Rd Sharma XII Vol 1 2020 for Class 12 Commerce Maths Chapter 5 - Determinants

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Rd Sharma XII Vol 1 2020 Solutions for Class 12 Commerce Maths Chapter 5 Determinants are provided here with simple step-by-step explanations. These solutions for Determinants are extremely popular among Class 12 Commerce students for Maths Determinants Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2020 Book of Class 12 Commerce Maths Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2020 Solutions. All Rd Sharma XII Vol 1 2020 Solutions for class Class 12 Commerce Maths are prepared by experts and are 100% accurate.

Page No 5.10:

Question 1:

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:
(i) $A = [\begin{matrix} 5 & 20 \\ 0 & - 1 \end{matrix}]$

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

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

(iv) $A = [\begin{matrix} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{matrix}]$

(v) $A = [\begin{matrix} 0 & 2 & 6 \\ 1 & 5 & 0 \\ 3 & 7 & 1 \end{matrix}]$

(vi) $A = [\begin{matrix} a & h & g \\ h & b & f \\ g & f & c \end{matrix}]$

(vii) $A = [\begin{matrix} 2 & - 1 & 0 & 1 \\ - 3 & 0 & 1 & - 2 \\ 1 & 1 & - 1 & 1 \\ 2 & - 1 & 5 & 0 \end{matrix}]$

Answer:

(i)
        $M_{11} = - 1 M_{21} = 20 C_{i j} = {(- 1)}^{i + j} M_{i j} C_{11} = {(- 1)}^{1 + 1} (- 1) = - 1 C_{21 =} {(- 1)}^{1 + 2} (20) = - 20 D = (- 1 \times 5) - (20 \times 0) = - 5$

(ii)
        $M_{11} = 3 M_{21} = 4 C_{i j} = {(- 1)}^{i + j} M_{i j} C_{11} = {(- 1)}^{1 + 1} M_{11} = 3 C_{21} = {(- 1)}^{2 + 1} M_{21} = - (4) = - 4 D = (3 \times - 1) - (4 \times 2) = - 3 - 8 = - 11$

(iii)
      $M_{11} = |\begin{matrix} - 1 & 2 \\ 5 & 2 \end{matrix}| = - 2 - 10 = - 12 M_{21 =} |\begin{matrix} - 3 & 2 \\ 5 & 2 \end{matrix}| = - 6 - 10 = - 16 M_{31 =} |\begin{matrix} - 3 & 2 \\ - 1 & 2 \end{matrix}| = - 6 + 2 = - 4 C_{11} = {(- 1)}^{1 + 1} M_{11} = - 12 C_{21 =} {(- 1)}^{2 + 1} M_{21} = - (- 16) = 16 C_{31} = {(- 1)}^{3 + 1} M_{31} = - 4 D = 1 (- 12) + 3 (8 - 6) + 2 (20 + 3) = - 12 + 6 + 46 = 40$

(iv)
      $M_{11} = |\begin{matrix} b & c a \\ c & a b \end{matrix}| = a b^{2} - c^{2} a = a (b^{2} - c^{2}) M_{21} = |\begin{matrix} a & b c \\ c & a b \end{matrix}| = a^{2} b - c^{2} b = b (a^{2} - c^{2}) M_{31 =} |\begin{matrix} a & b c \\ b & c a \end{matrix}| = a^{2} c - b^{2} c = c (a^{2} - b^{2}) C_{11} = {(- 1)}^{1 + 1} M_{11} = a (b^{2} - c^{2}) C_{21} = {(- 1)}^{2 + 1} M_{21} = - b (a^{2} - c^{2}) C_{31} = {(- 1)}^{3 + 1} M_{31} = c (a^{2} - b^{2}) D = 1 . a (b^{2} - c^{2}) - a (a b - c a) + b . c (c - b) = a b^{2} - a c^{2} - a^{2} b + a^{2} c + c^{2} b - b^{2} c = a^{2} (c - b) + b^{2} (a - c) + c^{2} (b - a)$

(v)
       $M_{11} = |\begin{matrix} 5 & 0 \\ 7 & 1 \end{matrix}| = 5 - 0 = 5 M_{21} = |\begin{matrix} 2 & 6 \\ 7 & 1 \end{matrix}| = 2 - 42 = - 40 M_{31} = |\begin{matrix} 2 & 6 \\ 5 & 0 \end{matrix}| = 0 - 30 = - 30 C_{11} = {(- 1)}^{1 + 1} M_{11} = 5 C_{21} = {(- 1)}^{2 + 1} M_{21} = - (- 40) C_{31} = {(- 1)}^{3 + 1} M_{31} = - 30 D = 0 (5 - 0) - 2 (1 - 0) + 6 (7 - 15) = - 2 - 48 = - 50$

(vi)
       $M_{11} = |\begin{matrix} b & f \\ f & c \end{matrix}| = b c - f^{2} M_{21 =} |\begin{matrix} h & g \\ f & c \end{matrix}| = h c - f g M_{31 =} |\begin{matrix} h & g \\ b & f \end{matrix}| = h f - g b C_{11} = {(- 1)}^{1 + 1} M_{11} = b c - f^{2} C_{21} = {(- 1)}^{2 + 1} M_{11} = - (h c - f g) = f g - h c C_{31} = {(- 1)}^{3 + 1} M_{11} = h f - g b D = a (b c - f^{2}) - h (h c - f g) + g (f h - b g) = a b c - a f^{2} - h^{2} c + f g h + f g h - b g^{2} = a b c + 2 h f g - a f^{2} - b g^{2} - c h^{2}$

(vii)

      $M_{11} = 0 (0 - 5) - 1 (0 + 1) - 2 (5 - 1) = - 1 - 8 = - 9 M_{21} = - 1 (0 - 5) + 1 (5 - 1) = 5 + 4 = 9 M_{31} = - 1 (0 + 10) + 1 (0 + 1) = - 10 + 1 = - 9 M_{41} = - 1 (1 - 2) + 1 (0 - 1) = 1 - 1 = 0 C_{11} = {(- 1)}^{1 + 1} M_{11} = - 9 C_{21} = {(- 1)}^{2 + 1} M_{21} = (- 1) \times 9 C_{31} = {(- 1)}^{3 + 1} M_{31} = - 9 C_{41} = {(- 1)}^{4 + 1} M_{41} = 0 D = 2 |\begin{matrix} 0 & 1 & - 2 \\ 1 & - 1 & 1 \\ - 1 & 5 & 0 \end{matrix}| + 1 |\begin{matrix} - 3 & 1 & - 2 \\ 1 & - 1 & 1 \\ 2 & 5 & 0 \end{matrix}| - 1 |\begin{matrix} - 3 & 0 & 1 \\ 1 & 1 & - 1 \\ 2 & - 1 & 5 \end{matrix}| = - 18 - 27 + 15 = 30$

Page No 5.10:

Question 2:

Evaluate the following determinants:

(i) $|\begin{matrix} x & - 7 \\ x & 5 x + 1 \end{matrix}|$

(ii) $|\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}|$

(iii) $|\begin{matrix} \cos 15 ° & \sin 15 ° \\ \sin 75 ° & \cos 75 ° \end{matrix}|$

(iv) $|\begin{matrix} a + i b & c + i d \\ - c + i d & a - i b \end{matrix}|$

Answer:

(i)
$∆ = x (5 x + 1) + 7 x = 5 x^{2} + x + 7 x = 5 x^{2} + 8 x$

(ii)
$∆ = \cos^{2} θ - (- \sin^{2} θ) = \cos^{2} θ + \sin^{2} θ = 1$

(iii)
$∆ = \cos 15 ° \cos 75 ° - \sin 15 ° \sin 75 ° = \cos 15 ° \cos 75 ° - \sin (90 ° - 75 °) \sin (90 ° - 15 °) [∵ \sin (90 ° - θ) = \cos θ] = \cos 15 ° \cos 75 ° - \cos 75 ° \cos 15 ° = \cos 15 ° \cos 75 ° - \cos 15 ° \cos 75 ° = 0$

(iv)
$∆ = a^{2} - i^{2} b^{2} - (i^{2} d^{2} - c^{2}) = a^{2} - i^{2} b^{2} - i^{2} d^{2} + c^{2} = a^{2} + c^{2} - i^{2} (b^{2} + d^{2}) [∵ i^{2} = - 1] = a^{2} + c^{2} + b^{2} + d^{2}$

Page No 5.10:

Question 3:

Evaluate ${|\begin{matrix} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{matrix}|}^{2} .$

Answer:

$Let A = |\begin{matrix} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{matrix}| = 2 (204 - 100) - 3 (156 - 75) + 7 (260 - 255) \Rightarrow A = 2 (104) - 3 (81) + 7 (5) \Rightarrow A = 208 - 243 + 35 \Rightarrow A = 243 - 243 = 0 ∴ |\begin{matrix} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{matrix}| = 0 \Rightarrow {|\begin{matrix} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{matrix}|}^{2} = 0^{2} = 0 [∵ \det A^{2} = {(\det A)}^{2}]$

Page No 5.10:

Question 4:

Show that $|\begin{matrix} \sin 10 ° & - \cos 10 ° \\ \sin 80 ° & \cos 80 ° \end{matrix}| = 1$

Answer:

$Let ∆ = |\begin{matrix} \sin 10 ° & - \cos 10 ° \\ \sin 80 ° & \cos 80 ° \end{matrix}| \Rightarrow ∆ = \sin 10 ° \cos 80 ° + \cos 10 ° \sin 80 ° = \sin 10 ° \cos (90 ° - 10 °) + \cos 10 ° \sin (90 ° - 10 °) [∵ \cos θ = \sin (90 - θ)] \Rightarrow ∆ = \sin 10 ° \sin 10 ° + \cos 10 ° \cos 10 ° = \sin^{2} 10 ° + \cos^{2} 10 ° [∵ \sin^{2} θ + c o s^{2} θ = 1] \Rightarrow ∆ = 1$

Page No 5.10:

Question 5:

Evaluate $|\begin{matrix} 2 & 3 & - 5 \\ 7 & 1 & - 2 \\ - 3 & 4 & 1 \end{matrix}|$ by two methods.

Answer:

Let $∆ = |\begin{matrix} 2 & 3 & - 5 \\ 7 & 1 & - 2 \\ - 3 & 4 & 1 \end{matrix}|$

First method

$∆ = {(- 1)}^{1 + 1} 2 (1 + 8) + {(- 1)}^{1 + 2} 3 (7 - 6) + {(- 1)}^{1 + 3} (- 5) (28 + 3) = 2 (1 + 8) - 3 (7 - 6) - 5 (28 + 3) = 18 - 3 - 155 = - 140$

Second method is the Sarus Method, where we adjoin the first two columns to the right to get

$\begin{matrix} 2 & 3 & - 5 & 2 & 3 \\ 7 & 1 & - 2 & 7 & 1 \\ - 3 & 4 & 1 & - 3 & 4 \end{matrix} \begin{matrix} \end{matrix} = (2 \times 1 \times 1 + 3 \times - 2 \times - 3 - 5 \times 7 \times 4) - (- 5 \times 1 \times - 3 + 2 \times - 2 \times 4 + 3 \times 7 \times 1) = (2 + 18 - 140) - (15 - 16 + 21) = - 120 - 20 = - 140$

Page No 5.10:

Question 6:

Evaluate $∆ = |\begin{matrix} 0 & \sin α & - \cos α \\ - \sin α & 0 & \sin β \\ \cos α & - \sin β & 0 \end{matrix}|$

Answer:

Let $∆ = |\begin{matrix} 0 & \sin α & - \cos α \\ - \sin α & 0 & \sin β \\ \cos α & - \sin β & 0 \end{matrix}|$

$∆ = {(- 1)}^{1 + 1} 0 (0 + \sin^{2} β) + {(- 1)}^{1 + 2} \sin α (0 - \sin β \cos α) + {(- 1)}^{1 + 3} (- \cos α) (\sin α \sin β - 0) [Expanding along R_{1}] = 0 (0 + \sin^{2} β) - \sin α (0 - \sin β \cos α) - \cos α (\sin α \sin β - 0) = \sin α \sin β \cos α - \sin α \sin β \cos α = 0$

Page No 5.10:

Question 7:

$∆ = |\begin{matrix} \cos α \cos β & \cos α \sin β & - \sin α \\ - \sin β & \cos β & 0 \\ \sin α \cos β & \sin α \sin β & \cos α \end{matrix}|$

Answer:

Given: $∆ = |\begin{matrix} \cos α \cos β & \cos α \sin β & - \sin α \\ - \sin β & \cos β & 0 \\ \sin α \cos β & \sin α \sin β & \cos α \end{matrix}|$

$\Rightarrow ∆ = {(- 1)}^{1 + 1} \cos α \cos β (\cos α \cos β - 0) + {(- 1)}^{1 + 2} \cos α \sin β (- \sin β \cos α - 0) + {(- 1)}^{1 + 3} (- \sin α) (- \sin^{2} β \sin α - \sin α \cos^{2} β) [Expanding along R_{1}] = \cos α \cos β (\cos α \cos β - 0) - \cos α \sin β (- \sin β \cos α - 0) - \sin α (- \sin^{2} β \sin α - \sin α \cos^{2} β) = \cos^{2} α \cos^{2} β + \cos^{2} α \sin^{2} β + \sin^{2} α \sin^{2} β + \sin^{2} α \cos^{2} β = \cos^{2} α (\cos^{2} β + \sin^{2} β) + \sin^{2} α (\sin^{2} β + \cos^{2} β) \Rightarrow ∆ = \cos^{2} α + \sin^{2} α [∵ \sin^{2} θ + \cos^{2} θ = 1] \Rightarrow ∆ = 1 [∵ \sin^{2} θ + \cos^{2} θ = 1]$

Page No 5.10:

Question 8:

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

Answer:

$Consider LHS A B = [\begin{matrix} 2 & 5 \\ 2 & 1 \end{matrix}] [\begin{matrix} 4 & - 3 \\ 2 & 5 \end{matrix}] = [\begin{matrix} 8 + 10 & - 6 + 25 \\ 8 + 2 & - 6 + 5 \end{matrix}] = [\begin{matrix} 18 & 19 \\ 10 & - 1 \end{matrix}] |A B| = - 18 - 190 = - 208 Consider RHS |A| = 2 - 10 = - 8 |B| = 20 - (- 6) = 26 |A| |B| = - 8 \times 26 = - 208 ∴ LHS = RHS$

Page No 5.10:

Question 9:

If A $[\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix}]$ , then show that |3 A| = 27 |A|.

Answer:

$A = [\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix}] \Rightarrow 3 A = [\begin{matrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{matrix}] [Multiplying each element of A by 3] \Rightarrow |\begin{matrix} 3 A \end{matrix}| = {(- 1)}^{1 + 1} 3 (36 - 0) + {(- 1)}^{1 + 2} 0 (0 - 0) + {(- 1)}^{1 + 3} 3 (0 - 0) = 3 (36 - 0) - 0 (0 - 0) + 3 (0 - 0) [Expanding along R_{1}] = 3 \times 36 = 108 . . . (1) \Rightarrow |\begin{matrix} A \end{matrix}| = {(- 1)}^{1 + 1} 1 (4 - 0) + {(- 1)}^{1 + 2} 0 (0 - 0) + {(- 1)}^{1 + 3} 1 (0 - 0) = 1 (4 - 0) - 0 (0 - 0) + 1 (0 - 0) = 4 [Expanding along R_{1}] \Rightarrow 27 |A| = 27 \times 4 = 108 . . . (2) ∴ |3 A| = 27 |A| [From eqs . (1) and (2)]$

Page No 5.10:

Question 10:

Find the values of x, if

(i) $|\begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix}| = |\begin{matrix} 2 x & 4 \\ 6 & x \end{matrix}|$

(ii) $|\begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix}| = |\begin{matrix} x & 3 \\ 2 x & 5 \end{matrix}|$

(iii) $|\begin{matrix} 3 & x \\ x & 1 \end{matrix}| = |\begin{matrix} 3 & 2 \\ 4 & 1 \end{matrix}|$

(iv) If $|\begin{matrix} 3 x & 7 \\ 2 & 4 \end{matrix}| = 10$ , find the value of x.

(v) $|\begin{matrix} x + 1 & x - 1 \\ x - 3 & x + 2 \end{matrix}| = |\begin{matrix} 4 & - 1 \\ 1 & 3 \end{matrix}|$

(vi) $|\begin{matrix} 2 x & 5 \\ 8 & x \end{matrix}| = |\begin{matrix} 6 & 5 \\ 8 & 3 \end{matrix}|$

Answer:

(i)
$Given : |\begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix}| = |\begin{matrix} 2 x & 4 \\ 6 & x \end{matrix}| \Rightarrow 2 - 20 = 2 x^{2} - 24 \Rightarrow - 18 = 2 x^{2} - 24 \Rightarrow 2 x^{2} = 6 \Rightarrow x^{2} = 3 \Rightarrow x = \pm \sqrt{3}$

(ii)
$Given : |\begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix}| = |\begin{matrix} x & 3 \\ 2 x & 5 \end{matrix}| \Rightarrow 10 - 12 = 5 x - 6 x \Rightarrow - 2 = - x \Rightarrow x = 2$

(iii)
$Given : |\begin{matrix} 3 & x \\ x & 1 \end{matrix}| = |\begin{matrix} 3 & 2 \\ 4 & 1 \end{matrix}| \Rightarrow 3 - x^{2} = 3 - 8 \Rightarrow - x^{2} = - 8 \Rightarrow x^{2} = 8 \Rightarrow x = \pm 2 \sqrt{2}$

(iv)
$Given : |\begin{matrix} 3 x & 7 \\ 2 & 4 \end{matrix}| = 10 \Rightarrow 12 x - 14 = 10 \Rightarrow 12 x = 24 \Rightarrow x = 2$

(v)
$Given : |\begin{matrix} x + 1 & x - 1 \\ x - 3 & x + 2 \end{matrix}| = |\begin{matrix} 4 & - 1 \\ 1 & 3 \end{matrix}| \Rightarrow (x + 1) (x + 2) - (x - 3) (x - 1) = 12 + 1 \Rightarrow x^{2} + 3 x + 2 - x^{2} + 4 x - 3 = 13 \Rightarrow 7 x - 1 = 13 \Rightarrow 7 x = 14 \Rightarrow x = 2$

(vi)
$Given : |\begin{matrix} 2 x & 5 \\ 8 & x \end{matrix}| = |\begin{matrix} 6 & 5 \\ 8 & 3 \end{matrix}| \Rightarrow 2 x^{2} - 40 = 18 - 40 \Rightarrow 2 x^{2} = 18 \Rightarrow x^{2} = 9 \Rightarrow x = \pm 3$

Page No 5.11:

Question 11:

Find the integral value of x, if $|\begin{matrix} x^{2} & x & 1 \\ 0 & 2 & 1 \\ 3 & 1 & 4 \end{matrix}| = 28 .$

Answer:

$Given : |\begin{matrix} x^{2} & x & 1 \\ 0 & 2 & 1 \\ 3 & 1 & 4 \end{matrix}| = 28 \Rightarrow x^{2} (8 - 1) - x (0 - 3) + 1 (0 - 6) \Rightarrow 8 x^{2} - x^{2} + 3 x - 6 = 28 \Rightarrow 7 x^{2} + 3 x - 6 = 28 \Rightarrow 7 x^{2} + 3 x - 34 = 0 \Rightarrow (7 x + 17) (x - 2) = 0 \Rightarrow x = 2$

Integral value of x is 2. Thus, $x = \frac{- 17}{7}$ is not an integer.

Page No 5.11:

Question 12:

For what value of x the matrix A is singular?

