8 - 1 and - 3 x = 4 - 2 \Rightarrow x = \frac{- 7}{3}, y = 7, y = 7 and x = - \frac{2}{3}$

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Question 29:

If $A = [\begin{matrix} 0 & 2 \\ 3 & - 4 \end{matrix}]$ and $k A = [\begin{matrix} 0 & 3 a \\ 2 b & 24 \end{matrix}]$ , then the values of k, a, b, are respectively
(a) −6, −12, −18
(b) −6, 4, 9
(c) −6, −4, −9
(d) −6, 12, 18

Answer:

(c) −6, −4, −9

$Given : A = [\begin{matrix} 0 & 2 \\ 3 & - 4 \end{matrix}] Here, k A = [\begin{matrix} 0 & 3 a \\ 2 b & 24 \end{matrix}] \Rightarrow k [\begin{matrix} 0 & 2 \\ 3 & - 4 \end{matrix}] = [\begin{matrix} 0 & 3 a \\ 2 b & 24 \end{matrix}] \Rightarrow [\begin{matrix} 0 & 2 k \\ 3 k & - 4 k \end{matrix}] = [\begin{matrix} 0 & 3 a \\ 2 b & 24 \end{matrix}] The corresponding elements of two equal matrices are equal . \Rightarrow 2 k = 3 a, 3 k = 2 b and - 4 k = 24 Now, - 4 k = 24 \Rightarrow k = - 6 Also, 2 k = 3 a and 3 k = 2 b \Rightarrow 2 (- 6) = 3 a and 3 (- 6) = 2 b [Using k = - 6] \Rightarrow - 12 = 3 a and - 18 = 2 b ∴ a = - 4 and b = - 9$

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Question 30:

If $I = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}], J = [\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}] and B = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]$ , then B equals
(a) I cos θ + J sin θ
(b) I sin θ + J cos θ
(c) I cos θ − J sin θ
(d) −I cos θ + J sin θ

Answer:

$(a) I \cos θ + J \sin θ Here, I \cos θ + J \sin θ = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \cos θ + [\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}] \sin θ = [\begin{matrix} \cos θ & 0 \\ 0 & \cos θ \end{matrix}] + [\begin{matrix} 0 & \sin θ \\ - \sin θ & 0 \end{matrix}] = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] = B$

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Question 31:

The trace of the matrix $A = [\begin{matrix} 1 & - 5 & 7 \\ < \end{matrix}]$