$(i) A = [\begin{matrix} 1 + x & 7 \\ 3 - x & 8 \end{matrix}]$

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

Answer:

(i) Matrix A will be singular if

$|A| = 0$
$|A| = |\begin{matrix} 1 + x & 7 \\ 3 - x & 8 \end{matrix}| = 0 \Rightarrow 8 + 8 x - 21 + 7 x = 0 \Rightarrow 15 x - 13 = 0 \Rightarrow 15 x = 13 \Rightarrow x = \frac{13}{15}$

(ii) Matrix A will be singular if

$|A| = 0$
$\Rightarrow (x - 1) [{(x - 1)}^{2} - 1] - 1 (x - 1 - 1) + 1 [1 - (x - 1)] = 0 \Rightarrow (x - 1) (x^{2} - 2 x) - 1 (x - 2) + 1 (2 - x) = 0 \Rightarrow x^{3} - 2 x^{2} - x^{2} + 2 x - x + 2 - x + 2 = 0 \Rightarrow x^{3} - 3 x^{2} + 4 = 0 \Rightarrow {(x - 2)}^{2} (x + 1) = 0 \Rightarrow x = 2 or x = - 1$

Page No 5.57:

Question 1:

Evaluate the following determinant:
(i) $|\begin{matrix} 1 & 3 & 5 \\ 2 & 6 & 10 \\ 31 & 11 & 38 \end{matrix}|$

(ii) $|\begin{matrix} 67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{matrix}|$

(iii) $|\begin{matrix} a & h & g \\ h & b & f \\ g & f & c \end{matrix}|$

(iv) $|\begin{matrix} 1 & - 3 & 2 \\ 4 & - 1 & 2 \\ 3 & 5 & 2 \end{matrix}|$

(v) $|\begin{matrix} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{matrix}|$

(vi) $|\begin{matrix} 6 & - 3 & 2 \\ 2 & - 1 & 2 \\ - 10 & 5 & 2 \end{matrix}|$

(vii) $|\begin{matrix} 1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9 \end{matrix}|$

(viii)

Answer:

$(i) ∆ = |\begin{matrix} 1 & 3 & 5 \\ 2 & 6 & 10 \\ 31 & 11 & 38 \end{matrix}| = 1 |\begin{matrix} 6 & 10 \\ 11 & 38 \end{matrix}| - 3 |\begin{matrix} 2 & 10 \\ 31 & 38 \end{matrix}| + 5 |\begin{matrix} 2 & 6 \\ 31 & 11 \end{matrix}| = 1 (228 - 110) - 3 (76 - 310) + 5 (22 - 186) = 1 (118) - 3 (- 234) + 5 (- 164) = 118 + 702 - 820 = 0 (ii) ∆ = |\begin{matrix} 67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{matrix}| = 67 (338 - 336) - 19 (1014 - 1134) + 21 (936 - 1053) = 67 (2) - 19 (- 120) + 21 (- 117) = 134 + 2280 - 2457 = - 43 (iii) ∆ = |\begin{matrix} a & h & g \\ h & b & f \\ g & f & c \end{matrix}| = a |\begin{matrix} b & f \\ f & c \end{matrix}| - h |\begin{matrix} h & f \\ g & c \end{matrix}| + g |\begin{matrix} h & b \\ g & f \end{matrix}| = a (b c - f^{2}) - h (h c - f g) + g (h f - g b) = a b c - a f^{2} - h^{2} c + f g h + f g h - g^{2} b = a b c + 2 f g h - a f^{2} - c h^{2} - b g^{2} (iv) ∆ = |\begin{matrix} 1 & - 3 & 2 \\ 4 & - 1 & 2 \\ 3 & 5 & 2 \end{matrix}| = 1 |\begin{matrix} - 1 & 2 \\ 5 & 2 \end{matrix}| - (- 3) |\begin{matrix} 4 & 2 \\ 3 & 2 \end{matrix}| + 2 |\begin{matrix} 4 & - 1 \\ 3 & 5 \end{matrix}| = 1 (- 2 - 10) + 3 (8 - 6) + 2 (20 + 3) = (- 12) + 6 + 46 = 40 (v) ∆ = |\begin{matrix} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{matrix}| = 1 |\begin{matrix} 9 & 16 \\ 16 & 25 \end{matrix}| - 4 |\begin{matrix} 4 & 16 \\ 9 & 25 \end{matrix}| + 9 |\begin{matrix} 4 & 9 \\ 9 & 16 \end{matrix}| = 1 (225 - 256) - 4 (100 - 144) + 9 (64 - 81) = 1 (- 31) - 4 (- 44) + 9 (- 17) = - 31 + 176 - 153 = - 8 (vi) ∆ = |\begin{matrix} 6 & - 3 & 2 \\ 2 & - 1 & 2 \\ - 10 & 5 & 2 \end{matrix}| = 6 (- 2 - 10) - (- 3) (4 + 20) + 2 (10 - 10) = - 72 + 72 + 0 = - 72 + 72 = 0 (vii) ∆ = |\begin{matrix} 1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9 \end{matrix}| = 1 |\begin{matrix} 9 & 27 & 1 \\ 27 & 1 & 3 \\ 1 & 3 & 9 \end{matrix}| - 3 |\begin{matrix} 3 & 27 & 1 \\ 9 & 1 & 3 \\ 27 & 3 & 9 \end{matrix}| + 9 |\begin{matrix} 3 & 9 & 1 \\ 9 & 27 & 3 \\ 27 & 1 & 9 \end{matrix}| - 27 |\begin{matrix} 3 & 9 & 27 \\ 9 & 27 & 1 \\ 27 & 1 & 3 \end{matrix}| = 1 \{9 (9 - 9) - 27 (243 - 3) + 1 (81 - 1)\} - 3 \{3 (9 - 9) - 27 (81 - 81) + 1 (27 - 27)\} + 9 \{3 (243 - 3) - 9 (81 - 81) + 1 (9 - 729)\} - 27 \{(81 - 1) - 9 (27 - 27) + 27 (9 - 729)\} = 1 \{0 - 6480 + 80\} - 3 \{0 - 0 + 0\} + 9 \{720 - 0 - 720\} - 27 \{80 - 0 - 19440\} = - 6400 + 522720 = 516320$

(viii)

Page No 5.57:

Question 2:

Without expanding, show that the values of each of the following determinants are zero:
(i) $|\begin{matrix} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \end{matrix}|$

(ii) $|\begin{matrix} 6 & - 3 & 2 \\ 2 & - 1 & 2 \\ - 10 & 5 & 2 \end{matrix}|$

(iii) $|\begin{matrix} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{matrix}|$

(iv) $|\begin{matrix} 1 / a & a^{2} & b c \\ 1 / b & b^{2} & a c \\ 1 / c & c^{2} & a b \end{matrix}|$

(v) $|\begin{matrix} a + b & 2 a + b & 3 a + b \\ 2 a + b & 3 a + b & 4 a + b \\ 4 a + b & 5 a + b & 6 a + b \end{matrix}|$

(vi) $|\begin{matrix} 1 & a & a^{2} - b c \\ 1 & b & b^{2} - a c \\ 1 & c & c^{2} - a b \end{matrix}|$

(vii) $|\begin{matrix} 49 & 1 & 6 \\ 39 & 7 & 4 \\ 26 & 2 & 3 \end{matrix}|$

(viii) $|\begin{matrix} 0 & x & y \\ - x & 0 & z \\ - y & - z & 0 \end{matrix}|$

(ix) $|\begin{matrix} 1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2 \end{matrix}|$

(x) $|\begin{matrix} 1^{2} & 2^{2} & 3^{2} & 4^{2} \\ 2^{2} & 3^{2} & 4^{2} & 5^{2} \\ 3^{2} & 4^{2} & 5^{2} & 6^{2} \\ 4^{2} & 5^{2} & 6^{2} & 7^{2} \end{matrix}|$

(xi) $|\begin{matrix} a & b & c \\ a + 2 x & b + 2 y & c + 2 z \\ x & y & z \end{matrix}|$

(xii) $|\begin{matrix} {(2^{x} + 2^{- x})}^{2} & {(2^{x} - 2^{- x})}^{2} & 1 \\ {(3^{x} + 3^{- x})}^{2} & {(3^{x} - 3^{- x})}^{2} & 1 \\ {(4^{x} + 4^{- x})}^{2} & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix}|$

(xiii) $|\begin{matrix} \sin α & \cos α & \cos (α + δ) \\ \sin β & \cos β & \cos (β + δ) \\ \sin γ & \cos γ & \cos (γ + δ) \end{matrix}|$

(xiv) $|\begin{matrix} \sin^{2} 23 ° & \sin^{2} 67 ° & \cos 180 ° \\ - \sin^{2} 67 ° & - \sin^{2} 23 ° & \cos^{2} 180 ° \\ \cos 180 ° & \sin^{2} 23 ° & \sin^{2} 67 ° \end{matrix}|$

(xv) $|\begin{matrix} \cos (x + y) & - \sin (x + y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ - \cos x & \sin x & - \cos y \end{matrix}|$

(xvi) $|\begin{matrix} \sqrt{23} + \sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15} + \sqrt{46} & 5 & \sqrt{10} \\ 3 + \sqrt{115} & \sqrt{15} & 5 \end{matrix}|$

(xvii) $|\begin{matrix} \sin^{2} A & \cot A & 1 \\ \sin^{2} B & \cot B & 1 \\ \sin^{2} C & \cot C & 1 \end{matrix}|, where A, B, C are the angles of ∆ A B C .$

Answer:

$(i) ∆ = |\begin{matrix} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \end{matrix}| = |\begin{matrix} 0 & 2 & 7 \\ 0 & 3 & 5 \\ 0 & 4 & 3 \end{matrix}| [Applying C_{1} \to C_{1} - 4 C_{2}] \Rightarrow ∆ = 0 (i i) ∆ = |\begin{matrix} 6 & - 3 & 2 \\ 2 & - 1 & 2 \\ - 10 & 5 & 2 \end{matrix}| = |\begin{matrix} 0 & - 3 & 2 \\ 0 & - 1 & 2 \\ 0 & 5 & 2 \end{matrix}| [Applying C_{1} \to C_{1} + 2 C_{2}] \Rightarrow ∆ = 0$

$(i i i) ∆ = |\begin{matrix} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{matrix}| = |\begin{matrix} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 13 & 17 & 5 \end{matrix}| [Applying R_{3} \to R_{3} - R_{1}] \Rightarrow ∆ = 0 (i v) ∆ = |\begin{matrix} \frac{1}{a} & a^{2} & b c \\ \frac{1}{b} & b^{2} & a c \\ \frac{1}{c} & c^{2} & a b \end{matrix}| = |\begin{matrix} 1 & a^{3} & a b c \\ 1 & b^{3} & a b c \\ 1 & c^{3} & a b c \end{matrix}| [Applying R_{1} \to a R_{1}, R_{2} \to b R_{2} and R_{3} \to c R_{3}] = a b c |\begin{matrix} 1 & a^{3} & 1 \\ 1 & b^{3} & 1 \\ 1 & c^{3} & 1 \end{matrix}| \Rightarrow ∆ = 0 (v) ∆ = |\begin{matrix} a + b & 2 a + b & 3 a + b \\ 2 a + b & 3 a + b & 4 a + b \\ 4 a + b & 5 a + b & 6 a + b \end{matrix}| = |\begin{matrix} a & a & a \\ 2 a & 2 a & 2 a \\ 4 a + b & 5 a + b & 6 a + b \end{matrix}| [Applying R_{1} \to R_{2} - R_{1} and R_{2} \to R_{3} - R_{2}] = 2 |\begin{matrix} a & a & a \\ a & a & a \\ 4 a + b & 5 a + b & 6 a + b \end{matrix}| = 0 (v i) ∆ = |\begin{matrix} 1 & a & a^{2} - b c \\ 1 & b & b^{2} - a c \\ 1 & c & c^{2} - a b \end{matrix}| = |\begin{matrix} 0 & a - b & a^{2} - b c - b^{2} + a c \\ 0 & b - c & b^{2} - a c - c^{2} + a b \\ 1 & c & c^{2} - a b \end{matrix}| [Applying R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}] = |\begin{matrix} 0 & a - b & (a - b) (a + b) + c (a - b) \\ 0 & b - c & (b - c) (b + c) + a (b - c) \\ 1 & c & c^{2} - a b \end{matrix}| = (a - b) (b - c) |\begin{matrix} 0 & 1 & (a + b + c) \\ 0 & 1 & (a + b + c) \\ 1 & c & c^{2} - a b \end{matrix}| \Rightarrow ∆ = 0 (v i i) ∆ = |\begin{matrix} 49 & 1 & 6 \\ 39 & 7 & 4 \\ 26 & 2 & 3 \end{matrix}| = |\begin{matrix} 1 & 1 & 6 \\ 7 & 7 & 4 \\ 2 & 2 & 3 \end{matrix}| [Applying C_{1} \to C_{1} - 8 C_{3}] \Rightarrow ∆ = 0 (v i i i) ∆ = |\begin{matrix} 0 & x & y \\ - x & 0 & z \\ - y & - z & 0 \end{matrix}| = \frac{x y z}{x y z} |\begin{matrix} 0 & x & y \\ - x & 0 & z \\ - y & - z & 0 \end{matrix}| = \frac{1}{x y z} |\begin{matrix} 0 & x z & y z \\ - x y & 0 & z y \\ - y x & - z x & 0 \end{matrix}| = \frac{1}{x y z} |\begin{matrix} - 2 x y & 0 & 2 y z \\ - x y & 0 & z y \\ - y x & - z x & 0 \end{matrix}| [Applying R_{1} \to R_{1} + R_{2} + R_{3}] = \frac{1}{x y z} |\begin{matrix} 0 & 0 & 0 \\ - x y & 0 & z y \\ - y x & - z x & 0 \end{matrix}| = 0 [Applying R_{1} \to R_{1} - 2 R_{2}] (i x) ∆ = |\begin{matrix} 1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2 \end{matrix}| = |\begin{matrix} 1 & 1 & 6 \\ 7 & 7 & 4 \\ 3 & 3 & 2 \end{matrix}| = 0 [Applying C_{2} \to C_{2} - 7 C_{3}]$

$x) ∆ = |\begin{matrix} 1^{2} & 2^{2} & 3^{2} & 4^{2} \\ 2^{2} & 3^{2} & 4^{2} & 5^{2} \\ 3^{2} & 4^{2} & 5^{2} & 6^{2} \\ 4^{2} & 5^{2} & 6^{2} & 7^{2} \end{matrix}| = |\begin{matrix} 1 & 4 & 9 & 16 \\ 4 & 9 & 16 & 25 \\ 9 & 16 & 25 & 36 \\ 16 & 25 & 36 & 49 \end{matrix}| = |\begin{matrix} 1 & 4 & 9 & 16 \\ 4 & 9 & 16 & 25 \\ 5 & 7 & 9 & 11 \\ 7 & 9 & 11 & 13 \end{matrix}| [Applying R_{3} \to R_{3} - R_{2} and R_{4} \to R_{4} - R_{3}] = |\begin{matrix} 1 & 4 & 9 & 16 \\ 4 & 9 & 16 & 25 \\ 7 & 9 & 11 & 13 \\ 7 & 9 & 11 & 13 \end{matrix}| = 0 [Applying R_{3} \to 2 + R_{3}] xi) ∆ = |\begin{matrix} a & b & c \\ a + 2 x & b + 2 y & c + 2 z \\ x & y & z \end{matrix}| = |\begin{matrix} a + 2 x & b + 2 y & c + 2 z \\ a + 2 x & b + 2 y & c + 2 z \\ x & y & z \end{matrix}| [Applying R_{1} \to R_{1} + 2 R_{3}] = |\begin{matrix} 0 & 0 & 0 \\ a + 2 x & b + 2 y & c + 2 z \\ x & y & z \end{matrix}| = 0 [Applying R_{1} \to R_{1} - R_{2}]$

(xii)
$|\begin{matrix} {(2^{x} + 2^{- x})}^{2} & {(2^{x} - 2^{- x})}^{2} & 1 \\ {(3^{x} + 3^{- x})}^{2} & {(3^{x} - 3^{- x})}^{2} & 1 \\ {(4^{x} + 4^{- x})}^{2} & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix}| = |\begin{matrix} (2^{2 x} + 2^{- 2 x} + 2) & (2^{2 x} + 2^{- 2 x} - 2) & 1 \\ (3^{2 x} + 3^{- 2 x} + 2) & (3^{2 x} + 3^{- 2 x} - 2) & 1 \\ (4^{2 x} + 4^{- 2 x} + 2) & (4^{2 x} + 4^{- 2 x} - 2) & 1 \end{matrix}| = |\begin{matrix} 4 & (2^{2 x} + 2^{- 2 x} - 2) & 1 \\ 4 & (3^{2 x} + 3^{- 2 x} - 2) & 1 \\ 4 & (4^{2 x} + 4^{- 2 x} - 2) & 1 \end{matrix}| [Applying C_{1} \to C_{1} - C_{2}] = 4 |\begin{matrix} 1 & (2^{2 x} + 2^{- 2 x} - 2) & 1 \\ 1 & (3^{2 x} + 3^{- 2 x} - 2) & 1 \\ 1 & (4^{2 x} + 4^{- 2 x} - 2) & 1 \end{matrix}| = 0$

(xiii)
$|\begin{matrix} \sin α & \cos α & \cos (α + δ) \\ \sin β & \cos β & \cos (β + δ) \\ \sin γ & \cos γ & \cos (γ + δ) \end{matrix}| = |\begin{matrix} \sin α \sin δ & \cos α \cos δ & \cos (α + δ) \\ \sin β \sin δ & \cos β \cos δ & \cos (β + δ) \\ \sin γ \sin δ & \cos γ \cos δ & \cos (γ + δ) \end{matrix}| [Applying C_{1} \to \sin δ C_{1} and C_{2} \to \cos δ C_{2}] = |\begin{matrix} \sin α \sin δ & \cos (α + δ) & \cos (α + δ) \\ \sin β \sin δ & \cos (β + δ) & \cos (β + δ) \\ \sin γ \sin δ & \cos (γ + δ) & \cos (γ + δ) \end{matrix}| [Applying C_{2} \to C_{2} - C_{1}] = 0$

(xiv)
$|\begin{matrix} \sin^{2} 23 ° & \sin^{2} 67 ° & \cos 180 ° \\ - \sin^{2} 67 ° & - \sin^{2} 23 ° & \cos^{2} 180 ° \\ \cos 180 ° & \sin^{2} 23 ° & \sin^{2} 67 ° \end{matrix}| = |\begin{matrix} \sin^{2} 23 ° & \sin^{2} (90 - 23) ° & - 1 \\ - \sin^{2} (90 - 23) ° & - \sin^{2} 23 ° & 1 \\ - 1 & \sin^{2} 23 ° & \sin^{2} (90 - 23) ° \end{matrix}| = |\begin{matrix} \sin^{2} 23 ° & \cos^{2} 23 ° & - 1 \\ - \cos^{2} 23 ° & - \sin^{2} 23 ° & 1 \\ - 1 & \sin^{2} 23 ° & \cos^{2} 23 ° \end{matrix}| = |\begin{matrix} \sin^{2} 23 ° + \cos^{2} 23 ° & \cos^{2} 23 ° & - 1 \\ - \cos^{2} 23 ° - \sin^{2} 23 ° & - \sin^{2} 23 ° & 1 \\ - 1 + \sin^{2} 23 ° & \sin^{2} 23 ° & \cos^{2} 23 ° \end{matrix}| [A p p l y i n g C_{1} \to C_{1} + C_{2}] = |\begin{matrix} 1 & 1 & - 1 \\ - 1 & - \sin^{2} 23 ° & 1 \\ - \cos^{2} 23 ° & \sin^{2} 23 ° & \cos^{2} 23 ° \end{matrix}| = (- 1) |\begin{matrix} - 1 & 1 & - 1 \\ 1 & - \sin^{2} 23 ° & 1 \\ \cos^{2} 23 ° & \sin^{2} 23 ° & \cos^{2} 23 ° \end{matrix}| = 0$

(xv)
$|\begin{matrix} \cos (x + y) & - \sin (x + y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ - \cos x & \sin x & - \cos y \end{matrix}| = \frac{1}{\sin y \cos y} |\begin{matrix} \cos (x + y) & - \sin (x + y) & \cos 2 y \\ \sin x \sin y & \cos x \sin y & \sin^{2} y \\ - \cos x \cos y & \sin x \cos y & - \cos^{2} y \end{matrix}| [Applying R_{2} \to \sin y R_{2} and R_{3} \to \cos y R_{3}] = \frac{1}{\sin y \cos y} |\begin{matrix} \cos (x + y) & - \sin (x + y) & \cos 2 y \\ \sin x \sin y - \cos x \cos y & \cos x \sin y + \sin x \cos y & \sin^{2} y - \cos^{2} y \\ - \cos x \cos y & \sin x \cos y & - \cos^{2} y \end{matrix}| [Applying R_{2} \to R_{2} + R_{3}] = \frac{- 1}{\sin y \cos y} |\begin{matrix} \cos (x + y) & - \sin (x + y) & \cos 2 y \\ \cos (x + y) & - \sin (x + y) & \cos 2 y \\ - \cos x \cos y & \sin x \cos y & - \cos^{2} y \end{matrix}| = 0$

(xvi)
$|\begin{matrix} \sqrt{23} + \sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15} + \sqrt{46} & 5 & \sqrt{10} \\ 3 + \sqrt{115} & \sqrt{15} & 5 \end{matrix}| = |\begin{matrix} \sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15} & 5 & \sqrt{10} \\ 3 & \sqrt{15} & 5 \end{matrix}| + |\begin{matrix} \sqrt{23} & \sqrt{5} & \sqrt{5} \\ \sqrt{46} & 5 & \sqrt{10} \\ \sqrt{115} & \sqrt{15} & 5 \end{matrix}| = \sqrt{3} |\begin{matrix} 1 & \sqrt{5} & \sqrt{5} \\ \sqrt{5} & 5 & \sqrt{10} \\ \sqrt{3} & \sqrt{15} & 5 \end{matrix}| + \sqrt{23} |\begin{matrix} 1 & \sqrt{5} & \sqrt{5} \\ \sqrt{2} & 5 & \sqrt{10} \\ \sqrt{5} & \sqrt{15} & 5 \end{matrix}| = \sqrt{3} \times \sqrt{5} |\begin{matrix} 1 & 1 & \sqrt{5} \\ \sqrt{5} & \sqrt{5} & \sqrt{10} \\ \sqrt{3} & \sqrt{3} & 5 \end{matrix}| + \sqrt{23} \times \sqrt{5} |\begin{matrix} 1 & \sqrt{5} & 1 \\ \sqrt{2} & 5 & \sqrt{2} \\ \sqrt{5} & \sqrt{15} & \sqrt{5} \end{matrix}| = 0 + 0 = 0$

(xvii)
$|\begin{matrix} \sin^{2} A & \cot A & 1 \\ \sin^{2} B & \cot B & 1 \\ \sin^{2} C & \cot C & 1 \end{matrix}| = |\begin{matrix} \sin^{2} A - \sin^{2} B & \cot A - \cot B & 0 \\ \sin^{2} B & \cot B & 1 \\ \sin^{2} C - \sin^{2} B & \cot C - \cot B & 0 \end{matrix}| [Applying R_{1} \to R_{1} - R_{2} and R_{3} \to R_{3} - R_{2}] = |\begin{matrix} \sin (A + B) \sin (A - B) & \frac{\cos A \sin B - \cos B \sin A}{\sin A \sin B} & 0 \\ \sin^{2} B & \cot B & 1 \\ \sin (C + B) \sin (C - B) & \frac{\cos C \sin B - \cos B \sin C}{\sin B \sin C} & 0 \end{matrix}| = |\begin{matrix} \sin (π - C) \sin (A - B) & \frac{- \sin (A - B)}{\sin A \sin B} & 0 \\ \sin^{2} B & c o t B & 1 \\ \sin (π - A) \sin (C - B) & \frac{- \sin (C - B)}{\sin B \sin C} & 0 \end{matrix}| [∵ A + B + C = π] = |\begin{matrix} \sin C \sin (A - B) & \frac{- \sin (A - B)}{\sin A \sin B} & 0 \\ \sin^{2} B & \frac{\cos B}{\sin B} & 1 \\ \sin A \sin (C - B) & \frac{- \sin (C - B)}{\sin B \sin C} & 0 \end{matrix}| = \frac{\sin (A - B) \sin (C - B)}{\sin B} |\begin{matrix} \sin C & \frac{- 1}{\sin A} & 0 \\ \sin^{2} B & \cos B & 1 \\ \sin A & \frac{- 1}{\sin C} & 0 \end{matrix}| = \frac{\sin (A - B) \sin (C - B)}{\sin B \sin A \sin C} |\begin{matrix} \sin C \sin A & - 1 & 0 \\ \sin^{2} B & \cos B & 1 \\ \sin A \sin C & - 1 & 0 \end{matrix}| [Applying R_{1} \to \sin A R_{1} and R_{3} \to \sin C R_{3}] = \frac{\sin (A - B) \sin (C - B)}{\sin B \sin A \sin C} |\begin{matrix} 0 & 0 & 0 \\ \sin^{2} B & \cos B & 1 \\ \sin A \sin C & - 1 & 0 \end{matrix}| [Applying R_{1} \to R_{1} - R_{3}] = 0$

Page No 5.58:

Question 3:

Evaluate :

$|\begin{matrix} a & b + c & a^{2} \\ b & c + a & b^{2} \\ c & a + b & c^{2} \end{matrix}|$

Answer:

$∆ = |\begin{matrix} a & b + c & a^{2} \\ b & c + a & b^{2} \\ c & a + b & c^{2} \end{matrix}| When a = b, the first two rows become identical . Hence, a - b is a factor . Similarly, when b = c the second and third rows become identical . So, b - c is also a factor . Also, when c = a, the third and first rows become identical . Hence, c - a is also a factor . The product of diagonal elements, a (c + a) c^{2} is 4 . So, the other factor should be a linear in a, b and c . It should also remain unaltered when any two letters are changed . Let this factor be λ (a + b + c) . Here, λ is a constant . To find this, we have a = 0, b = 1, c = 2 |\begin{matrix} 0 & 3 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & 4 \end{matrix}| = λ (a - b) (b - c) (c - a) (a + b + c) |\begin{matrix} 0 & 3 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & 4 \end{matrix}| = λ (0 - 1) (1 - 2) (2 - 1) (0 + 1 + 2) \Rightarrow - 6 = 6 λ \Rightarrow λ = - 1 Thus, |\begin{matrix} a & b + c & a^{2} \\ b & c + a & b^{2} \\ c & a + b & c^{2} \end{matrix}| = - ((a + b + c)) (a - b) (b - c) (c - a)$

Page No 5.58:

Question 4:

Evaluate :

$|\begin{matrix} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{matrix}|$

Answer:

$∆ = |\begin{matrix} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{matrix}| When a = b, the first two rows become identical . Hence, a - b i s a f a c t o r . Similarly, when b = c and c = a, the second and third and third and first rows become identical . Hence, b - c and c - a are also factors . The degree of product of the diagonal elements is 3 . Hence, there are no other factors . |\begin{matrix} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{matrix}| = λ (a - b) (b - c) (c - a) [Where λ is a constant] |\begin{matrix} 1 & 0 & 2 \\ 1 & 1 & 0 \\ 1 & 2 & 0 \end{matrix}| = 2 λ [Putting a = 0, b = 1 and c = 2 to find λ] \Rightarrow 2 = 2 λ \Rightarrow λ = 1 H e n c e, |\begin{matrix} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{matrix}| = (a - b) (b - c) (c - a)$

Page No 5.58:

Question 5:

Evaluate :

$|\begin{matrix} x + λ & x & x \\ x & x + λ & x \\ x & x & x + λ \end{matrix}|$

Answer:

$∆ = |\begin{matrix} x + λ & x & x \\ x & x + λ & x \\ x & x & x + λ \end{matrix}| = |\begin{matrix} λ & 0 & x \\ - λ & λ & x \\ 0 & - λ & x + λ \end{matrix}| [Applying C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3}] = |\begin{matrix} λ & 0 & x \\ - λ & 0 & 2 x + λ \\ 0 & - λ & x + λ \end{matrix}| [Applying R_{1} \to R_{2} + R_{3}] = λ |\begin{matrix} 0 & 2 x + λ \\ - λ & x + λ \end{matrix}| + x |\begin{matrix} - λ & 0 \\ 0 & - λ \end{matrix}| = λ [λ (2 x + λ)] + x λ^{2} = λ^{2} (2 x + λ + λ^{2} x) = 3 λ^{2} x + λ^{3} = λ^{2} (3 x + λ)$

Page No 5.58:

Question 6:

Evaluate :

$|\begin{matrix} a & b & c \\ c & a & b \\ b & c & a \end{matrix}|$

Answer:

$∆ = |\begin{matrix} a & b & c \\ c & a & b \\ b & c & a \end{matrix}| = a (a^{2} - b c) - b (c a - b^{2}) + c (c^{2} - b a) = a^{3} - a b c - b c a + b^{3} + c^{3} - a b c = a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$

Page No 5.58:

Question 7:

Evaluate the following:

$|\begin{matrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{matrix}|$

Answer:

Let $∆ = |\begin{matrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{matrix}|$ .
$∆ = |\begin{matrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{matrix}| = |\begin{matrix} x - 1 & 1 - x & 0 \\ 1 & x & 1 \\ 0 & 1 - x & x - 1 \end{matrix}| [Applying R_{1} \to R_{1} - R_{2} and R_{3} \to R_{3} - R_{2}] = {(x - 1)}^{2} |\begin{matrix} 1 & - 1 & 0 \\ 1 & x & 1 \\ 0 & - 1 & 1 \end{matrix}| = {(x - 1)}^{2} |\begin{matrix} 1 & - 1 & 0 \\ 1 & x + 1 & 1 \\ 0 & 0 & 1 \end{matrix}| [Applying C_{2} \to C_{2} + C_{3}] = {(x - 1)}^{2} (x + 1 + 1) [Expanding along last row] = {(x - 1)}^{2} (x + 2) ∴ ∆ = {(x - 1)}^{2} (x + 2)$

Page No 5.58:

Question 8:

Evaluate the following:

$|\begin{matrix} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{matrix}|$

Answer:

Let $∆ = |\begin{matrix} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{matrix}|$ .
$∆ = |\begin{matrix} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{matrix}| = x^{2} y^{2} z^{2} |\begin{matrix} 0 & x & x \\ y & 0 & y \\ z & z & 0 \end{matrix}| [Taking x^{2} common from C_{1}, y^{2} common from C_{2} and z^{2} common from C_{3}] = x^{3} y^{3} z^{3} |\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}| [Taking x common from R_{1}, y common from R_{2} and z common from R_{3}] = x^{3} y^{3} z^{3} |\begin{matrix} 0 & 0 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & 0 \end{matrix}| [Applying C_{2} \to C_{2} - C_{3}] = x^{3} y^{3} z^{3} (1 + 1) [Expanding along first row] = 2 x^{3} y^{3} z^{3} ∴ ∆ = 2 x^{3} y^{3} z^{3}$

Page No 5.58:

Question 9:

Evaluate the following:

$|\begin{matrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{matrix}|$

Answer:

Let $∆ = |\begin{matrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{matrix}|$ .
$∆ = |\begin{matrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{matrix}| = |\begin{matrix} a + x + y + z & y & z \\ a + x + y + z & a + y & z \\ a + x + y + z & y & a + z \end{matrix}| [Applying C_{1} \to C_{1} + C_{2} + C_{3}] = (a + x + y + z) |\begin{matrix} 1 & y & z \\ 1 & a + y & z \\ 1 & y & a + z \end{matrix}| [Taking (a + x + y + z) common from C_{1}] = (a + x + y + z) |\begin{matrix} 1 & y & z \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix}| [Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}] = (a + x + y + z) a^{2} [Expanding along first column] ∴ ∆ = (a + x + y + z) a^{2}$

Page No 5.58:

Question 10:

$If ∆ = |\begin{matrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{matrix}|, ∆_{1} = |\begin{matrix} 1 & 1 & 1 \\ y z & z x & x y \\ x & y & z \end{matrix}|, then prove that ∆ + ∆_{1} = 0 .$

Answer:

$∆ + ∆_{1} = |\begin{matrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{matrix}| + |\begin{matrix} 1 & 1 & 1 \\ y z & z x & x y \\ x & y & z \end{matrix}| = |\begin{matrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{matrix}| + |\begin{matrix} 1 & y z & x \\ 1 & z x & y \\ 1 & x y & z \end{matrix}| [Interchanging rows and coloumns in ∆_{1}] = |\begin{matrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{matrix}| - |\begin{matrix} 1 & x & y z \\ 1 & y & z x \\ 1 & z & x y \end{matrix}| [Applying C_{2} \leftrightarrow C_{3} in ∆_{1}] = |\begin{matrix} 1 & x & x^{2} \\ 0 & y - x & y^{2} - x^{2} \\ 0 & z - x & z^{2} - x^{2} \end{matrix}| - |\begin{matrix} 1 & x & y z \\ 0 & y - x & z x - y z \\ 0 & z - x & x y - y z \end{matrix}| [Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}] = (y - x) (z - x) |\begin{matrix} 1 & x & x^{2} \\ 0 & 1 & y + x \\ 0 & 1 & z + x \end{matrix}| - (y - x) (z - x) |\begin{matrix} 1 & x & y z \\ 0 & 1 & - z \\ 0 & 1 & - y \end{matrix}| [Taking (y - x) common from R_{2} and (z - x) common from R_{3}] = (y - x) (z - x) (z + x - y - x) - (y - x) (z - x) (- y + z) [Expanding along first column] = (y - x) (z - x) (z - y) (1 - 1) = 0 ∴ ∆ + ∆_{1} = 0 .$

Page No 5.58:

Question 11:

Prove that :

$|\begin{matrix} a & b & c \\ a - b & b - c & c - a \\ b + c & c + a & a + b \end{matrix}| = a^{3} + b^{3} + c^{3} - 3 a b c$

Answer:

$∆ = |\begin{matrix} a & b & c \\ a - b & b - c & c - a \\ b + c & c + a & a + b \end{matrix}| = |\begin{matrix} a & b & c \\ a - b & b - c & c - a \\ a + b + c & c + a + b & a + b + c \end{matrix}| [Applying R_{3} \to R_{3} + R_{2}] = (a + b + c) |\begin{matrix} a & b & c \\ a - b & b - c & c - a \\ 1 & 1 & 1 \end{matrix}| [Taking (a + b + c) common] = (a + b + c) |\begin{matrix} a & b & c \\ b & c & a \\ 1 & 1 & 1 \end{matrix}| [Applying R_{2} \to R_{1} - R_{2}] = (a + b + c) |\begin{matrix} a - b & b - c & c \\ b - c & c - a & a \\ 0 & 0 & 1 \end{matrix}| [C_{1} \to C_{1} - C_{2} and C_{2} \to C_{2} - C_{3}] = (a + b + c) [- 1 \{(a - b) (c - a) - {(b - c)}^{2}\}] = (a + b + c) [- \{a c - b c - a^{2} + a b - b^{2} - c^{2} + 2 b c\}] = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) = a^{3} + b^{3} + c^{3} - 3 a b c$

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Question 12:

Prove that :

$|\begin{matrix} b + c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c \end{matrix}| = 3 a b c - a^{3} - b - c^{3}$

Answer:

$Let LHS = Δ = |b + c a - b a c + a b - c b a + b c - a c| Δ = (b + c) |b - c b c - a c| - (a - b) |c + a b a + b c| + a |c + a b - c a + b c - a| [Expanding] = (b + c) \{b c - c^{2} - b c + a b\} - (a - b) \{c^{2} + a c - a b - b^{2}\} + a \{c^{2} - a^{2} - a b + a c - b^{2} + b c\} = b c^{2} - c^{3} + a b c - a c^{2} - a^{2} c + a^{2} b + a b^{2} + b c^{2} + a b c - a b^{2} - b^{3} + a c^{2} - a^{3} - a^{2} c - a b^{2} + a b c \Rightarrow Δ = 3 a b c - a^{3} - b^{3} - c^{3} [Simplyfying] = RHS$

Page No 5.58:

Question 13:

Prove that :

$|\begin{matrix} a + b & b + c & c + a \\ b + c & c + a & a + b \\ c + a & a + b & b + c \end{matrix}| = 2 |\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}|$

Answer:

$Let Δ = |a + b b + c c + a b + c c + a a + b c + a a + b b + c| Using the property of determinants that if each element of a row or column is expressed as the sum of two or more quantities, the determinant is expressed as the sum of two or more determinants, we get Δ = |a b c b c a c a b| + |b c a c a b a b c| = |a b c b c a c a b| + (- 1) |a c b b a c c b a| [Applying C_{1} \leftrightarrow C_{3} in second determinant to get negative value of the deteminant] = |a b c b c a c a b| + (- 1) (- 1) |a b c b c a c a b| [Applying C_{2} \leftrightarrow C_{3}] = 2 |a b c b c a c a b| = RHS$

Page No 5.58:

Question 14:

Prove that :

$|\begin{matrix} a + b + 2 c & a & b \\ c & b + c + 2 a & b \\ c & a & c + a + 2 b \end{matrix}| = 2 {(a + b + c)}^{3}$

Answer:

$Let LHS = Δ = |a + b + 2 c a b c b + c + 2 a b c a c + a + 2 b| \Rightarrow ∆ = |2 a + 2 b + 2 c a b 2 a + 2 b + 2 c b + c + 2 a b 2 a + 2 b + 2 c a c + a + 2 b| [Applying C_{1} \to C_{1} + C_{2} + C_{3}] = 2 (a + b + c) |1 a b 1 b + c + 2 a b 1 a c + a + 2 b| [Taking out 2 (a + b + c) common from C_{1}] ∆ = 2 (a + b + c) |1 a b 0 b + c + a 0 0 - b - c - a c + a + b| [Applying R_{2} \to R_{2} - R_{1} and R_{2} \to R_{2} - R_{3}] = 2 (a + b + c) (a + b + c) (a + b + c) |1 a b 0 1 0 0 - 1 1| [Taking out (a + b + c) common from R_{2} and R_{3}] = 2 {(a + b + c)}^{3} \{1 (1 - 0)\} [Expanding along C_{1}] = 2 {(a + b + c)}^{3} = RHS$

Page No 5.59:

Question 15:

Prove that :

$|\begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix}| = {(a + b + c)}^{3}$

Answer:

$Let LHS = ∆ = |a - b - c 2 a 2 a 2 b b - c - a 2 b 2 c 2 c c - a - b| \Rightarrow ∆ = |a + b + c a + b + c a + b + c 2 b b - c - a 2 b 2 c 2 c c - a - b| [Applying R_{1} \to R_{1} + R_{2} + R_{3}] = (a + b + c) |1 1 1 2 b b - c - a 2 b 2 c 2 c c - a - b| = (a + b + c) |0 1 1 b + c + a b - c - a 2 b 0 2 c c - a - b| [Applying C_{1} \to C_{1} - C_{2}] = (a + b + c) \{(a + b + c) \times |1 1 2 c c - a - b|\} [Expanding along C_{1}] = {(a + b + c)}^{3} = RHS$

Page No 5.59:

Question 16:

Prove that :

$|\begin{matrix} 1 & b + c & b^{2} + c^{2} \\ 1 & c + a & c^{2} + a^{2} \\ 1 & a + b & a^{2} + b^{2} \end{matrix}| = (a - b) (b - c) (c - a)$

Answer:

$Let LHS = ∆ = |1 b + c b^{2} + c^{2} 1 c + a c^{2} + a^{2} 1 a + b a^{2} + b^{2}| \Rightarrow ∆ = |0 (b + c) - (c + a) (b^{2} + c^{2}) - (c^{2} + a^{2}) 0 (c + a) - (a + b) (c^{2} + a^{2}) - (a^{2} + b^{2}) 1 a + b a^{2} + b^{2}| [Applying R_{1} \to R_{1} - R_{2} and R_{2} \to R_{2} - R_{3}] = |0 b - a b^{2} - a^{2} 0 c - b c^{2} - b^{2} 1 a + b a^{2} + b^{2}| = {(- 1)}^{2} |0 a - b a^{2} - b^{2} 0 b - c b^{2} - c^{2} 1 a + b a^{2} + b^{2}| [Taking out (- 1) common from R_{1} and R_{2}] = (a - b) (b - c) |0 1 a + b 0 1 b + c 1 a + b a^{2} + b^{2}| = (a - b) (b - c) \{1 \times |1 a + b 1 b + c|\} [Expanding along C_{1}] = (a - b) (b - c) (c - a) = RHS$

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Question 17:

Prove that :

$|\begin{matrix} a & a + b & a + 2 b \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{matrix}| = 9 (a + b) b^{2}$

Answer:

$Let LHS = ∆ = |a a + b a + 2 b a + 2 b a a + b a + b a + 2 b a| Δ = |3 a + 3 b 3 a + 3 b 3 a + 3 b a + 2 b a a + b a + b a + 2 b a| [Applying R_{1} \to R_{1} + R_{2} + R_{3}] = 3 (a + b) |1 1 1 a + 2 b a a + b a + b a + 2 b a| [Taking out 3 (a + b) c o m m o n from R_{1}] = 3 (a + b) |0 0 1 2 b - b a + b - b 2 b a| [Applying C_{1} \to C_{1} - C_{2} and C_{2} \to C_{2} - C_{3}] = 3 (a + b) b^{2} |0 0 1 2 - 1 a + b - 1 2 a| [Taking out b c o m m o n from C_{1} and C_{2}] = 3 (a + b) b^{2} \times 3 = 9 (a + b) b^{2} = RHS$

Page No 5.59:

Question 18:

Prove that :

$|\begin{matrix} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{matrix}| = |\begin{matrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{matrix}|$

Answer:

$Let LHS = ∆ = |1 a b c 1 b c a 1 c a b| = \frac{1}{abc} |a a^{2} a b c b b^{2} b c a c c^{2} a b c| [Applying R_{1} \to a R_{1}, R_{2} \to b R_{2} and R_{3} \to c R_{3} and then dividing it by a b c] = \frac{a b c}{a b c} |a a^{2} 1 b b^{2} 1 c c^{2} 1| [Taking out a b c common from C_{3}] = (- 1) |1 a^{2} a 1 b^{2} b 1 c^{2} c| [Interchanging C_{3} and C_{1} to get - ve value of original determinant] = (- 1) (- 1) |1 a a^{2} 1 b^{} b^{2} 1 c c^{2}| [Applying C_{2} \leftrightarrow C_{3}] = |1 a a^{2} 1 b^{} b^{2} 1 c c| = RHS$

Page No 5.59:

Question 19:

Prove that :

$|\begin{matrix} z & x & y \\ z^{2} & x^{2} & y^{2} \\ z^{4} & x^{4} & y^{4} \end{matrix}| = |\begin{matrix} x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \end{matrix}| = |\begin{matrix} x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z \end{matrix}| = x y z (x - y) (y - z) (z - x) (x + y + z) .$

Answer:

$Let Δ_{1} = |z x y z^{2} x^{2} y^{2} z^{4} x^{4} y^{4}|, Δ_{2} = |x y z x^{2} y^{2} z^{2} x^{4} y^{4} z^{4}|, Δ_{3} = |x^{2} y^{2} z^{2} x^{4} y^{4} z^{4} x y z| and Δ_{4} = x y z (x - y) (y - z) (z - x) (x + y + z) Now, Δ_{1 =} |z x y z^{2} x^{2} y^{2} z^{4} x^{4} y^{4}| Using the property that if two rows (or columns) of a determinant are interchanged, the value of the determinant becomes negetive, we get \Rightarrow Δ_{1} = (- 1) |x z y x^{2} z^{2} y^{2} x^{4} z^{4} y^{4}| [∵ C_{1} \leftrightarrow C_{2}] = (- 1) (- 1) |x y z x^{2} y^{2} z^{2} x^{4} y^{4} z^{4}| [∵ C_{2} \leftrightarrow C_{3}] = |x y z x^{2} y^{2} z^{2} x^{4} y^{4} z^{4}| = Δ_{2} . . . (1) = (- 1) |x^{2} y^{2} z^{2} x y z x^{4} y^{4} z^{4}| [Applying R_{1} \leftrightarrow R_{2}] = (- 1) (- 1) |x^{2} y^{2} z^{2} x^{4} y^{4} z^{4} x y^{} z| [Applying R_{2} \leftrightarrow R_{3}] = |x^{2} y^{2} z^{2} x^{4} y^{4} z^{4} x y^{} z| = Δ_{3} . . . (2) Thus, Δ_{1} = Δ_{2} = Δ_{3} [From eqs . (1) and (2)]$

$∆_{2} = |x y z x^{2} y^{2} z^{2} x^{4} y^{4} z^{4}| = x y z |1 1 1 x y^{} z x^{3} y^{3} z^{3}| [Taking out common facto r x from C_{1,} y from C_{2} and z from C_{3}] = xyz |0 0 1 x - y y - z^{} z x^{3} - y^{3} y^{3} - z^{3} z^{3}| [Applying C \to C_{1} - C_{2} and C_{2} \to C_{2} - C_{3}] = xyz (x - y) (y - z) |0 0 1 1 1 z x^{2} + 2 x y + y^{2} y^{2} + 2 y z + z^{2} z^{3}| [∵ (a^{3} - b^{3}) = (a - b) (a^{2} + ab + b^{2})] [Taking out common factor (x - y) from C_{1} and (y - z) from C_{2}] = xyz (x - y) (y - z) \{1 \times |1 1 x^{2} + x y + y^{2} y^{2} + y z + z^{2}|\} [Expanding along R_{1}] = xyz (x - y) (y - z) \{y^{2} + y z + z^{2} - x^{2} - x y - y^{2}\} = xyz (x - y) (y - z) \{y z - x y + z^{2} - x^{2}\} = xyz (x - y) (y - z) \{y (z - x) + (z - x) (z + x)\} = xyz (x - y) (y - z) (z - x) (y + x + z) = xyz (x - y) (y - z) (z - x) (x + y + z) = ∆_{4} Thus, ∆_{1} = ∆_{2} = ∆_{3} = ∆_{4}$

Page No 5.59:

Question 20:

Prove that :

$|\begin{matrix} {(b + c)}^{2} & a^{2} & b c \\ {(c + a)}^{2} & b^{2} & c a \\ {(a + b)}^{2} & c^{2} & a b \end{matrix}| = (a - b) (b - c) (c - a) (a + b + c) (a^{2} + b^{2} + c^{2})$

Answer:

$Let LHS = Δ = |{(b + c)}^{2} a^{2} b c {(c + a)}^{2} b^{2} c a {(a + b)}^{2} c^{2} a b| = |{(b + c)}^{2} - {(c + a)}^{2} a^{2} - b^{2} bc - ca {(c + a)}^{2} - {(a + b)}^{2} b^{2} - c^{2} c a - a b {(a + b)}^{2} c^{2} a b| [Applying R_{1} \to R_{1} - R_{2} and R_{2} \to R_{2} - R_{3}] = |(b - a) (b + 2 c + a) (a + b) (a - b) c (b - a) (c - b) (b + 2 a + c) (b - c) (b + c) a (c - b) {(a + b)}^{2} c^{2} a b| = (a - b) (b - c) |- (b + 2 c + a) a + b - c - (b + 2 a + c) b + c - a {(a + b)}^{2} c^{2} a b| [A pplying x^{2} - y^{2} = (x + y) (x - y) and taking out (a - b) co m m o n from R_{1} and (b - c) from R_{2}] = (a - b) (b - c) |- 2 (b + c + a) a + b - c - 2 (b + a + c) b + c - a {(a + b)}^{2} - c^{2} c^{2} a b| [Applying C_{1} \to C_{1} - C_{2}] = (a - b) (b - c) |- 2 (b + c + a) a + b - c - 2 (b + a + c) b + c - a (a + b + c) (a + b - c) c^{2} a b| [Applying x^{2} - y^{2} = (x + y) (x - y) in C_{1}] = (a - b) (b - c) (a + b + c) |- 2 a + b - c - 2 b + c - a (a + b - c) c^{2} a b| [Taking out (a + b + c) common from C_{1}] = (a - b) (b - c) (a + b + c) |- 2 a + b - c 0 c - a c - a (a + b - c) c^{2} a b| [Applying R_{2} \to R_{2} - R_{1}] = (a - b) (b - c) (a + b + c) (c - a) |- 2 a + b - c 0 1 1 (a + b - c) c^{2} a b| [Taking out (c - a) common from R_{2}] = (a - b) (b - c) (a + b + c) (c - a) |- 2 a + b + c - c 0 0 1 (a + b - c) c^{2} - ab a b| [Applying C_{2} \to C_{2} - C_{3}] = (a - b) (b - c) (a + b + c) (c - a) \{(- 1) |- 2 a + b + c (a + b - c) c^{2} - ab|\} [Expanding along R_{2}] = - (a - b) (b - c) (a + b + c) (c - a) \{- 2 c^{2} + 2 ab - a^{2} - b^{2} - 2 ab + c^{2}\} = - (a - b) (b - c) (a + b + c) (c - a) (- a^{2} - b^{2} - c^{2}) = (a - b) (b - c) (a + b + c) (c - a) (a^{2} + b^{2} + c^{2}) = RHS$

Page No 5.59:

Question 21:

Prove that :

$|\begin{matrix} (a + 1) (a + 2) & a + 2 & 1 \\ (a + 2) (a + 3) & a + 3 & 1 \\ (a + 3) (a + 4) & a + 4 & 1 \end{matrix}| = - 2$

Answer:

$Let LHS = Δ = |(a + 1) (a + 2) a + 2 1 (a + 2) (a + 3) a + 3 1 (a + 3) (a + 4) a + 4 1| = |(a + 1) (a + 2) - (a + 2) (a + 3) (a + 2) - (a + 3) 0 (a + 2) (a + 3) - (a + 3) (a + 4) (a + 3) - (a + 4) 0 (a + 3) (a + 4) (a + 4) 1| [Applying C_{1} \to C_{1} - C_{2} and C_{2} \to C_{2} - C_{3}] = |- 2 (a + 2) - 1 0 - 2 (a + 3) - 1 0 (a + 3) (a + 4) (a + 4) 1| = \{1 \times |- 2 (a + 2) - 1 - 2 (a + 3) - 1|\} [Expanding along C_{3}] = 4 + 2 a - 2 a - 6 = - 2 = RHS$

Hence proved.

Page No 5.59:

Question 22:

Prove that :

$|\begin{matrix} a^{2} & a^{2} - {(b - c)}^{2} & b c \\ b^{2} & b^{2} - {(c - a)}^{2} & c a \\ c^{2} & c^{2} - {(a - b)}^{2} & a b \end{matrix}| = (a - b) (b - c) (c - a) (a + b + c) (a^{2} + b^{2} + c^{2})$

Answer:

$Let LHS = Δ = |a^{2} a^{2} - {(b - c)}^{2} bc b^{2} b^{2} - {(c - a)}^{2} ca c^{2} c^{2} - {(a - b)}^{2} ab|$

$\Rightarrow ∆ = |a^{2} - {(b - c)}^{2} bc b^{2} - {(c - a)}^{2} ca c^{2} - {(a - b)}^{2} ab| [Applying C_{2} \to C_{2} - C_{1}] = (- 1) |a^{2} {(b - c)}^{2} bc b^{2} {(c - a)}^{2} ca c^{2} {(a - b)}^{2} ab| = - |a^{2} b^{2} + c^{2} bc b^{2} c^{2} + a^{2} ca c^{2} a^{2} + b^{2} ab| [Applying C_{2} \to C_{2} - 2 C_{1}]$

$= - |a^{2} + b^{2} + c^{2} b^{2} + c^{2} bc b^{2} + c^{2} + a^{2} c^{2} + a^{2} ca c^{2} + a^{2} + b^{2} a^{2} + b^{2} ab| [Applying C_{1} \to C_{1} + C_{2}] = - (a^{2} + b^{2} + c^{2}) |1 b^{2} + c^{2} bc 1 c^{2} + a^{2} ca 1 a^{2} + b^{2} ab|$

$= - (a^{2} + b^{2} + c^{2}) |1 b^{2} + c^{2} bc 0 (c^{2} + a^{2}) - (b^{2} + c^{2}) ca - bc 0 (a^{2} + b^{2}) - (b^{2} + c^{2}) ab - bc| [Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}] = ((a^{2} + b^{2} + c^{2})) |1 b^{2} + c^{2} bc 0 a^{2} - b^{2} c (a - b) 0 a^{2} - c^{2} b (a - c)| = - (a^{2} + b^{2} + c^{2}) (a - b^{}) (a - c) |1 b^{2} + c^{2} bc 0 a + b^{} c 0 a^{} + c^{} b| [Taking (a - b) common from R_{2} and (a - c) common from R_{3}] = (a^{2} + b^{2} + c^{2}) (a - b^{}) (c - a) \times \{1 \times |a + b^{} c a^{} + c^{} b|\} [∵ (c - a) = - (a - c)] [Expanding along C_{1}] = (a^{2} + b^{2} + c^{2}) (a - b^{}) (c - a) (ab + b^{2} - ac - c^{2})$

$= (a^{2} + b^{2} + c^{2}) (a - b) (c - a) \{a (b - c) + (b + c) (b - c)\} = (a - b) (c - a) (b - c) (a + b + c) (a^{2} + b^{2} + c^{2})$

= RHS

Hence proved.

Page No 5.59:

Question 23:

Prove that :

$|\begin{matrix} 1 & a^{2} + b c & a^{3} \\ 1 & b^{2} + c a & b^{3} \\ 1 & c^{2} + a b & c^{3} \end{matrix}| = - (a - b) (b - c) (c - a) (a^{2} + b^{2} + c^{2})$

Answer:

$Let LHS = Δ = |1 a^{2} + bc a^{3} 1 b^{2} + ca b^{3} 1 c^{2} + ab c^{3}| \Rightarrow Δ = |0 (a^{2} + bc) - (b^{2} + ca) a^{3} - b^{3} 0 (b^{2} + ca) - (c^{2} + ab) b^{3} - c^{3} 1 c^{2} + ab c| [Applying R_{1} \to R_{1} - R_{2} and R_{2} \to R_{2} - R_{3}]$

$= |0 a^{2} - b^{2} - ca + bc a^{3} - b^{3} 0 b^{2} - c^{2} - ab + ca b^{3} - c^{3} 1 c^{2} + ab c^{3}| = |0 (a - b) (a + b - c) (a - b) (a^{2} + ab + b^{2}) 0 (b - c) (b + c - a) (b - c) (b^{2} + bc + a^{2}) 1 c^{2} + ab c^{3}|$

$= (a - b) (b - c) |0 a + b - c a^{2} + ab + b^{2} 0 (b + c - a) (b^{2} + bc + c^{2}) 1 c^{2} + ab c^{3}| [Taking out (a - b) common from R_{1} and (b - c) from R_{2}] = (a - b) (b - c) |0 a + b - c a^{2} + ab + b^{2} 0 (b + c - a) - (a + b - c) (b^{2} + bc + c^{2}) - (a^{2} + ab + b^{2}) 1 c^{2} + ab c^{3}| [Applying R_{2} \to R_{2} - R_{1}]$

$= (a - b) (b - c) |0 a + b - c a^{2} + ab + b^{2} 0 2 (c - a) b (c - a) + (c^{2} - a^{2}) 1 c^{2} + ab c^{3}| = (a - b) (b - c) (c - a) |0 a + b - c a^{2} + ab + b^{2} 0 2 a + b + c 1 c^{2} + ab c^{3}| = (a - b) (b - c) (c - a) \times \{1 \times |a + b - c a^{2} + ab + b^{2} 2 a + b + c|\} [Expanding along C_{1}]$

$= (a - b) (b - c) (c - a) \times \{{(a + b)}^{2} - c^{2} - (2 a^{2} + 2 ab + 2 b^{2})\} = (a - b) (b - c) (c - a) \{{(a + b)}^{2} - c^{2} - {(a + b)}^{2} - (a^{2} + b^{2})\} = - (a - b) (b - c) (c - a) (a^{2} + b^{2} + c^{2}) = RHS$

Hence proved.

Page No 5.59:

Question 24:

Prove that :

$|\begin{matrix} a^{2} & b c & a c + c^{2} \\ a^{2} + a b & b^{2} & a c \\ a b & b^{2} + b c & c^{2} \end{matrix}| = 4 a^{2} b^{2} c^{2}$

Answer:

$Let LHS = Δ = |a^{2} b c a c + c^{2} a^{2} + a b b^{2} a c a b b^{2} + b c c^{2}| Δ = a b c |a c a + c a + b b a b b + c c| [Taking out a, b and c common from C_{1}, C_{2} and C_{3}]$

$= a b c |a c 0 a + b b - 2 b b b + c - 2 b| [Applying C_{3} \to C_{3} - C_{2} - C_{1}] = (a b c) (- 2 b) |a c 0 a + b b 1 b b + c 1| [Taking (- 2 b) common from C_{3}] = (a b c) (- 2 b) |a c 0 a - c 0 b b + c 1| [Applying R_{2} \to R_{2} - R_{3}] = (a b c) (- 2 b) \times 1 |a c a - c| [Exp anding along C_{3}] = (a b c) (- 2 b) (- 2 ac) = 4 a^{2} b^{2} c^{2} = RHS$

Page No 5.59:

Question 25:

Prove that :

$|\begin{matrix} x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4 \end{matrix}| = 16 (3 x + 4)$

Answer:

$Let LHS = Δ = |x + 4 x x x x + 4 x x x x + 4| = |3 x + 4 3 x + 4 3 x + 4 x x + 4 x x x x + 4| [Applying R_{1} \to R_{1} + R_{2} + R_{3}] = (3 x + 4) |1 1 1 x x + 4 x x x x + 4| [Taking out (3 x + 4) common from R_{1}] = (3 x + 4) |1 0 0 x 4 0 x 0 4| [Applying C_{2} \to C_{2} - C_{1} and C_{3} \to C_{3} - C_{1}] = (3 x + 4) (4^{2}) [Expanding along R_{1}] = 16 (3 x + 4) = RHS$

Page No 5.59:

Question 26:

Prove that :

$|\begin{matrix} 1 & 1 + p & 1 + p + q \\ 2 & 3 + 2 p & 4 + 3 p + 2 q \\ 3 & 6 + 3 p & 10 + 6 p + 3 q \end{matrix}| = 1$

Answer:

$Let LHS = Δ = |1 1 + p 1 + p + q 2 3 + 2 p 4 + 3 p + 2 q 3 6 + 3 p 10 + 6 p + 3 q| = |1 1 1 + p 2 3 4 + 3 p 3 6 10 + 6 p| + |1 p q 2 2 p 2 q 3 3 p 3 q| = |1 1 1 2 3 4 3 6 10| + |1 1 p 2 3 3 p 3 6 6 p| + (pq) |1 1 1 2 2 2 3 3 3| [Taking out p q common from last determinant] = |1 1 1 2 3 4 3 6 10| + (p) |1 1 1 2 3 3 3 6 6| + 0 [Taking out p common from second determinant] = |1 1 1 2 3 4 3 6 10| + 0 [∵ Value of determinant with two identical columns is zero] = |1 0 0 2 1 2 3 3 7| [Applying C_{2} \to C_{2} - C_{1} and C_{3} \to C_{3} - C 1] = \{1 \times |\begin{matrix} 1 & 2 \\ 3 & 7 \end{matrix}|\} [Expanding along R_{1}] = 7 - 6 = 1 = RHS$

Page No 5.59:

Question 27:

Prove that :

$|\begin{matrix} a & b - c & c - b \\ a - c & b & c - a \\ a - b & b - a & c \end{matrix}| = (a + b - c) (b + c - a) (c + a - b)$

Answer:

$Let LHS = Δ = |a b - c c - b a - c b c - a a - b b - a c| Δ = |a 0 c - b + a a - c b + c - a 0 a - b b + c - a c + a - b| [Applying C_{2} \to C_{2} + C_{3} and C_{3} \to C_{1} + C_{3}] = (b + c - a) (c + a - b) |a 0 1 a - c 1 0 a - b 1 1| [Taking out common factor from C_{2} and C_{3}] = (b + c - a) (c + a - b) \{(a \times |1 0 1 1|) + (1 \times |a - c 1 a - b 1|)\} [Expanding along R_{1}] = (a + b - c) (b + c - a) (c + a - b) = RHS$

Page No 5.60:

Question 28:

Prove that $|\begin{matrix} a^{2} & 2 a b & b^{2} \\ b^{2} & a^{2} & 2 a b \\ 2 a b & b^{2} & a^{2} \end{matrix}| = {(a^{3} + b^{3})}^{2}$

Answer:

$Let LHS = ∆ = |a^{2} 2 a b b^{2} b^{2} a^{2} 2 a b 2 a b b^{2} a^{2}| = a^{2} |a^{2} 2 a b b^{2} a^{2}| - (2 a b) |b^{2} 2 a b 2 a b a^{2}| + b^{2} |b^{2} a^{2} 2 a b b^{2}| [Expanding] = a^{2} (a^{4} - 2 a b^{3}) - (2 a b) (b^{2} a^{2} - 4 a^{2} b^{2}) + b^{2} (b^{4} - 2 a^{3} b) = a^{6} - 2 a^{3} b^{3} - 2 a^{3} b^{3} + 8 a^{3} b^{3} + b^{6} - 2 a^{3} b^{3} = a^{6} + 2 a^{3} b^{3} + b^{6} = {(a^{3})}^{2} + 2 a^{3} b^{3} + {(b^{3})}^{2} = {(a^{3} + b^{3})}^{2} = RHS$

Hence proved.

Page No 5.60:

Question 29:

Prove that $|\begin{matrix} a^{2} + 1 & a b & a c \\ a b & b^{2} + 1 & b c \\ c a & c b & c^{2} + 1 \end{matrix}| = 1 + a^{2} + b^{2} + c^{2}$

Answer:

$Let LHS = Δ = |a^{2} + 1 a b a c a b b^{2} + 1 b c c a c b c^{2} + 1| = (a b c) |a + \frac{1}{a} b c a b + \frac{1}{b} c a b c + \frac{1}{c}| [Taking out a, b and c common from R_{1}, R_{2} and R_{3}] = (a b c) |a + \frac{1}{a} b c - \frac{1}{a} \frac{1}{b} 0 - \frac{1}{a} 0 \frac{1}{c}| [Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}] = (a b c) (\frac{1}{a b c}) |a^{2} + 1 b^{2} c^{2} - 1 1 0 - 1 0 1| [Applying C_{1} \to a C_{1}, C_{2} \to b C_{2} and C_{3} \to c C_{3}] = |a^{2} + 1 b^{2} c^{2} - 1 1 0 - 1 0 1| = (- 1) |b^{2} c^{2} 1 0| + (1) |a^{2} + 1 b^{2} - 1 1| [Expanding along R_{3}] = (- 1) (- c^{2}) + (a^{2} + 1 + b^{2}) = (a^{2} + 1 + b^{2} + c^{2}) = (a^{2} + b^{2} + c^{2} + 1) = RHS$

Page No 5.60:

Question 30:

$|\begin{matrix} 1 & a & a^{2} \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{matrix}| = {(a^{3} - 1)}^{2}$

Answer:

$Let LHS = Δ = |1 a a^{2} a^{2} 1 a a a^{2} 1| Δ = |1 + a^{2} + a 1 + a^{2} + a 1 + a^{2} + a a^{2} 1 a a a^{2} 1| [Applyng R_{1} \to R_{1} + R_{2} + R_{2}] = (1 + a^{2} + a) |1 1 1 a^{2} 1 a a a^{2} 1| [Applying C_{2} \to C_{2} - C_{1} and C_{3} \to C_{3} - C_{1}] = (1 + a^{2} + a) |1 0 0 a^{2} 1 - a^{2} a - a^{2} a a^{2} - a 1 - a| = (1 + a^{2} + a) |1 0 0 a^{2} (1 - a) (1 + a) a (1 - a) a a (a - 1) 1 - a| = (1 + a^{2} + a) (a - 1) (a - 1) |1 0 0 a^{2} - (1 + a) - a a a - 1| [Taking out (a - 1) common from C_{2} and C_{3}] = (a^{3} - 1) \{(a - 1) |1 0 0 a - (1 + a) - a a a - 1|\} [∵ (1 + a^{2} + a) (a - 1) = (a^{3} - 1)] = (a^{3} - 1) \{(a - 1) (1 + a^{} + a^{2})\} = (a^{3} - 1) (a^{3} - 1) = {(a^{3} - 1)}^{2} = RHS$

Hence proved.

Page No 5.60:

Question 31:

$|\begin{matrix} a + b + c & - c & - b \\ - c & a + b + c & - a \\ - b & - a & a + b + c \end{matrix}| = 2 (a + b) (b + c) (c + a)$

Answer:

$Let LHS = ∆ = |a + b + c - c - b - c a + b + c - a - b - a a + b + c| = |a - c - b b a + b + c - a c - a a + b + c| [Applying C_{1} \to C_{1} + C_{2} + C_{3}] = |a + b a + b - (a + b) b + c b + c b + c c - a a + b + c| [Applying R_{1} \to R_{1} + R_{2} and R_{2} \to R_{2} + R_{3}] = (a + b) (b + c) |1 1 - 1 1 1 1 c - a a + b + c| [Taking out common factor from R_{1} and R_{2}] = (a + b) (b + c) |0 0 - 2 1 1 1 c - a a + b + c| [Applying R_{1} \to R_{1} - R_{2}] = (a + b) (b + c) \{(- 2) (- a - c)\} [Expanding along R_{1}] = 2 (a + b) (b + c) (c + a) = RHS$

Hence proved.

Page No 5.60:

Question 32:

$|\begin{matrix} b + c & a & a \\ b & c + a & b \\ c & c & a + b \end{matrix}| = 4 a b c$

Answer:

$∆ = |\begin{matrix} b + c & a & a \\ b & c + a & b \\ c & c & a + b \end{matrix}| = |\begin{matrix} 0 & - 2 c & - 2 b \\ b & c + a & b \\ c & c & a + b \end{matrix}| [Applying R_{1} \to R_{1} - (R_{2} + R_{3})] = |\begin{matrix} 0 & - 2 c & - 2 b \\ b & c + a - b & 0 \\ c & 0 & a + b - c \end{matrix}| [Applying C_{2} \to C_{2} - C_{1} and C_{3} \to C_{3} - C_{1}] = 0 |\begin{matrix} c + a - b & 0 \\ 0 & a + b - c \end{matrix}| - (- 2 c) |\begin{matrix} b & 0 \\ c & a + b - c \end{matrix}| - 2 b |\begin{matrix} b & c + a - b \\ c & 0 \end{matrix}| [Expanding along R_{1}] = 2 c [b (a + b - c) - 0] - 2 b [0 - c (c + a - b)] = 2 b c [a + b - c] - 2 b c [b - c - a] = 2 b c [(a + b - c) - (b - c - a)] = 4 a b c$

Hence proved.

Page No 5.60:

Question 33:

$|\begin{matrix} b^{2} + c^{2} & a b & a c \\ b a & c^{2} + a^{2} & b c \\ c a & c b & a^{2} + b^{2} \end{matrix}| = 4 a^{2} b^{2} c^{2}$

Answer:

$∆ = |\begin{matrix} b^{2} + c^{2} & a b & a c \\ b a & c^{2} + a^{2} & b c \\ c a & c b & a^{2} + b^{2} \end{matrix}|$

$|\begin{matrix} a (b^{2} + c^{2}) & a^{2} b & a^{2} c \\ b^{2} a & b (c^{2} + a^{2}) & b^{2} c \\ c^{2} a & c^{2} b & c (a^{2} + b^{2}) \end{matrix}| [Multiplying the three rows by a, b and c]$

$= \frac{a b c}{a b c} |\begin{matrix} b^{2} + c^{2} & a^{2} & a^{2} \\ b^{2} & c^{2} + a^{2} & b^{2} \\ c^{2} & c^{2} & a^{2} + b^{2} \end{matrix}| [Taking out a, b a n d c common from the three columns] = |\begin{matrix} 2 (b^{2} + c^{2}) & 2 (a^{2} + c^{2}) & 2 (a^{2} + b^{2}) \\ b^{2} & c^{2} + a^{2} & b^{2} \\ c^{2} & c^{2} & a^{2} + b^{2} \end{matrix}| [Applying R_{1} \to R_{1} + R_{2} + R_{3}] = 2 |\begin{matrix} b^{2} + c^{2} & a^{2} + c^{2} & a^{2} + b^{2} \\ - c^{2} & 0 & - a^{2} \\ - b^{2} & - a^{2} & 0 \end{matrix}| [Taking out 2 common from the three columns and then applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}] = 2 |\begin{matrix} 0 & c^{2} & b^{2} \\ - c^{2} & 0 & - a^{2} \\ - b^{2} & - a^{2} & 0 \end{matrix}| [Applying R_{1} \to R_{1} + R_{2} + R_{3}] = 2 {[- c^{2} (- a^{2} b^{2})] + [b^{2} (c^{2} a^{2})]} [Expanding along R_{1}] = 4 a^{2} b^{2} c^{2}$

Page No 5.60:

Question 34:

$|\begin{matrix} 0 & b^{2} a & c^{2} a \\ a^{2} b & 0 & c^{2} b \\ a^{2} c & b^{2} c & 0 \end{matrix}| = 2 a^{3} b^{3} c^{3}$

Answer:

$∆ = |\begin{matrix} 0 & b^{2} a & c^{2} a \\ a^{2} b & 0 & c^{2} b \\ a^{2} c & b^{2} c & 0 \end{matrix}| = \frac{1}{a b c} |\begin{matrix} 0 & b^{3} a & c^{3} a \\ a^{3} b & 0 & c^{3} b \\ a^{3} c & b^{3} c & 0 \end{matrix}| [Multiplying the three columns by a, b and c] = \frac{a b c}{a b c} |\begin{matrix} 0 & b^{3} & c^{3} \\ a^{3} & 0 & c^{3} \\ a^{3} & b^{3} & 0 \end{matrix}| [Taking out a, b and c common from the three rows] = b^{3} |\begin{matrix} a^{3} & c^{3} \\ a^{3} & 0 \end{matrix}| + c^{3} |\begin{matrix} a^{3} & 0 \\ a^{3} & b^{3} \end{matrix}| = 2 a^{3} b^{3} c^{3} [Expanding along R_{1}]$

Page No 5.60:

Question 35:

Prove that $|\begin{matrix} \frac{a^{2} + b^{2}}{c} & c & c \\ a & \frac{b^{2} + c^{2}}{a} & a \\ b & b & \frac{c^{2} + a^{2}}{b} \end{matrix}| = 4 a b c$

Answer:

$∆ = |\begin{matrix} \frac{a^{2} + b^{2}}{c} & c & c \\ a & \frac{b^{2} + c^{2}}{a} & a \\ b & b & \frac{c^{2} + a^{2}}{b} \end{matrix}| = \frac{1}{a b c} |\begin{matrix} a^{2} + b^{2} & c^{2} & c^{2} \\ a^{2} & b^{2} + c^{2} & a^{2} \\ b^{2} & b^{2} & c^{2} + a^{2} \end{matrix}| [Multiplying R_{1}, R_{2} and R_{3} by c, a and b and then dividing by a b c] = \frac{1}{a b c} |\begin{matrix} a^{2} + b^{2} & c^{2} - a^{2} - b^{2} & c^{2} - a^{2} - b^{2} \\ a^{2} & b^{2} + c^{2} - a^{2} & 0 \\ b^{2} & 0 & c^{2} + a^{2} - b^{2} \end{matrix}| [Applying C_{2} \to C_{2} - C_{1} and C_{3} \to C_{3} - C_{1}] = \frac{1}{a b c} |\begin{matrix} 0 & - 2 b^{2} & - 2 a^{2} \\ a^{2} & b^{2} + c^{2} - a^{2} & 0 \\ b^{2} & 0 & c^{2} + a^{2} - b^{2} \end{matrix}| [Applying R_{1} \to R_{1} - R_{2} - R_{3}] = \frac{1}{a b c} [- a^{2} |\begin{matrix} - 2 b^{2} & - 2 a^{2} \\ 0 & c^{2} + a^{2} - b^{2} \end{matrix}| + b^{2} |\begin{matrix} - 2 b^{2} & - 2 a^{2} \\ b^{2} + c^{2} - a^{2} & 0 \end{matrix}| [Expanding along C_{1}] = \frac{1}{a b c} [- a^{2} \{- 2 b^{2} (c^{2} + a^{2} - b^{2})\} + b^{2} \{0 + 2 a^{2} (b^{2} + c^{2} - a^{2})\}] = \frac{1}{a b c} [- a^{2} \{- 2 b^{2} c^{2} - 2 b^{2} a^{2} + 2 b^{4}\} + b^{2} \{2 a^{2} b^{2} + 2 a^{2} c^{2} - 2 a^{4}\}] = \frac{1}{a b c} [2 a^{2} b^{2} c^{2} + 2 a^{4} b^{2} - 2 a^{2} b^{4} + 2 a^{2} b^{4} + 2 a^{2} b^{2} c^{2} - 2 a^{4} b^{2}] = \frac{1}{a b c} 4 a^{2} b^{2} c^{2} = 4 a b c$

Hence proved.

Page No 5.60:

Question 36:

Prove that $|\begin{matrix} - b c & b^{2} + b c & c^{2} + b c \\ a^{2} + a c & - a c & c^{2} + a c \\ a^{2} + a b & b^{2} + a b & - a b \end{matrix}| = {(a b + b c + c a)}^{3}$

Answer:

$∆ = |\begin{matrix} - b c & b^{2} + b c & c^{2} + b c \\ a^{2} + a c & - a c & c^{2} + a c \\ a^{2} + a b & b^{2} + a b & - a b \end{matrix}| = \frac{1}{a b c} |\begin{matrix} - a b c & a b^{2} + a b c & a c^{2} + a b c \\ a^{2} b + a b c & - a b c & c^{2} b + a b c \\ a^{2} c + a b c & b^{2} c + a b c & - a b c \end{matrix}| [Applying R_{1} \to a R_{1}, R_{2} \to b R_{2} and R_{3} \to c R_{3} and then dividing by a b c] = \frac{a b c}{a b c} |\begin{matrix} - b c & a b + a c & a c + a b \\ a b + b c & - a c & c b + a b \\ a c + b c & b c + a c & - a b \end{matrix}| [Taking out a, b and c common from the three columns] |\begin{matrix} a b + b c + c a & a b + b c + c a & a b + b c + c a \\ a b + b c & - a c & c b + a b \\ a c + b c & b c + a c & - a b \end{matrix}| [Applying R_{1} \to R_{1} + R_{2} + R_{3}] = (a b + b c + c a) |\begin{matrix} 1 & 1 & 1 \\ a b + b c & - a c & c b + a b \\ a c + b c & b c + a c & - a b \end{matrix}| = (a b + b c + c a) |\begin{matrix} 0 & 0 & 1 \\ 0 & - (a b + b c + a c) & c b + a b \\ a c + b c + a b & b c + a c + a b & - a b \end{matrix}| [Applying C_{1} \to C_{1} - C_{3} and C_{2} \to C_{2} - C_{3}] = (a b + b c + c a) |\begin{matrix} 0 & - (a b + b c + a c) \\ a c + b c + a b & b c + a c + a b \end{matrix}| = (a b + b c + c a) (a b + b c + a c)^{2} = (a b + b c + c a)^{3}$

Hence proved.

Page No 5.60:

Question 37:

Prove the following identities:
$|\begin{matrix} x + λ & 2 x & 2 x \\ 2 x & x + λ & 2 x \\ 2 x & 2 x & x + λ \end{matrix}| = (5 x + λ) {(λ - x)}^{2}$

Answer:

$LHS : |\begin{matrix} x + λ & 2 x & 2 x \\ 2 x & x + λ & 2 x \\ 2 x & 2 x & x + λ \end{matrix}| = |\begin{matrix} x + λ & 2 x & 2 x \\ 2 x - x - λ & x + λ - 2 x & 0 \\ 2 x - x - λ & 0 & x + λ - 2 x \end{matrix}| [Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}] = |\begin{matrix} x + λ & 2 x & 2 x \\ - (λ - x) & λ - x & 0 \\ - (λ - x) & 0 & λ - x \end{matrix}| = {(λ - x)}^{2} |\begin{matrix} x + λ & 2 x & 2 x \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{matrix}| [Taking (λ - x) common from R_{2} and (λ - x) common from R_{3}] = {(λ - x)}^{2} [- 1 (- 2 x) + 1 (x + λ + 2 x)] [Expanding along last row] = {(λ - x)}^{2} (λ + 5 x) = RHS ∴ |\begin{matrix} x + λ & 2 x & 2 x \\ 2 x & x + λ & 2 x \\ 2 x & 2 x & x + λ \end{matrix}| = {(λ - x)}^{2} (λ + 5 x)$

Page No 5.60:

Question 38:

Using properties of determinants prove that

$|\begin{matrix} x + 4 & 2 x & 2 x \\ 2 x & x + 4 & 2 x \\ 2 x & 2 x & x + 4 \end{matrix}| = (5 x + 4) {(4 - x)}^{2}$

Answer:

$∆ = |\begin{matrix} x + 4 & 2 x & 2 x \\ 2 x & x + 4 & 2 x \\ 2 x & 2 x & x + 4 \end{matrix}| = |\begin{matrix} 5 x + 4 & 5 x + 4 & 5 x + 4 \\ 2 x & x + 4 & 2 x \\ 2 x & 2 x & x + 4 \end{matrix}| [Applying R_{1} \to R_{1} + R_{2} + R_{3}] = 5 x + 4 |\begin{matrix} 1 & 1 & 1 \\ 2 x & x + 4 & 2 x \\ 2 x & 2 x & x + 4 \end{matrix}| [Take out 5 x + 4 common from R_{1}] = 5 x + 4 |\begin{matrix} 1 & 0 & 0 \\ 2 x & 4 - x & 0 \\ 2 x & 0 & 4 - x \end{matrix}| [Applying C_{2} \to C_{2} - C_{1} and C_{3} \to C_{3} - C_{1}] = 5 x + 4 (4 - x)^{2} [Expanding along R_{1}]$

Hence proved.

Page No 5.60:

Question 39:

Prove the following identities:

$|\begin{matrix} y + z & z & y \\ z & z + x & x \\ y & x & x + y \end{matrix}| = 4 x y z$

Answer:

$LHS : |\begin{matrix} y + z & z & y \\ z & z + x & x \\ y & x & x + y \end{matrix}| = |\begin{matrix} y + z - z - y & z - z - x - x & y - x - x - y \\ z & z + x & x \\ y & x & x + y \end{matrix}| [Applying R_{1} \to R_{1} - R_{2} - R_{3}] = |\begin{matrix} 0 & - 2 x & - 2 x \\ z & z + x & x \\ y & x & x + y \end{matrix}| = - 2 x |\begin{matrix} 0 & 1 & 1 \\ z & z + x & x \\ y & x & x + y \end{matrix}| [Taking - 2 x common from R_{1}] = - 2 x |\begin{matrix} 0 & 0 & 1 \\ z & z & x \\ y & - y & x + y \end{matrix}| [Applying C_{2} \to C_{2} - C_{3}] = - 2 x (- z y - z y) [Expanding along first row] = 4 x y z = RHS ∴ |\begin{matrix} y + z & z & y \\ z & z + x & x \\ y & x & x + y \end{matrix}| = 4 x y z$

Page No 5.61:

Question 40:

$|\begin{matrix} - a (b^{2} + c^{2} - a^{2}) & 2 b^{3} & 2 c^{3} \\ 2 a^{3} & - b (c^{2} + a^{2} - b^{2}) & 2 c^{3} \\ 2 a^{3} & 2 b^{3} & - c (a^{2} + b^{2} - c^{2}) \end{matrix}| = a b c {(a^{2} + b^{2} + c^{2})}^{3}$

Answer:

$∆ = |\begin{matrix} - a (b^{2} + c^{2} - a^{2}) & 2 b^{3} & 2 c^{3} \\ 2 a^{3} & - b (c^{2} + a^{2} - b^{2}) & 2 c^{3} \\ 2 a^{3} & 2 b^{3} & - c (a^{2} + b^{2} - c^{2}) \end{matrix}| = a b c |\begin{matrix} - b^{2} - c^{2} + a^{2} & 2 b^{2} & 2 c^{2} \\ 2 a^{2} & - c^{2} - a^{2} + b^{2} & 2 c^{2} \\ 2 a^{2} & 2 b^{2} & - a^{2} - b^{2} + c^{2} \end{matrix}| [Taking out a, b and c common from C_{1}, C_{2} and C_{3}] = a b c |\begin{matrix} a^{2} + b^{2} + c^{2} & 2 b^{2} & 2 c^{2} \\ a^{2} + b^{2} + c^{2} & - c^{2} - a^{2} + b^{2} & 2 c^{2} \\ a^{2} + b^{2} + c^{2} & 2 b^{2} & - a^{2} - b^{2} + c^{2} \end{matrix}| [Applying C_{1} \to C_{1} + C_{2} + C_{3}] = a b c (a^{2} + b^{2} + c^{2}) |\begin{matrix} 1 & 2 b^{2} & 2 c^{2} \\ 1 & - c^{2} - a^{2} + b^{2} & 2 c^{2} \\ 1 & 2 b^{2} & - a^{2} - b^{2} + c^{2} \end{matrix}| [Taking out a^{2} + b^{2} + c common from C_{1}] = a b c (a^{2} + b^{2} + c^{2}) |\begin{matrix} 1 & 2 b^{2} & 2 c^{2} \\ 0 & - c^{2} - a^{2} - b^{2} & 0 \\ 0 & 0 & - a^{2} - b^{2} - c^{2} \end{matrix}| [Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}] = a b c (a^{2} + b^{2} + c^{2})^{3} [Expanding]$

Hence proved.

Page No 5.61:

Question 41:

Evaluate the following determinant:

(i) $|\begin{matrix} 1 + a & 1 & 1 \\ 1 & 1 + a & 1 \\ 1 & 1 & 1 + a \end{matrix}| = a^{3} + 3 a^{2}$

(ii) $|\begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{matrix}| = {(a - 1)}^{3}$

Answer:

(ii) To Prove: $|\begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{matrix}| = {(a - 1)}^{3}$

$LHS = |\begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{matrix}| Applying R_{1} \to R_{1} - R_{2} = |\begin{matrix} a^{2} + 2 a - 2 a - 1 & 2 a + 1 - a - 2 & 1 - 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{matrix}| = |\begin{matrix} a^{2} - 1 & a - 1 & 0 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{matrix}| Taking (a - 1) common from R_{1} = (a - 1) |\begin{matrix} a + 1 & 1 & 0 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{matrix}| Applying R_{2} \to R_{2} - R_{3} = (a - 1) |\begin{matrix} a + 1 & 1 & 0 \\ 2 a + 1 - 3 & a + 2 - 3 & 1 - 1 \\ 3 & 3 & 1 \end{matrix}| = (a - 1) |\begin{matrix} a + 1 & 1 & 0 \\ 2 a - 2 & a - 1 & 0 \\ 3 & 3 & 1 \end{matrix}| Taking (a - 1) common from R_{2} = {(a - 1)}^{2} |\begin{matrix} a + 1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{matrix}| Expanding through C_{3} = {(a - 1)}^{2} [1 (1 (a + 1) - 2)] = {(a - 1)}^{2} [1 (a + 1 - 2)] = {(a - 1)}^{2} (a - 1) = {(a - 1)}^{3} = RHS$

Hence, $|\begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{matrix}| = {(a - 1)}^{3}$ .

Page No 5.61:

Question 42:

Prove the following identities:

$|\begin{matrix} 2 y & y - z - x & 2 y \\ 2 z & 2 z & z - x - y \\ x - y - z & 2 x & 2 x \end{matrix}| = {(x + y + z)}^{3}$

Answer:

$LHS = |\begin{matrix} 2 y & y - z - x & 2 y \\ 2 z & 2 z & z - x - y \\ x - y - z & 2 x & 2 x \end{matrix}| = |\begin{matrix} 2 y + 2 z + x - y - z & y - z - x + 2 z + 2 x & 2 y + z - x - y + 2 x \\ 2 z & 2 z & z - x - y \\ x - y - z & 2 x & 2 x \end{matrix}| [Applying R_{1} \to R_{1} + R_{2} + R_{3}] = |\begin{matrix} x + y + z & x + y + z & x + y + z \\ 2 z & 2 z & z - x - y \\ x - y - z & 2 x & 2 x \end{matrix}| = (x + y + z) |\begin{matrix} 1 & 1 & 1 \\ 2 z & 2 z & z - x - y \\ x - y - z & 2 x & 2 x \end{matrix}| [Taking (x + y + z) common from R_{1}] = (x + y + z) |\begin{matrix} 0 & 1 & 1 \\ 0 & 2 z & z - x - y \\ - x - y - z & 2 x & 2 x \end{matrix}| [Applying C_{1} \to C_{1} - C_{2}] = {(x + y + z)}^{2} |\begin{matrix} 0 & 1 & 1 \\ 0 & 2 z & z - x - y \\ - 1 & 2 x & 2 x \end{matrix}| [Taking (x + y + z) common from C_{1}] = {(x + y + z)}^{2} [- 1 (z - x - y - 2 z)] [Expanding along first column] = {(x + y + z)}^{3} = RHS ∴ |\begin{matrix} 2 y & y - z - x & 2 y \\ 2 z & 2 z & z - x - y \\ x - y - z & 2 x & 2 x \end{matrix}| = {(x + y + z)}^{3}$

Page No 5.61:

Question 43:

Show that $|\begin{matrix} y + z & x & y \\ z + x & z & x \\ x + y & y & z \end{matrix}| = (x + y + z) {(x - z)}^{2}$

Answer:

$Let ∆ = |y + z x y z + x z x x + y y z| \Rightarrow ∆ = |2 (x + y + z) x + y + z x + y + z z + x z x x + y y z| [Applying R_{1} \to R_{1} + R_{2} + R_{3}] = (x + y + z) |2 1 1 z + x z x x + y y z| = (x + y + z) |0 1 1 0 z x x - z y z| [Applying C_{1} \to C_{1} - C_{2} - C_{3}] = (x + y + z) \{(x - z) \times |\begin{matrix} 1 & 1 \\ z & x \end{matrix}|\} [Expanding along C_{1}] = (x + y + z) {(x - z)}^{2}$

Page No 5.61:

Question 44:

Prove the following identities:

$|\begin{matrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{matrix}| = a^{2} (a + x + y + z)$

Answer:

$LHS = |\begin{matrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{matrix}| = |\begin{matrix} a + x + y + z & y & z \\ a + x + y + z & a + y & z \\ a + x + y + z & y & a + z \end{matrix}| [Applying C_{1} \to C_{1} + C_{2} + C_{3}] = (a + x + y + z) |\begin{matrix} 1 & y & z \\ 1 & a + y & z \\ 1 & y & a + z \end{matrix}| [Taking (a + x + y + z) common from C_{1}] = (a + x + y + z) |\begin{matrix} 1 & y & z \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix}| [Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}] = (a + x + y + z) a^{2} [Expanding along first column] = a^{2} (a + x + y + z) = RHS ∴ |\begin{matrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{matrix}| = a^{2} (a + x + y + z)$

Page No 5.61:

Question 45:

Prove the following identities:

$|\begin{matrix} a^{3} & 2 & a \\ b^{3} & 2 & b \\ c^{3} & 2 & c \end{matrix}| = 2 (a - b) (b - c) (c - a) (a + b + c) \begin{matrix} \end{matrix}$

Answer:

$LHS = |\begin{matrix} a^{3} & 2 & a \\ b^{3} & 2 & b \\ c^{3} & 2 & c \end{matrix}| = |\begin{matrix} a^{3} & 2 & a \\ b^{3} - a^{3} & 0 & b - a \\ c^{3} - a^{3} & 0 & c - a \end{matrix}| [Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}] = - (a - b) (c - a) |\begin{matrix} a^{3} & 2 & a \\ b^{2} + a^{2} + a b & 0 & 1 \\ c^{2} + a^{2} + a c & 0 & 1 \end{matrix}| [Taking (b - a) common from R_{2} and (c - a) common from R_{3}] = - (a - b) (c - a) |\begin{matrix} a^{3} & 2 & a \\ b^{2} - c^{2} + a b - a c & 0 & 0 \\ c^{2} + a^{2} + a c & 0 & 1 \end{matrix}| [Applying R_{2} \to R_{2} - R_{3}] = - (a - b) (c - a) |\begin{matrix} a^{3} & 2 & a \\ (b - c) (a + b + c) & 0 & 0 \\ c^{2} + a^{2} + a c & 0 & 1 \end{matrix}| = - (a - b) (c - a) (b - c) (a + b + c) |\begin{matrix} a^{3} & 2 & a \\ 1 & 0 & 0 \\ c^{2} + a^{2} + a c & 0 & 1 \end{matrix}| [Taking (b - c) (a + b + c) common from R_{2}] = - (a - b) (c - a) (b - c) (a + b + c) (- 2) [Expanding along second column] = 2 (a - b) (c - a) (b - c) (a + b + c) = RHS ∴ |\begin{matrix} a^{3} & 2 & a \\ b^{3} & 2 & b \\ c^{3} & 2 & c \end{matrix}| = 2 (a - b) (b - c) (c - a) (a + b + c) \begin{matrix} \end{matrix}$

Page No 5.61:

Question 46:

Without expanding, prove that

$|\begin{matrix} a & b & c \\ x & y & z \\ p & q & r \end{matrix}| = |\begin{matrix} x & y & z \\ p & q & r \\ a & b & c \end{matrix}| = |\begin{matrix} y & b & q \\ x & a & p \\ z & c & r \end{matrix}|$

Answer:

$|\begin{matrix} x & y & z \\ p & q & r \\ a & b & c \end{matrix}| R_{2} \leftrightarrow R_{3} = - |\begin{matrix} x & y & z \\ a & b & c \\ p & q & r \end{matrix}| R_{1} \leftrightarrow R_{2} = |\begin{matrix} a & b & c \\ x & y & z \\ p & q & r \end{matrix}| \begin{matrix} \end{matrix} |\begin{matrix} y & b & q \\ x & a & p \\ z & c & r \end{matrix}| = |\begin{matrix} y & x & z \\ b & a & c \\ q & p & r \end{matrix}| C_{1} \leftrightarrow C_{2} = - |\begin{matrix} x & y & z \\ a & b & c \\ p & q & r \end{matrix}| R_{1} \leftrightarrow R_{2} = |\begin{matrix} a & b & c \\ x & y & z \\ p & q & r \end{matrix}|$

Hence proved.

Page No 5.61:

Question 47:

Show that $|\begin{matrix} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{matrix}| = 0 where a, b, c are in A . P .$

Answer:

Given: a, b, c are in A.P.

$2 b = a + c$

$∆ = |\begin{matrix} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{matrix}| [Applying R_{2} = 2 R_{2}] ∆ = \frac{1}{2} |\begin{matrix} x + 1 & x + 2 & x + a \\ 2 x + 4 & 2 x + 6 & 2 x + 2 b \\ x + 3 & x + 4 & x + c \end{matrix}| ∆ = \frac{1}{2} |\begin{matrix} x + 1 & x + 2 & x + a \\ 0 & 0 & 0 \\ x + 3 & x + 4 & x + c \end{matrix}| [∵ 2 b = a + c] [Applying R_{2} \to R_{2} - (R_{1} + R_{3})] ∆ = 0$

Page No 5.61:

Question 48:

Show that $|\begin{matrix} x - 3 & x - 4 & x - α \\ x - 2 & x - 3 & x - β \\ x - 1 & x - 2 & x - γ \end{matrix}| = 0, where α, β, γ are in AP .$

Answer:

Given:α, β, γ areinA.P.

Now,
$2 β = α + γ$

$∆ = |\begin{matrix} x - 3 & x - 4 & x - α \\ x - 2 & x - 3 & x - β \\ x - 1 & x - 2 & x - γ \end{matrix}| ∆ = \frac{1}{2} |\begin{matrix} x - 3 & x - 4 & x - α \\ 2 x - 4 & 2 x - 6 & 2 x - 2 β \\ x - 1 & x - 2 & x - γ \end{matrix}| [Applying R_{2} \to 2 R_{2}] ∆ = \frac{1}{2} |\begin{matrix} x - 3 & x - 4 & x - α \\ 0 & 0 & - 2 β + α + γ \\ x - 1 & x - 2 & x - γ \end{matrix}| [∵ 2 β = α + γ] [Applying R_{2} \to R_{2} - (R_{1} + R_{3})] ∆ = \frac{1}{2} |\begin{matrix} x - 3 & x - 4 & x - α \\ 0 & 0 & 0 \\ x - 1 & x - 2 & x - γ \end{matrix}| ∆ = 0$

Page No 5.61:

Question 49:

If a, b, c are real numbers such that $|\begin{matrix} b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{matrix}| = 0$ , then show that either $a + b + c = 0 or, a = b = c$ .

Answer:

$Let Δ = |b + c c + a a + b c + a a + b b + c a + b b + c c + a| = |2 (a + b + c) 2 (a + b + c) 2 (a + b + c) c + a a + b b + c a + b b + c c + a| [Applying R_{1} \to R_{1} + R_{2} + R_{3}] = 2 (a + b + c) |1 1 1 c + a a + b b + c a + b b + c c + a| = 2 (a + b + c) |1 0 0 c + a b - c b - a a + b c - a c - b| [Applying C_{2} \to C_{2} - C_{1} and C_{3} \to C_{3} - C_{1}] = 2 (a + b + c) \{1 |\begin{matrix} b - c & b - a \\ c - a & c - b \end{matrix}|\} = 2 (a + b + c) \{(b - c) (c - b) - (b - a) (c - a)\} = - 2 (a + b + c) \{a^{2} + b^{2} + c^{2} - a b - b c - c a\} = - (a + b + c) \{2 a^{2} + 2 b^{2} + 2 c^{2} - 2 a b - 2 b c - 2 c a\} = - (a + b + c) \{{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}\} But Δ = 0 [Given] \Rightarrow - (a + b + c) \{{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}\} = 0 \Rightarrow Either (a + b + c) = 0 or {(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2} = 0 \Rightarrow (a + b + c) = 0 or a = b = c Hence proved .$

Page No 5.61:

Question 50:

$If |\begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix}| = 0, find the value of \frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}, p \neq a, q \neq b, r \neq c .$

Answer:

Let $∆ = |\begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix}|$ .
Now,
$∆ = |\begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix}| = |\begin{matrix} p & b & c \\ 0 & q - b & c - r \\ a & b & r \end{matrix}| [Applying R_{2} \to R_{2} - R_{3}] = p [r (q - b) - b (c - r)] + a [b (c - r) - c (q - b)] [Expanding along first column] = p r (q - b) + p b (r - c) - a b (r - c) - a c (q - b) = (p r - a c) (q - b) + b (p - a) (r - c) Since, ∆ = 0 . ∴ (p r - a c) (q - b) + b (p - a) (r - c) = 0 \Rightarrow \frac{p r - a c}{(p - a) (r - c)} + \frac{b}{q - b} = 0 \Rightarrow \frac{p r - a r + a r - a c}{(p - a) (r - c)} + \frac{b}{q - b} = 0 \Rightarrow \frac{r (p - a) + a (r - c)}{(p - a) (r - c)} + \frac{b}{q - b} = 0 \Rightarrow \frac{r}{r - c} + \frac{a}{p - a} + \frac{b}{q - b} = 0 \Rightarrow \frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = \frac{p}{p - a} + \frac{q}{q - b} - \frac{a}{p - a} - \frac{b}{q - b} \Rightarrow \frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = \frac{p - a}{p - a} + \frac{q - b}{q - b} \Rightarrow \frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = 2 Hence, the value of \frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} is 2 .$

Page No 5.61:

Question 51:

Show that x = 2 is a root of the equation

$|\begin{matrix} x & - 6 & - 1 \\ 2 & - 3 x & x - 3 \\ - 3 & 2 x & x + 2 \end{matrix}| = 0$ and solve it completely.

Answer:

$Let ∆ = |\begin{matrix} x & - 6 & - 1 \\ 2 & - 3 x & x - 3 \\ - 3 & 2 x & x + 2 \end{matrix}| = |\begin{matrix} x & - 6 & - 1 \\ 2 & - 3 x & x - 3 \\ - 3 - x & 2 x + 6 & x + 3 \end{matrix}| [Applying R_{3} \to R_{3} - R_{1}] = (x + 3) |\begin{matrix} x & - 6 & - 1 \\ 2 & - 3 x & x - 3 \\ - 1 & 2 & 1 \end{matrix}| = (x + 3) |\begin{matrix} x - 2 & 3 x - 6 & - x + 2 \\ 2 & - 3 x & x - 3 \\ - 1 & 2 & 1 \end{matrix}| [Applying R_{1} \to R_{1} - R_{2}] = (x + 3) (x - 2) |\begin{matrix} 1 & 3 & - 1 \\ 2 & - 3 x & x - 3 \\ - 1 & 2 & 1 \end{matrix}| = (x + 3) (x - 2) |\begin{matrix} 1 & 3 & 0 \\ 2 & - 3 x & x - 1 \\ - 1 & 2 & 0 \end{matrix}| [Applying C_{3} \to C_{3} + C_{1}] = (x + 3) (x - 2) (x - 1) |\begin{matrix} 1 & 3 & 0 \\ 2 & - 3 x & 1 \\ - 1 & 2 & 0 \end{matrix}| = (x + 3) (x - 2) (x - 1) \{- 1 |\begin{matrix} 1 & 3 \\ - 1 & 2 \end{matrix}|\} [Expanding along C_{3}] = - 5 (x + 3) (x - 2) (x - 1) x = 2, - 3, 1$

Page No 5.61:

Question 52:

Solve the following determinant equations:

(i) $|\begin{matrix} x + a & b & c \\ a & x + b & c \\ a & b & x + c \end{matrix}| = 0$

(ii) $|\begin{matrix} x + a & x & x \\ x & x + a & x \\ x & x & x + a \end{matrix}| = 0, a \neq 0$

(iii) $|\begin{matrix} 3 x - 8 & 3 & 3 \\ 3 & 3 x - 8 & 3 \\ 3 & 3 & 3 x - 8 \end{matrix}| = 0$

(iv) $|\begin{matrix} 1 & x & x^{2} \\ 1 & a & a^{2} \\ 1 & b & b^{2} \end{matrix}| = 0, a \neq b$

(v) $|\begin{matrix} x + 1 & 3 & 5 \\ 2 & x + 2 & 5 \\ 2 & 3 & x + 4 \end{matrix}| = 0$

(vi) $|\begin{matrix} 1 & x & x^{3} \\ 1 & b & b^{3} \\ 1 & c & c^{3} \end{matrix}| = 0, b \neq c$

(vii) $|\begin{matrix} 15 - 2 x & 11 - 3 x & 7 - x \\ 11 & 17 & 14 \\ 10 & 16 & 13 \end{matrix}| = 0$

(viii) $|\begin{matrix} 1 & 1 & x \\ p + 1 & p + 1 & p + x \\ 3 & x + 1 & x + 2 \end{matrix}| = 0$

(ix) $|\begin{matrix} 3 & - 2 & \sin (3 θ) \\ - 7 & 8 & \cos (2 θ) \\ - 11 & 14 & 2 \end{matrix}| = 0$

(x) $|\begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 + x \end{matrix}| = 0$

Answer:

(x) Given: $|\begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 + x \end{matrix}| = 0$

$LHS = |\begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 + x \end{matrix}| Applying R_{2} \to R_{2} - R_{1} = |\begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x - 4 + x & 4 - x - 4 - x & 4 + x - 4 - x \\ 4 + x & 4 + x & 4 + x \end{matrix}| = |\begin{matrix} 4 - x & 4 + x & 4 + x \\ 2 x & - 2 x & 0 \\ 4 + x & 4 + x & 4 + x \end{matrix}| Taking (2 x) common from R_{2} = (2 x) |\begin{matrix} 4 - x & 4 + x & 4 + x \\ 1 & - 1 & 0 \\ 4 + x & 4 + x & 4 + x \end{matrix}| Applying R_{3} \to R_{3} - R_{1} = (2 x) |\begin{matrix} 4 - x & 4 + x & 4 + x \\ 1 & - 1 & 0 \\ 4 + x - 4 + x & 4 + x - 4 - x & 4 + x - 4 - x \end{matrix}| = (2 x) |\begin{matrix} 4 - x & 4 + x & 4 + x \\ 1 & - 1 & 0 \\ 2 x & 0 & 0 \end{matrix}| Expanding through C_{3} = (2 x) [(4 + x) (1 \times 0 + 1 \times 2 x)] = (2 x) [(4 + x) (2 x)] = {(2 x)}^{2} (4 + x) Thus, |\begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 + x \end{matrix}| = {(2 x)}^{2} (4 + x) But it is given that, |\begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 + x \end{matrix}| = 0 \Rightarrow {(2 x)}^{2} (4 + x) = 0 \Rightarrow {(2 x)}^{2} = 0 or (4 + x) = 0 \Rightarrow x = 0 or x = - 4$

Hence, x = 0, −4.

Page No 5.62:

Question 53:

If $a, b$ and $c$ are all non-zero and $|\begin{matrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{matrix}| =$ 0, then prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} +$ 1 $=$ 0.

Answer:

We have,

$|\begin{matrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{matrix}| =$ 0

$C_{1} \to C_{1} - C_{2} |\begin{matrix} a & 1 & 1 \\ - b & 1 + b & 1 \\ 0 & 1 & 1 + c \end{matrix}| = 0 C_{2} \to C_{2} - C_{3} |\begin{matrix} a & 0 & 1 \\ - b & b & 1 \\ 0 & - c & 1 + c \end{matrix}| = 0 Expanding along R_{1}, we get a (b + b c + c) + 1 (b c) = 0 \Rightarrow a b + a b c + a c + b c = 0 Dividing by a b c, we get \frac{1}{c} + 1 + \frac{1}{b} + \frac{1}{a} = 0 ∴ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 1 = 0$

Page No 5.62:

Question 54:

If $|\begin{matrix} a & b - y & c - z \\ a - x & b & c - z \\ a - x & b - y & c \end{matrix}| =$ 0, then using properties of determinants, find the value of $\frac{a}{x} + \frac{b}{y} + \frac{c}{z}$ , where $x, y, z \neq$ 0.

Answer:

$|\begin{matrix} a & b - y & c - z \\ a - x & b & c - z \\ a - x & b - y & c \end{matrix}| =$ 0

$R_{1} \to R_{1} - R_{2} |\begin{matrix} x & - y & 0 \\ a - x & b & c - z \\ a - x & b - y & c \end{matrix}| = 0 R_{2} \to R_{2} - R_{3} \Rightarrow |\begin{matrix} x & - y & 0 \\ 0 & y & - z \\ a - x & b - y & c \end{matrix}| = 0 Expanding along first row, we get x (y c + z b - z y) + y (0 - z a + z x) = 0 \Rightarrow x y c + x z b - x y z + z y a - x y z = 0 x y c + x z b - 2 x y z + z y a = 0 Dividing by xyz, we get \frac{c}{z} + \frac{b}{y} - 2 + \frac{a}{x} = 0 ∴ \frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 2$

Page No 5.62:

Question 55:

Using properties of determinants, prove that $|\begin{matrix} 1 & 1 & 1 + 3 x \\ 1 + 3 y & 1 & 1 \\ 1 & 1 + 3 z & 1 \end{matrix}| = 9 (3 x y z + x y + y z + z x) .$

Answer:

$Let ∆ = |\begin{matrix} 1 & 1 & 1 + 3 x \\ 1 + 3 y & 1 & 1 \\ 1 & 1 + 3 z & 1 \end{matrix}| Applying R_{2} \to R_{2} - R_{1,} R_{3} \to R_{3} - R_{1} \Rightarrow ∆ = |\begin{matrix} 1 & 1 & 1 + 3 x \\ 3 y & 0 & - 3 x \\ 0 & 3 z & - 3 x \end{matrix}|$
Expanding along R₁,we get
$\begin{array}{rcl} ∆ & = & 1 (0 + 9 x z) - 1 (- 9 x y - 0) + (1 + 3 x) (9 y z - 0) \\ = & 9 x z + 9 x y + 9 y z + 27 x y z \\ = & 9 (3 x y z + x y + y z + z x) \end{array}$

Page No 5.71:

Question 1:

Find the area of the triangle with vertices at the points:
(i) (3, 8), (−4, 2) and (5, −1)
(ii) (2, 7), (1, 1) and (10, 8)
(iii) (−1, −8), (−2, −3) and (3, 2)
(iv) (0, 0), (6, 0) and (4, 3).

Answer:

(i)
$∆ = \frac{1}{2} |\begin{matrix} 3 & 8 & 1 \\ - 4 & 2 & 1 \\ 5 & - 1 & 1 \end{matrix}| ∆ = \frac{1}{2} |\begin{matrix} 3 & 8 & 1 \\ - 7 & - 6 & 0 \\ 5 & - 1 & 1 \end{matrix}| [Applying R_{2} \to R_{2} - R_{1}] ∆ = \frac{1}{2} |\begin{matrix} 3 & 8 & 1 \\ - 7 & - 6 & 0 \\ 2 & - 9 & 0 \end{matrix}| [Applying R_{3} \to R_{3} - R_{1}] ∆ = \frac{1}{2} |\begin{matrix} - 7 & - 6 \\ 2 & - 9 \end{matrix}| ∆ = \frac{1}{2} (63 + 12) ∆ = \frac{1}{2} (75) = \frac{75}{2} square units$

(ii)
$∆ = \frac{1}{2} |\begin{matrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{matrix}| ∆ = \frac{1}{2} |\begin{matrix} 2 & 7 & 1 \\ - 1 & - 6 & 0 \\ 10 & 8 & 1 \end{matrix}| [Applying R_{2} \to R_{2} - R_{1}] ∆ = \frac{1}{2} |\begin{matrix} 2 & 7 & 1 \\ - 1 & - 6 & 0 \\ 8 & 1 & 0 \end{matrix}| [Applying R_{3} \to R_{3} - R_{1}] ∆ = \frac{1}{2} |\begin{matrix} - 1 & - 6 \\ 8 & 1 \end{matrix}| ∆ = \frac{1}{2} (- 1 + 48) ∆ = \frac{1}{2} (47) = \frac{47}{2} square units$

(iii)
$∆ = \frac{1}{2} |\begin{matrix} - 1 & - 8 & 1 \\ - 2 & - 3 & 1 \\ 3 & 2 & 1 \end{matrix}| ∆ = \frac{1}{2} |\begin{matrix} - 1 & - 8 & 1 \\ - 1 & 5 & 0 \\ 3 & 2 & 1 \end{matrix}| [Applying R_{2} \to R_{2} - R_{1}] ∆ = \frac{1}{2} |\begin{matrix} - 1 & - 8 & 1 \\ - 1 & 5 & 0 \\ 4 & 10 & 0 \end{matrix}| [Applying R_{3} \to R_{3} - R_{1}] ∆ = \frac{1}{2} |\begin{matrix} - 1 & 5 \\ 4 & 10 \end{matrix}| ∆ = \frac{1}{2} |- 10 - 20| ∆ = \frac{1}{2} (30) = 15 square units$

(iv)
$∆ = \frac{1}{2} |\begin{matrix} 0 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{matrix}| ∆ = \frac{1}{2} |\begin{matrix} 0 & 0 & 1 \\ 6 & 0 & 0 \\ 4 & 3 & 1 \end{matrix}| [Applying R_{2} \to R_{2} - R_{1}] ∆ = \frac{1}{2} |\begin{matrix} 0 & 0 & 1 \\ 6 & 0 & 0 \\ 4 & 3 & 0 \end{matrix}| [Applying R_{3} \to R_{3} - R_{1}] ∆ = \frac{1}{2} |\begin{matrix} 6 & 0 \\ 4 & 3 \end{matrix}| ∆ = \frac{1}{2} (18 - 0) ∆ = \frac{1}{2} (18) = 9 square units$

Page No 5.71:

Question 2:

Using determinants show that the following points are collinear:
(i) (5, 5), (−5, 1) and (10, 7)
(ii) (1, −1), (2, 1) and (4, 5)
(iii) (3, −2), (8, 8) and (5, 2)
(iv) (2, 3), (−1, −2) and (5, 8)

Answer:

(i) If the points (5, 5), (−5, 1) and (10, 7) are collinear, then

$∆ = |\begin{matrix} 5 & 5 & 1 \\ - 5 & 1 & 1 \\ 10 & 7 & 1 \end{matrix}| = 0 = |\begin{matrix} 5 & 5 & 1 \\ - 10 & - 4 & 0 \\ 10 & 7 & 1 \end{matrix}| [Applying R_{2} \to R_{2} - R_{1}] = |\begin{matrix} 5 & 5 & 1 \\ - 10 & - 4 & 0 \\ 5 & 2 & 0 \end{matrix}| [Applyin g R_{3} \to R_{3} - R_{1}] = |\begin{matrix} - 10 & - 4 \\ 5 & 2 \end{matrix}| = - 20 + 20 = 0$

Thus, these points are colinear.

(ii) If the points (1, −1), (2, 1) and (4, 5) are collinear, then

$∆ = |\begin{matrix} 1 & - 1 & 1 \\ 2 & 1 & 1 \\ 4 & 5 & 1 \end{matrix}| = 0 = |\begin{matrix} 1 & - 1 & 1 \\ 1 & 2 & 0 \\ 4 & 5 & 1 \end{matrix}| [Applying R_{2} \to R_{2} - R_{1}] = |\begin{matrix} 1 & - 1 & 1 \\ 1 & 2 & 0 \\ 3 & 6 & 0 \end{matrix}| [Applying R_{3} \to R_{3} - R_{1}] = |\begin{matrix} 1 & 2 \\ 3 & 6 \end{matrix}| = 6 - 6 = 0$

Thus, these points are collinear.

(iii) If the points (3, −2), (8, 8) and (5, 2) are collinear, then

$∆ = |\begin{matrix} 3 & - 2 & 1 \\ 8 & 8 & 1 \\ 5 & 2 & 1 \end{matrix}| = 0 = |\begin{matrix} 3 & - 2 & 1 \\ 5 & 10 & 0 \\ 5 & 2 & 1 \end{matrix}| [Applying R_{2} \to R_{2} - R_{1}] = |\begin{matrix} 3 & - 2 & 1 \\ 5 & 10 & 0 \\ 2 & 4 & 0 \end{matrix}| [Applying R_{3} \to R_{3} - R_{1}] = |\begin{matrix} 5 & 10 \\ 2 & 4 \end{matrix}| = 20 - 20 = 0$

Thus the points are colinear.

(iv) If the points (2, 3), (−1, −2) and (5, 8) are collinear, then

$∆ = |\begin{matrix} 2 & 3 & 1 \\ - 1 & - 2 & 1 \\ 5 & 8 & 1 \end{matrix}| = 0 = |\begin{matrix} 2 & 3 & 1 \\ - 3 & - 5 & 0 \\ 5 & 8 & 1 \end{matrix}| [Applying R_{2} \to R_{2} - R_{1}] = |\begin{matrix} 2 & 3 & 1 \\ - 3 & - 5 & 0 \\ 3 & 5 & 0 \end{matrix}| [Applying R_{3} \to R_{3} - R_{1}] = |\begin{matrix} - 3 & - 5 \\ 3 & 5 \end{matrix}| = - 15 + 15 = 0$

Thus the points are colinear.

Page No 5.71:

Question 3:

If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab.

Answer:

If the points (a, 0), (0, b) and (1, 1) are collinear, then

$|\begin{matrix} a & 0 & 1 \\ 0 & b & 1 \\ 1 & 1 & 1 \end{matrix}| = 0 \Rightarrow |\begin{matrix} a & 0 & 1 \\ - a & b & 0 \\ 1 & 1 & 1 \end{matrix}| = 0 [Applying R_{2} \to R_{2} - R_{1}] \Rightarrow |\begin{matrix} a & 0 & 1 \\ - a & b & 0 \\ 1 - a & 1 & 0 \end{matrix}| = 0 [Applying R_{3} \to R_{3} - R_{1}] \Rightarrow ∆ = |\begin{matrix} - a & b \\ 1 - a & 1 \end{matrix}| = 0 \Rightarrow - a - b (1 - a) = 0 \Rightarrow a + b = a b$

Page No 5.71:

Question 4:

Using determinants prove that the points (a, b), (a', b') and (a − a', b − b') are collinear if ab' = a'b.

Answer:

$|\begin{matrix} a & b & 1 \\ a' & b' & 1 \\ a - a' & b - b' & 1 \end{matrix}| \Rightarrow ∆ = |\begin{matrix} a & b & 1 \\ a' - a & b' - b & 0 \\ a - a' & b - b' & 1 \end{matrix}| [Applying R_{2} \to R_{2} - R_{1}] \Rightarrow ∆ = |\begin{matrix} a & b & 1 \\ a' - a & b' - b & 0 \\ - a' & - b' & 0 \end{matrix}| [Applying R_{3} \to R_{3} - R_{1}] \Rightarrow ∆ = |\begin{matrix} a' - a & b' - b \\ - a' & - b' \end{matrix}| \Rightarrow ∆ = - b' (a' - a) + a' (b' - b) = - b' a' + b' a + a' b' - a' b = b' a - a' b$

If the points are collinear, then ∆ = 0. So,
ab' − a'b = 0

Thus, ab' = a'b

Page No 5.71:

Question 5:

Find the value of $λ$ so that the points (1, −5), (−4, 5) and $(λ, 7)$ are collinear.

Answer:

If the points (1, −5), (−4, 5) and $(λ, 7)$ are collinear, then

$|\begin{matrix} 1 & - 5 & 1 \\ - 4 & 5 & 1 \\ λ & 7 & 1 \end{matrix}| = 0 \Rightarrow |\begin{matrix} 1 & - 5 & 1 \\ - 5 & 10 & 0 \\ λ & 7 & 1 \end{matrix}| = 0 [Applying R_{2} \to R_{2} - R_{1}] \Rightarrow |\begin{matrix} 1 & - 5 & 1 \\ - 5 & 10 & 0 \\ λ - 1 & 12 & 0 \end{matrix}| = 0 [Applying R_{3} \to R_{3} - R_{1}] \Rightarrow ∆ = |\begin{matrix} - 5 & 10 \\ λ - 1 & 12 \end{matrix}| = 0 \Rightarrow - 60 - 10 (λ - 1) = 0 \Rightarrow - 60 - 10 λ + 10 = 0 \Rightarrow - 10 λ = 50 \Rightarrow λ = - 5$

Page No 5.71:

Question 6:

Find the value of x if the area of ∆ is 35 square cms with vertices (x, 4), (2, −6) and (5, 4).

Answer:

$∆ = \frac{1}{2} |\begin{matrix} x & 4 & 1 \\ 2 & - 6 & 1 \\ 5 & 4 & 1 \end{matrix}| = \pm 35 = \frac{1}{2} |\begin{matrix} x & 4 & 1 \\ 2 - x & - 10 & 0 \\ 5 & 4 & 1 \end{matrix}| = \pm 35 [Applying R_{2} \to R_{2} - R_{1}] = \frac{1}{2} |\begin{matrix} x & 4 & 1 \\ 2 - x & - 10 & 0 \\ 5 - x & 0 & 0 \end{matrix}| = \pm 35 [Applying R_{3} \to R_{3} - R_{1}] = \frac{1}{2} |\begin{matrix} 2 - x & - 10 \\ 5 - x & 0 \end{matrix}| = \pm 35 = 0 + 10 (5 - x) = \pm 70 \Rightarrow 50 - 10 x = 70 or 50 - 10 x = - 70 \Rightarrow - 10 x = 20 or - 10 x = - 120 \Rightarrow x = - 2 or x = 12$

Page No 5.71:

Question 7:

Using determinants, find the area of the triangle whose vertices are (1, 4), (2, 3) and (−5, −3). Are the given points collinear?

Answer:

$∆ = \frac{1}{2} |\begin{matrix} 1 & 4 & 1 \\ 2 & 3 & 1 \\ - 5 & - 3 & 1 \end{matrix}| = \frac{1}{2} |\begin{matrix} 1 & 4 & 1 \\ 1 & - 1 & 0 \\ - 5 & - 3 & 1 \end{matrix}| [Applying R_{2} \to R_{2} - R_{1}] = \frac{1}{2} |\begin{matrix} 1 & 4 & 1 \\ 1 & - 1 & 0 \\ - 6 & - 7 & 0 \end{matrix}| [Applying R_{3} \to R_{3} - R_{1}] = \frac{1}{2} |\begin{matrix} 1 & - 1 \\ - 6 & - 7 \end{matrix}| = \frac{1}{2} (- 7 - 6) = \frac{13}{2} square units [∵ Area cannot be negative]$

Therefore, (1, 4), (2, 3) and (−5, −3) are not collinear because, $|\begin{matrix} 1 & 4 & 1 \\ 2 & 3 & 1 \\ - 5 & - 3 & 1 \end{matrix}|$ is not equal to 0.

Page No 5.71:

Question 8:

Using determinants, find the area of the triangle with vertices (−3, 5), (3, −6), (7, 2).

Answer:

Given:
Vertices of triangle: (− 3, 5), (3, − 6) and (7, 2)

$Area of the triangle = ∆ = \frac{1}{2} |\begin{matrix} - 3 & 5 & 1 \\ 3 & - 6 & 1 \\ 7 & 2 & 1 \end{matrix}| = \frac{1}{2} |\begin{matrix} - 3 & 5 & 1 \\ 6 & - 11 & 0 \\ 7 & 2 & 1 \end{matrix}| [Applying R_{2} \to R_{2} - R_{1}] = \frac{1}{2} |\begin{matrix} - 3 & 5 & 1 \\ 6 & - 11 & 0 \\ 10 & - 3 & 0 \end{matrix}| [Applying R_{3} \to R_{3} - R_{1}] = \frac{1}{2} |\begin{matrix} 6 & - 11 \\ 10 & - 3 \end{matrix}| = \frac{1}{2} (- 18 + 110) = 46 square units$

Page No 5.71:

Question 9:

Using determinants, find the value of k so that the points (k, 2 − 2 k), (−k + 1, 2k) and (−4 − k, 6 − 2k) may be collinear.

Answer:

If the points (k, 2 − 2 k), (− k + 1, 2k) and (− 4 − k, 6 − 2k) are collinear, then

$∆ = |\begin{matrix} k & 2 - 2 k & 1 \\ - k + 1 & 2 k & 1 \\ - 4 - k & 6 - 2 k & 1 \end{matrix}| = 0 \Rightarrow |\begin{matrix} k & 2 - 2 k & 1 \\ - 2 k + 1 & 4 k - 2 & 0 \\ - 4 - k & 6 - 2 k & 1 \end{matrix}| = 0 [Applying R_{2} \to R_{2} - R_{1}] \Rightarrow |\begin{matrix} k & 2 - 2 k & 1 \\ - 2 k + 1 & 4 k - 2 & 0 \\ - 4 - 2 k & 4 & 0 \end{matrix}| = 0 [Applying R_{3} \to R_{3} - R_{1}] \Rightarrow |\begin{matrix} - 2 k + 1 & 4 k - 2 \\ - 4 - 2 k & 4 \end{matrix}| = 0 \Rightarrow - 8 k + 4 + 16 k - 8 + 8 k^{2} - 4 k = 0 \Rightarrow 8 k^{2} + 4 k - 4 = 0 \Rightarrow (8 k - 4) (k + 1) = 0 \Rightarrow k = - 1 or k = \frac{1}{2}$

Page No 5.71:

Question 10:

If the points (x, −2), (5, 2), (8, 8) are collinear, find x using determinants.

Answer:

If the points (x, −2), (5, 2), (8, 8) are collinear, then

$|\begin{matrix} x & - 2 & 1 \\ 5 & 2 & 1 \\ 8 & 8 & 1 \end{matrix}| = 0 ∆ = |\begin{matrix} x & - 2 & 1 \\ 5 & 2 & 1 \\ 8 & 8 & 1 \end{matrix}| ∆ = |\begin{matrix} x & - 2 & 1 \\ 5 - x & 4 & 0 \\ 8 & 8 & 1 \end{matrix}| [Applying R_{2} \to R_{2} - R_{1}] = |\begin{matrix} x & - 2 & 1 \\ 5 - x & 4 & 0 \\ 8 - x & 10 & 0 \end{matrix}| [Applying R_{3} \to R_{3} - R_{1}] = |\begin{matrix} 5 - x & 4 \\ 8 - x & 10 \end{matrix}| = 50 - 10 x - 32 + 4 x = 18 - 6 x ∆ = 18 - 6 x ∆ = 0 [Given] \Rightarrow 18 - 6 x = 0 \Rightarrow x = 3$

Page No 5.72:

Question 11:

If the points (3, −2), (x, 2), (8, 8) are collinear, find x using determinant.

Answer:

If the points (3, −2), (x, 2) and (8, 8) are collinear, then

$|\begin{matrix} 3 & - 2 & 1 \\ x & 2 & 1 \\ 8 & 8 & 1 \end{matrix}| = 0 ∆ = |\begin{matrix} 3 & - 2 & 1 \\ x & 2 & 1 \\ 8 & 8 & 1 \end{matrix}| = |\begin{matrix} 3 & - 2 & 1 \\ x - 3 & 4 & 0 \\ 8 & 8 & 1 \end{matrix}| [Applying R_{2} \to R_{2} - R_{1}] = |\begin{matrix} 3 & - 2 & 1 \\ x - 3 & 4 & 0 \\ 5 & 10 & 0 \end{matrix}| [Applying R_{3} \to R_{3} - R_{1}] = |\begin{matrix} x - 3 & 4 \\ 5 & 10 \end{matrix}| = 10 x - 30 - 20 ∆ = 10 x - 50 ∆ = 0 [Given] \Rightarrow 10 x - 50 = 0 \Rightarrow 10 x = 50 \Rightarrow x = 5$

Page No 5.72:

Question 12:

Using determinants, find the equation of the line joining the points
(i) (1, 2) and (3, 6)
(ii) (3, 1) and (9, 3)

Answer:

(i)
Given: A = (1, 2) and B = (3, 6)

Let the point P be (x, y). So,
Area of triangle ABP = 0

$\Rightarrow ∆ = \frac{1}{2} |\begin{matrix} 1 & 2 & 1 \\ 3 & 6 & 1 \\ x & y & 1 \end{matrix}| = 0 \Rightarrow 1 (6 - y) - 2 (3 - x) + 1 (3 y - 6 x) = 0 \Rightarrow 6 - y - 6 + 2 x + 3 y - 6 x = 0 \Rightarrow 2 y - 4 x = 0 \Rightarrow y = 2 x$

(ii)
Given: A = (3, 1) and B = (9, 3)

Let the point P be (x, y). So,
Area of triangle ABP = 0

$\Rightarrow ∆ = \frac{1}{2} |\begin{matrix} 3 & 1 & 1 \\ 9 & 3 & 1 \\ x & y & 1 \end{matrix}| = 0 \Rightarrow 3 (3 - y) - 1 (9 - x) + 1 (9 y - 3 x) = 0 \Rightarrow 9 - 3 y - 9 + x + 9 y - 3 x = 0 \Rightarrow - 2 x + 6 y = 0 \Rightarrow x = 3 y$

Page No 5.72:

Question 13:

Find values of k, if area of triangle is 4 square units whose vertices are
(i) (k, 0), (4, 0), (0, 2)
(ii) (−2, 0), (0, 4), (0, k)

Answer:

$(i) If the area of a triangle with vertices (k, 0), (4, 0) and (0, 2) is 4 square units, then Δ = \frac{1}{2} |k 0 1 4 0 1 0 2 1| = \frac{1}{2} \{(2) \times |k 1 4 1|\} [Expanding along C_{2}] = (k - 4) Since area is always + ve, we take its absolute value, which is given as 4 square units . \Rightarrow (k - 4) = \pm 4 \Rightarrow (k - 4) = 4 or (k - 4) = - 4 \Rightarrow k - 4 = 4 or k - 4 = - 4 \Rightarrow k = 8 or k = 0 \Rightarrow k = 8, 0 (ii) If the area of a triangle with vertices (- 2, 0) (0, 4) and (0, k) is 4 square units, then ∆_{1} = \frac{1}{2} |- 2 0 1 0 4 1 0 k 1| = \frac{1}{2} \{- 2 \times |4 1 k 1|\} [Expanding along C_{1}] = - (4 - k) Since area is always + ve, we take its absolute value, which is given as 4 square units . \Rightarrow - (4 - k) = \pm 4 \Rightarrow - (4 - k) = \pm 4 \Rightarrow - (4 - k) = 4 or - (4 - k) = - 4 \Rightarrow k = 4 + 4 or k = - 4 + 4 \Rightarrow k = 8 or k = 0$

Page No 5.84:

Question 1:

x − 2y = 4
−3x + 5y = −7

Answer:

$Given : x - 2 y = 4 - 3 x + 5 y = - 7 Using the properties of determinants, we get D = |1 - 2 - 3 5| = 5 - 6 = - 1 \neq 0 D_{1} = |4 - 2 - 7 5| = 20 - 14 = 6 D_{2} = |1 4 - 3 - 7| = - 7 + 12 = 5 Using Cra mer' s Rule, we get x = \frac{D_{1}}{D} = \frac{6}{- 1} = - 6 y = \frac{D_{2}}{D} = \frac{5}{- 1} = - 5 ∴ x = - 6 and y = - 5$

Page No 5.84:

Question 2:

2x − y = 1
7x − 2y = −7

Answer:

$Given : 2 x - y = 1 7 x - 2 y = - 7 Using Crammer' s Rule, we get D = |2 - 1 7 - 2| = - 4 + 7 = 3 D_{1} = |1 - 1 - 7 - 2| = - 2 - 7 = - 9 D_{2} = |2 1 7 - 7| = - 14 - 7 = - 21 Now, x = \frac{D_{1}}{D} = \frac{- 9}{3} = - 3 y = \frac{D_{2}}{D} = \frac{- 21}{3} = - 7 ∴ x = - 3 and y = - 7$

Page No 5.84:

Question 3:

2x − y = 17
3x + 5y = 6

Answer:

$Given : 2 x - y = 17 3 x + 5 y = 6 Using Cramers Rule, we get D = |2 - 1 3 5| = 10 + 3 = 13 D_{1} = |17 - 1 6 5| = 85 + 6 = 91 D_{2} = |2 17 3 6| = 12 - 51 = - 39 Now, x = \frac{D_{1}}{D} = \frac{91}{13} = 7 y = \frac{D_{2}}{D} = \frac{- 39}{13} = - 3 ∴ x = 7 and y = - 3$

Page No 5.84:

Question 4:

3x + y = 19
3x − y = 23

Answer:

$Given : 3 x + y = 19 3 x - y = 23 Using Cramer' s Rule, we get D = |3 1 3 - 1| = - 3 - 3 = - 6 D_{1} = |19 1 23 - 1| = - 19 - 23 = - 42 D_{2} = |3 19 3 23| = (3 \times 23) - (3 \times 19) = 3 \times 4 = 12 Now, x = \frac{D_{1}}{D} = \frac{- 42}{- 6} = 7 y = \frac{D_{2}}{D} = \frac{12}{- 6} = - 2 ∴ x = 7 and y = - 2$

Page No 5.84:

Question 5:

2x − y = − 2
3x + 4y = 3

Answer:

$Given : 2 x - y = - 2 3 x + 4 y = 3 Using Cramer' s Rule, we get D = |2 - 1 3 4| = 8 + 3 = 11 D_{1} = |- 2 - 1 3 4| = - 8 + 3 = - 5 D_{2} = |2 - 2 3 3| = 6 + 6 = 12 Now, x = \frac{D_{1}}{D} = \frac{- 5}{11} y = \frac{D_{2}}{D} = \frac{12}{11} ∴ x = - \frac{5}{11} and y = \frac{12}{11}$

Page No 5.84:

Question 6:

3x + ay = 4
2x + ay = 2, a ≠ 0

Answer:

$Given : 3 x + a y = 4 2 x + a y = 2 Using Cramer' s rule, we get D = |3 a 2 a| = 3 a - 2 a = a D_{1} = |4 a 2 a| = 4 a - 2 a = 2 a D_{2} = |3 4 2 2| = 6 - 8 = - 2 Now, x = \frac{D_{1}}{D} = \frac{2 a}{a} = 2 y = \frac{D_{2}}{D} = \frac{- 2}{a} = - \frac{2}{a} ∴ x = 2 and y = - \frac{2}{a}$

Page No 5.84:

Question 7:

2x + 3y = 10
x + 6y = 4

Answer:

$Given : 2 x + 3 y = 10 x + 6 y = 4 Using Cramer' s Rule, we get D = |2 3 1 6| = 12 - 3 = 9 D_{1} = |10 3 4 6| = 60 - 12 = 48 D_{2} = |2 10 1 4| = 8 - 10 = - 2 Now, x = \frac{D_{1}}{D} = \frac{48}{9} = \frac{16}{3} y = \frac{D_{2}}{D} = \frac{- 2}{9} ∴ x = \frac{16}{3} and y = \frac{- 2}{9}$

Page No 5.84:

Question 8:

5x + 7y = − 2
4x + 6y = − 3

Answer:

$Given : 5 x + 7 y = - 2 4 x + 6 y = - 3 Using Cramer' s Rule, we get D = |5 7 4 6| = 30 - 28 = 2 D_{1} = |- 2 7 - 3 6| = - 12 + 21 = 9 D_{2} = |5 - 2 4 - 3| = - 15 + 8 = - 7 Now, x = \frac{D_{1}}{D} = \frac{9}{2} y = \frac{D_{2}}{D} = \frac{- 7}{2} ∴ x = \frac{9}{2} and y = \frac{- 7}{2}$

Page No 5.84:

Question 9:

9x + 5y = 10
3y − 2x = 8

Answer:

$Given : 9 x + 5 y = 10 3 y - 2 x = 8 Rearranging the second equation, the two equations can be written as 9 x + 5 y = 10 - 2 x + 3 y = 8 Now, D = |9 5 - 2 3| = 27 + 10 = 37 D_{1} = |10 5 8 3| = 30 - 40 = - 10 D_{2} = |9 10 - 2 8| = 72 + 20 = 92 Using Cramer' s rule, we get x = \frac{D_{1}}{D} = \frac{- 10}{37} y = \frac{D_{2}}{D} = \frac{92}{37} ∴ x = \frac{- 10}{37} and y = \frac{92}{37}$

Page No 5.84:

Question 10:

x + 2y = 1
3x + y = 4

Answer:

Given: x + 2y = 1
3x + y = 4

$D = |\begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix}| = - 5 D_{1} = |\begin{matrix} 1 & 2 \\ 4 & 1 \end{matrix}| = - 7 D_{2} = |\begin{matrix} 1 & 1 \\ 3 & 4 \end{matrix}| = 1 Now, x = \frac{D_{1}}{D} = \frac{7}{5} y = \frac{D_{2}}{D} = - \frac{1}{5} ∴ x = \frac{7}{5} and y = - \frac{1}{5}$

Page No 5.84:

Question 11:

3x + y + z = 2
2x − 4y + 3z = − 1
4x + y − 3z = − 11

Answer:

Given: 3x + y + z = 2
2x − 4y + 3z = − 1
4x + y − 3z = − 11

$D = |\begin{matrix} 3 & 1 & 1 \\ 2 & - 4 & 3 \\ 4 & 1 & - 3 \end{matrix}| = 3 (12 - 3) - 2 (- 3 - 1) + 4 (3 + 4) = 27 + 8 + 28 = 63 D_{1} = |\begin{matrix} 2 & 1 & 1 \\ - 1 & - 4 & 3 \\ - 11 & 1 & - 3 \end{matrix}| = 2 (12 - 3) + 1 (- 3 - 1) - 11 (3 + 4) = 18 - 4 - 77 = - 63 D_{2} = |\begin{matrix} 3 & 2 & 1 \\ 2 & - 1 & 3 \\ 4 & - 11 & - 3 \end{matrix}| = 3 (3 + 33) - 2 (- 6 + 11) + 4 (6 + 1) = 108 - 10 + 28 = 126 D_{3} = |\begin{matrix} 3 & 1 & 2 \\ 2 & - 4 & - 1 \\ 4 & 1 & - 11 \end{matrix}| = 3 (44 + 1) - 2 (- 11 - 2) + 4 (- 1 + 8) = 135 + 26 + 28 = 189 Now, x = \frac{D_{1}}{D} = \frac{- 63}{63} = - 1 y = \frac{D_{2}}{D} = \frac{126}{63} = 2 z = \frac{D_{3}}{D} = \frac{189}{63} = 3 ∴ x = - 1, y = 2 and z = 3$

Page No 5.84:

Question 12:

x − 4y − z = 11
2x − 5y + 2z = 39
− 3x + 2y + z = 1

Answer:

Given: x − 4y − z = 11
2x − 5y + 2z = 39
− 3x + 2y + z = 1

$D = |\begin{matrix} 1 & - 4 & - 1 \\ 2 & - 5 & 2 \\ - 3 & 2 & 1 \end{matrix}| = 1 (- 5 - 4) - (- 4) (2 + 6) + (- 1) (4 - 15) = 1 (- 9) - (- 4) (8) + (- 1) (- 11) = 34 D_{1} = |\begin{matrix} 11 & - 4 & - 1 \\ 39 & - 5 & 2 \\ 1 & 2 & 1 \end{matrix}| = 11 (- 5 - 4) - (- 4) (39 - 2) + (- 1) (78 + 5) = 11 (- 9) - (- 4) (37) + (- 1) (83) = - 34 D_{2} = |\begin{matrix} 1 & 11 & - 1 \\ 2 & 39 & 2 \\ - 3 & 1 & 1 \end{matrix}| = 1 (39 - 2) - 11 (2 + 6) + (- 1) (2 + 117) = 1 (37) - 11 (8) + (- 1) (119) = - 170 D_{3} = |\begin{matrix} 1 & - 4 & 11 \\ 2 & - 5 & 39 \\ - 3 & 2 & 1 \end{matrix}| = 1 (- 5 - 78) - (- 4) (2 + 117) + 11 (4 - 15) = 1 (- 83) - (- 4) (119) + 11 (- 11) = 272 Now, x = \frac{D_{1}}{D} = \frac{- 34}{34} = - 1 y = \frac{D_{2}}{D} = \frac{- 170}{34} = - 5 z = \frac{D_{3}}{D} = \frac{272}{34} = 8 ∴ x = −1, y = -5 and z = 8$

Page No 5.84:

Question 13:

6x + y − 3z = 5
x + 3y − 2z = 5
2x + y + 4z = 8

Answer:

Given: 6x + y − 3z = 5
x + 3y − 2z = 5
2x + y + 4z = 8

$D = |\begin{matrix} 6 & 1 & - 3 \\ 1 & 3 & - 2 \\ 2 & 1 & 4 \end{matrix}| = 6 (12 + 2) - 1 (4 + 4) - 3 (1 - 6) = 6 (14) - 1 (8) - 3 (- 5) = 91 D_{1} = |\begin{matrix} 5 & 1 & - 3 \\ 5 & 3 & - 2 \\ 8 & 1 & 4 \end{matrix}| = 5 (12 + 2) - 1 (20 + 16) - 3 (5 - 24) = 5 (14) - 1 (36) - 3 (- 19) = 91 D_{2} = |\begin{matrix} 6 & 5 & - 3 \\ 1 & 5 & - 2 \\ 2 & 8 & 4 \end{matrix}| = 6 (20 + 16) - 5 (4 + 4) - 3 (8 - 10) = 6 (36) - 5 (8) - 3 (- 2) = 182 D_{3} = |\begin{matrix} 6 & 1 & 5 \\ 1 & 3 & 5 \\ 2 & 1 & 8 \end{matrix}| = 6 (24 - 5) - 1 (8 - 10) + 5 (1 - 6) = 6 (19) - 1 (- 2) + 5 (- 5) = 91 Now, x = \frac{D_{1}}{D} = \frac{91}{91} = 1 y = \frac{D_{2}}{D} = \frac{182}{91} = 2 z = \frac{D_{3}}{D} = \frac{91}{91} = 1 ∴ x = 1, y = 2 and z = 1$

Page No 5.84:

Question 14:

x+ y = 5
y + z = 3
x + z = 4

Answer:

These equations can be written as
x + y + 0z = 5
0x + y + z = 3
x + 0y + z = 4

$D = |\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{matrix}| = 1 (1 - 0) - 1 (0 - 1) + 0 (0 - 1) = 1 (1) - 1 (- 1) + 0 = 2 D_{1} = |\begin{matrix} 5 & 1 & 0 \\ 3 & 1 & 1 \\ 4 & 0 & 1 \end{matrix}| = 5 (1 - 0) - 1 (3 - 4) + 0 (0 - 4) = 5 (1) - 1 (- 1) = 6 D_{2} = |\begin{matrix} 1 & 5 & 0 \\ 0 & 3 & 1 \\ 1 & 4 & 1 \end{matrix}| = 1 (3 - 4) - 5 (0 - 1) + 0 (0 - 4) = 1 (- 1) - 5 (- 1) = 4 D_{3} = |\begin{matrix} 1 & 1 & 5 \\ 0 & 1 & 3 \\ 1 & 0 & 4 \end{matrix}| = 1 (4 - 0) - 1 (0 - 3) + 5 (0 - 1) = 1 (4) - 1 (- 3) + 5 (- 1) = 2 Now, x = \frac{D_{1}}{D} = \frac{6}{2} = 3 y = \frac{D_{2}}{D} = \frac{4}{2} = 2 z = \frac{D_{3}}{D} = \frac{2}{2} = 1 ∴ x = 3, y = 2 and z = 1$

Page No 5.84:

Question 15:

2y − 3z = 0
x + 3y = − 4
3x + 4y = 3

Answer:

These equations can be written as
0x + 2y − 3z = 0
x + 3y + 0z = − 4
3x + 4y + 0z = 3

$D = |\begin{matrix} 0 & 2 & - 3 \\ 1 & 3 & 0 \\ 3 & 4 & 0 \end{matrix}| = 0 (0 - 0) - 2 (0 - 0) - 3 (4 - 9) = 15 D_{1} = |\begin{matrix} 0 & 2 & - 3 \\ - 4 & 3 & 0 \\ 3 & 4 & 0 \end{matrix}| = 0 (0 - 0) - 2 (0 - 0) - 3 (- 16 - 9) = 75 D_{2} = |\begin{matrix} 0 & 0 & - 3 \\ 1 & - 4 & 0 \\ 3 & 3 & 0 \end{matrix}| = 0 (0 - 0) - 0 (0 - 0) - 3 (3 + 12) = - 45 D_{3} = |\begin{matrix} 0 & 2 & 0 \\ 1 & 3 & - 4 \\ 3 & 4 & 3 \end{matrix}| = 0 (9 + 16) - 2 (3 + 12) - 0 (4 - 9) = - 30 Now, x = \frac{D_{1}}{D} = \frac{75}{15} = 5 y = \frac{D_{2}}{D} = \frac{- 45}{15} = - 3 z = \frac{D_{3}}{D} = \frac{- 30}{15} = - 2 ∴ x = 5, y = - 3 and z = - 2$

Page No 5.84:

Question 16:

5x − 7y + z = 11
6x − 8y − z = 15
3x + 2y − 6z = 7

Answer:

Given: 5x − 7y + z = 11
6x − 8y − z = 15
3x + 2y − 6z = 7

$D = |\begin{matrix} 5 & - 7 & 1 \\ 6 & - 8 & - 1 \\ 3 & 2 & - 6 \end{matrix}| = 5 (48 + 2) + 7 (- 36 + 3) + 1 (12 + 24) = 5 (50) + 7 (- 33) + 1 (36) = 55 D_{1} = |\begin{matrix} 11 & - 7 & 1 \\ 15 & - 8 & - 1 \\ 7 & 2 & - 6 \end{matrix}| = 11 (48 + 2) + 7 (- 90 + 7) + 1 (30 + 56) = 11 (50) + 7 (- 83) + 1 (86) = 55 D_{2} = |\begin{matrix} 5 & 11 & 1 \\ 6 & 15 & - 1 \\ 3 & 7 & - 6 \end{matrix}| = 5 (- 90 + 7) - 11 (- 36 + 3) + 1 (42 - 45) = 5 (- 83) - 11 (- 33) + 1 (- 3) = - 55 D_{3} = |\begin{matrix} 5 & - 7 & 11 \\ 6 & - 8 & 15 \\ 3 & 2 & 7 \end{matrix}| = 5 (- 56 - 30) + 7 (42 - 45) + 11 (12 + 24) = 5 (- 86) + 7 (- 3) + 11 (36) = - 55 Now, x = \frac{D_{1}}{D} = \frac{55}{55} = 1 y = \frac{D_{2}}{D} = \frac{- 55}{55} = - 1 z = \frac{D_{3}}{D} = \frac{- 55}{55} = - 1 ∴ x = 1, y = - 1 and z = - 1$

Page No 5.84:

Question 17:

2x − 3y − 4z = 29
− 2x + 5y − z = − 15
3x − y + 5z = − 11

Answer:

Given: 2x − 3y − 4z = 29
− 2x + 5y − z = − 15
3x − y + 5z = − 11

$D = |\begin{matrix} 2 & - 3 & - 4 \\ - 2 & 5 & - 1 \\ 3 & - 1 & 5 \end{matrix}| = 2 (25 - 1) + 3 (- 10 + 3) - 4 (2 - 15) = 2 (24) + 3 (- 7) - 4 (- 13) = 79 D_{1} = |\begin{matrix} 29 & - 3 & - 4 \\ - 15 & 5 & - 1 \\ - 11 & - 1 & 5 \end{matrix}| = 29 (25 - 1) + 3 (- 75 - 11) - 4 (15 + 55) = 29 (24) + 3 (- 86) - 4 (70) = 158 D_{2} = |\begin{matrix} 2 & 29 & - 4 \\ - 2 & - 15 & - 1 \\ 3 & - 11 & 5 \end{matrix}| = 2 (- 75 - 11) - 29 (- 10 + 3) - 4 (22 + 45) = 2 (- 86) - 29 (- 7) - 4 (67) = - 237 D_{3} = |\begin{matrix} 2 & - 3 & 29 \\ - 2 & 5 & - 15 \\ 3 & - 1 & - 11 \end{matrix}| = 2 (- 55 - 15) + 3 (22 + 45) + 29 (2 - 15) = 2 (- 70) + 3 (67) + 29 (- 13) = - 316 Now, x = \frac{D_{1}}{D} = \frac{158}{79} = 2 y = \frac{D_{2}}{D} = \frac{- 237}{79} = - 3 z = \frac{D_{3}}{D} = \frac{- 316}{79} = - 4 ∴ x = 2, y = - 3 and z = - 4$

Page No 5.84:

Question 18:

x + y = 1
x + z = − 6
x − y − 2z = 3

Answer:

These equations can be written as
x+ y + 0z = 1
x + 0y + z = − 6
x − y − 2z = 3

$D = |\begin{matrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & - 1 & - 2 \end{matrix}| = 1 (0 + 1) - 1 (- 2 - 1) + 0 (- 1 - 0) = 4 D_{1} = |\begin{matrix} 1 & 1 & 0 \\ - 6 & 0 & 1 \\ 3 & - 1 & - 2 \end{matrix}| = 1 (0 + 1) - 1 (12 - 3) + 0 (6 - 0) = - 8 D_{2} = |\begin{matrix} 1 & 1 & 0 \\ 1 & - 6 & 1 \\ 1 & 3 & - 2 \end{matrix}| = 1 (12 - 3) - 1 (- 2 - 1) + 0 (3 + 6) = 12 D_{3} = |\begin{matrix} 1 & 1 & 1 \\ 1 & 0 & - 6 \\ 1 & - 1 & 3 \end{matrix}| = 1 (0 - 6) - 1 (3 + 6) + 1 (- 1 - 0) = - 16 Now, x = \frac{D_{1}}{D} = \frac{- 8}{4} = - 2 y = \frac{D_{2}}{D} = \frac{12}{4} = 3 z = \frac{D_{3}}{D} = \frac{- 16}{4} = - 4 ∴ x = - 2, y = 3 and z = - 4$

Page No 5.84:

Question 19:

x + y + z + 1 = 0
ax + by + cz + d = 0
a²x + b²y + x²z + d² = 0

Answer:

$These equations can be written as x + y + z = - 1 a x + b y + c z = - d a^{2} x + b^{2} y + x^{2} z = - d^{2} D = |\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix}| = |\begin{matrix} 1 & 0 & 0 \\ a & a - b & b - c \\ a^{2} & a^{2} - b^{2} & b^{2} - c^{2} \end{matrix}| [Applying C_{2} \to C_{1} - C_{2}, C_{3} \to C_{2} - C_{3}] Taking (b - a) and (c - a) common from C_{1} and C_{2}, respectively, we get = (a - b) (b - c) |\begin{matrix} 1 & 0 & 0 \\ a & 1 & 1 \\ a^{2} & a + b & b + c \end{matrix}| = (a - b) (b - c) (c - a) \dots (1) D_{1} = |\begin{matrix} - 1 & 1 & 1 \\ - d & b & c \\ - d^{2} & b^{2} & c^{2} \end{matrix}| = - |\begin{matrix} 1 & 1 & 1 \\ d & b & c \\ d^{2} & b^{2} & c^{2} \end{matrix}| D_{1} = - (d - b) (b - c) (c - d) [Replacing a by d in eq . (1)] D_{2} = |\begin{matrix} 1 & - 1 & 1 \\ a & - d & c \\ a^{2} & - d^{2} & c^{2} \end{matrix}| = - |\begin{matrix} 1 & 1 & 1 \\ a & d & c \\ a^{2} & d^{2} & c^{2} \end{matrix}| D_{2} = - (a - d) (d - c) (c - a) [Replacing b by d in eq . (1)] D_{3} = |\begin{matrix} 1 & 1 & - 1 \\ a & b & - d \\ a^{2} & b^{2} & - d^{2} \end{matrix}| = - |\begin{matrix} 1 & 1 & 1 \\ a & b & d \\ a^{2} & b^{2} & d^{2} \end{matrix}| D_{3} = - (a - b) (b - d) (d - a) [Replacing c by d in eq . (1)] Thus, x = \frac{D_{1}}{D} = - \frac{(d - b) (b - c) (c - d)}{(a - b) (b - c) (c - a)} y = \frac{D_{2}}{D} = - \frac{(a - d) (d - c) (c - a)}{(a - b) (b - c) (c - a)} z = \frac{D_{3}}{D} = - \frac{(a - b) (b - d) (d - a)}{(a - b) (b - c) (c - a)}$

Page No 5.84:

Question 20:

x + y + z + w = 2
x − 2y + 2z + 2w = − 6
2x + y − 2z + 2w = − 5
3x − y + 3z − 3w = − 3

Answer:

$D = |\begin{matrix} 1 & 1 & 1 & 1 \\ 1 & - 2 & 2 & 2 \\ 2 & 1 & - 2 & 2 \\ 3 & - 1 & 3 & - 3 \end{matrix}| 1 |\begin{matrix} - 2 & 2 & 2 \\ 1 & - 2 & 2 \\ - 1 & 3 & - 3 \end{matrix}| - 1 |\begin{matrix} 1 & 2 & 2 \\ 2 & - 2 & 2 \\ 3 & 3 & - 3 \end{matrix}| + 1 |\begin{matrix} 1 & - 2 & 2 \\ 2 & 1 & 2 \\ 3 & - 1 & - 3 \end{matrix}| - 1 |\begin{matrix} 1 & - 2 & 2 \\ 2 & 1 & - 2 \\ 3 & - 1 & 3 \end{matrix}| = 1 [- 2 (6 - 6) - 2 (- 3 + 2) + 2 (3 - 2)] - 1 [1 (6 - 6) - 2 (- 6 - 6) + 2 (6 + 6)] + 1 [1 (- 3 + 2) + 2 (- 6 - 6) + 2 (- 2 - 3)] - 1 [1 (3 - 2) + 2 (6 + 6) + 2 (- 2 - 3)] = 4 - 48 - 35 - 15 = - 94 D_{1} = |\begin{matrix} 2 & 1 & 1 & 1 \\ - 6 & - 2 & 2 & 2 \\ - 5 & 1 & - 2 & 2 \\ - 3 & - 1 & 3 & - 3 \end{matrix}| 2 |\begin{matrix} - 2 & 2 & 2 \\ 1 & - 2 & 2 \\ - 1 & 3 & - 3 \end{matrix}| - 1 |\begin{matrix} - 6 & 2 & 2 \\ - 5 & - 2 & 2 \\ - 3 & 3 & - 3 \end{matrix}| + 1 |\begin{matrix} - 6 & - 2 & 2 \\ - 5 & 1 & 2 \\ - 3 & - 1 & - 3 \end{matrix}| - 1 |\begin{matrix} - 6 & - 2 & 2 \\ - 5 & 1 & - 2 \\ - 3 & - 1 & 3 \end{matrix}| = 2 [- 2 (6 - 6) - 2 (- 3 + 2) + 2 (3 - 2)] - 1 [- 6 (6 - 6) - 2 (15 + 6) + 2 (- 15 - 6)] + 1 [- 6 (- 3 + 2) + 2 (15 + 6) + 2 (5 + 3)] - 1 [- 6 (3 - 2) + 2 (- 15 - 6) + 2 (5 + 3)] = 188 D_{2} = |\begin{matrix} 1 & 2 & 1 & 1 \\ 1 & - 6 & 2 & 2 \\ 2 & - 5 & - 2 & 2 \\ 3 & - 3 & 3 & - 3 \end{matrix}| 1 |\begin{matrix} - 6 & 2 & 2 \\ - 5 & - 2 & 2 \\ - 3 & 3 & - 3 \end{matrix}| - 2 |\begin{matrix} 1 & 2 & 2 \\ 2 & - 2 & 2 \\ 3 & 3 & - 3 \end{matrix}| + 1 |\begin{matrix} 1 & - 6 & 2 \\ 2 & - 5 & 2 \\ 3 & - 3 & - 3 \end{matrix}| - 1 |\begin{matrix} 1 & - 6 & 2 \\ 2 & - 5 & - 2 \\ 3 & - 3 & 3 \end{matrix}| 1 [- 6 (6 - 6) - 2 (15 + 6) + 2 (- 15 - 6)] - 2 [1 (6 - 6) - 2 (- 6 - 6) + 2 (6 + 6)] + 1 [1 (15 + 6) + 6 (- 6 - 6) + 2 (- 6 + 15)] - 1 [1 (- 15 - 6) - 6 (6 + 6) + 2 (- 6 + 15)] = 1 D_{3} = |\begin{matrix} 1 & 1 & 2 & 1 \\ 1 & - 2 & - 6 & 2 \\ 2 & 1 & - 5 & 2 \\ 3 & - 1 & - 3 & - 3 \end{matrix}| 1 |\begin{matrix} - 2 & - 6 & 2 \\ 1 & - 5 & 2 \\ - 1 & - 3 & - 3 \end{matrix}| - 1 |\begin{matrix} 1 & - 6 & 2 \\ 2 & - 5 & 2 \\ 3 & - 3 & - 3 \end{matrix}| + 2 |\begin{matrix} 1 & - 2 & 2 \\ 2 & 1 & 2 \\ 3 & - 1 & - 3 \end{matrix}| - 1 |\begin{matrix} 1 & - 2 & - 6 \\ 2 & 1 & - 5 \\ 3 & - 1 & - 3 \end{matrix}| = 1 [- 2 (15 + 6) + 6 (- 3 + 2) + 2 (- 3 - 5)] - 1 [1 (15 + 6) + 6 (- 6 - 6) + 2 (- 6 + 15)] + 2 [1 (- 3 + 2) + 2 (- 6 - 6) + 2 (- 2 - 3)] - 1 [1 (- 3 - 5) + 2 (- 6 + 15) - 6 (- 2 - 3)] = - 141 D_{4} = |\begin{matrix} 1 & 1 & 1 & 2 \\ 1 & - 2 & 2 & - 6 \\ 2 & 1 & - 2 & - 5 \\ 3 & - 1 & 3 & - 3 \end{matrix}| 1 |\begin{matrix} - 2 & 2 & - 6 \\ 1 & - 2 & - 5 \\ - 1 & 3 & - 3 \end{matrix}| - 1 |\begin{matrix} 1 & 2 & - 6 \\ 2 & - 2 & - 5 \\ 3 & 3 & - 3 \end{matrix}| + 1 |\begin{matrix} 1 & - 2 & - 6 \\ 2 & 1 & - 5 \\ 3 & - 1 & - 3 \end{matrix}| - 2 |\begin{matrix} 1 & - 2 & 2 \\ 2 & 1 & - 2 \\ 3 & - 1 & - 3 \end{matrix}| 1 [- 2 (6 + 15) - 2 (- 3 - 5) - 6 (3 - 2)] - 1 [1 (6 + 15) - 2 (- 6 + 15) - 6 (6 + 6)] + 1 [1 (- 3 - 5) + 2 (- 6 + 15) - 6 (- 2 - 3)] - 2 [1 (- 3 - 2) + 2 (- 6 + 6) + 2 (- 2 - 3)] = 47 Thus, x = \frac{D_{1}}{D} = \frac{188}{- 94} = - 2 y = \frac{D_{2}}{D} = \frac{- 282}{- 94} = 3 z = \frac{D_{3}}{D} = \frac{- 141}{- 94} = 1.5 w = \frac{D_{4}}{D} = \frac{47}{- 94} = - 0.5$

Page No 5.84:

Question 21:

2x − 3z + w = 1
x − y + 2w = 1
− 3y + z + w = 1
x + y + z = 1

Answer:

$D = |\begin{matrix} 2 & 0 & - 3 & 1 \\ 1 & - 1 & 0 & 2 \\ 0 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{matrix}| 2 |\begin{matrix} - 1 & 0 & 2 \\ - 3 & 1 & 1 \\ 1 & 1 & 0 \end{matrix}| - 0 - 3 |\begin{matrix} 1 & - 1 & 2 \\ 0 & - 3 & 1 \\ 1 & 1 & 0 \end{matrix}| - 1 |\begin{matrix} 1 & - 1 & 0 \\ 0 & - 3 & 1 \\ 1 & 1 & 1 \end{matrix}| = 2 [- 1 (0 - 1) - 0 (0 - 1) + 2 (- 3 - 1)] - 3 [1 (0 - 1) + 1 (0 - 1) + 2 (0 + 3)] - 1 [1 (- 3 - 1) + 1 (0 - 1) + 0 (0 + 3)] = - 21 D_{1} = |\begin{matrix} 1 & 0 & - 3 & 1 \\ 1 & - 1 & 0 & 2 \\ 1 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{matrix}| 1 |\begin{matrix} - 1 & 0 & 2 \\ - 3 & 1 & 1 \\ 1 & 1 & 0 \end{matrix}| - 0 - 3 |\begin{matrix} 1 & - 1 & 2 \\ 1 & - 3 & 1 \\ 1 & 1 & 0 \end{matrix}| - 1 |\begin{matrix} 1 & - 1 & 0 \\ 1 & - 3 & 1 \\ 1 & 1 & 1 \end{matrix}| = 1 [- 1 (0 - 1) - 0 (0 - 1) + 2 (- 3 - 1)] - 3 [1 (0 - 1) + 1 (0 - 1) + 2 (1 + 3)] - 1 [1 (- 3 - 1) + 1 (0 - 1) + 2 (1 + 3)] = - 21 D_{2} = |\begin{matrix} 2 & 1 & - 3 & 1 \\ 1 & 1 & 0 & 2 \\ 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{matrix}| = 2 |\begin{matrix} 1 & 0 & 2 \\ 1 & 1 & 1 \\ 1 & 1 & 0 \end{matrix}| - 1 |\begin{matrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{matrix}| + (- 3) |\begin{matrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{matrix}| - 1 |\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}| 2 [1 (0 - 1) + 2 (1 - 1)] - 1 [1 (0 - 1) + 2 (0 - 1)] - 3 [1 (0 - 1) - 1 (0 - 1) + 2 (0 - 1)] - 1 [1 (1 - 1) - 1 (0 - 1)] = 6 D_{3} = |\begin{matrix} 2 & 0 & 1 & 1 \\ 1 & - 1 & 1 & 2 \\ 0 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{matrix}| = 2 |\begin{matrix} - 1 & 1 & 2 \\ - 3 & 1 & 1 \\ 1 & 1 & 0 \end{matrix}| - 0 + 1 |\begin{matrix} 1 & - 1 & 2 \\ 0 & - 3 & 1 \\ 1 & 1 & 0 \end{matrix}| - 1 |\begin{matrix} 1 & - 1 & 1 \\ 0 & - 3 & 1 \\ 1 & 1 & 1 \end{matrix}| = 2 [- 1 (0 - 1) - 1 (0 - 1) + 2 (- 3 - 1)] + 1 [1 (0 - 1) + 1 (0 - 1) + 2 (0 + 3)] - 1 [1 (- 3 - 1) + 1 (0 - 1) + 1 (0 + 3)] = - 6 D_{4} = |\begin{matrix} 2 & 0 & - 3 & 1 \\ 1 & - 1 & 0 & 1 \\ 0 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{matrix}| = 2 |\begin{matrix} - 1 & 0 & 1 \\ - 3 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}| - 0 - 3 |\begin{matrix} 1 & - 1 & 1 \\ 0 & - 3 & 1 \\ 1 & 1 & 1 \end{matrix}| - 1 |\begin{matrix} 1 & - 1 & 0 \\ 0 & - 3 & 1 \\ 1 & 1 & 1 \end{matrix}| = 2 [- 1 (1 - 1) + 1 (- 3 - 1)] - 3 [1 (- 3 - 1) + 1 (0 - 1) + 1 (0 + 3)] - 1 [1 (- 3 - 1) + 1 (0 - 1)] = 3 So, by Cramer' s rule, we obtain x = \frac{D_{1}}{D} = \frac{21}{21} = 1 y = \frac{D_{2}}{D} = \frac{6}{- 21} = - \frac{2}{7} z = \frac{D_{3}}{D} = \frac{- 6}{- 21} = \frac{2}{7} w = \frac{D_{4}}{D} = \frac{3}{- 21} = - \frac{1}{7} Hence, x = 1, y = - \frac{2}{7}, z = \frac{2}{7}, w = - \frac{1}{7}$

Page No 5.84:

Question 22:

2x − y = 5
4x − 2y = 7

Answer:

Given: 2x − y = 5
4x − 2y = 7

$D = |\begin{matrix} 2 & - 1 \\ 4 & - 2 \end{matrix}| = - 4 + 4 = 0 D_{1} = |\begin{matrix} 5 & - 1 \\ 7 & - 2 \end{matrix}| = - 10 + 7 = - 3 D_{2 =} |\begin{matrix} 2 & 5 \\ 4 & 7 \end{matrix}| = 14 - 20 = - 6$

Here, D₁and D₂are non-zero, but D is zero. Thus, the given system of linear equations is inconsistent.

Page No 5.84:

Question 23:

3x + y = 5
− 6x − 2y = 9

Answer:

Given: 3x + y = 5
− 6x − 2y = 9

$D = |\begin{matrix} 3 & 1 \\ - 6 & - 2 \end{matrix}| = - 6 + 6 = 0 D_{1 =} |\begin{matrix} 5 & 1 \\ 9 & - 2 \end{matrix}| = - 10 - 9 = - 19 D_{2} = |\begin{matrix} 3 & 5 \\ - 6 & 9 \end{matrix}| = 27 + 30 = 57$

Here, D₁and D₂are non-zero, but D is zero. Thus, the system of linear equations is inconsistent.

Page No 5.84:

Question 24:

3x − y + 2z = 3
2x + y + 3z = 5
x − 2y − z = 1

Answer:

Given: 3x − y + 2z = 3
2x + y + 3z = 5
x − 2y − z = 1

$D = |\begin{matrix} 3 & - 1 & 2 \\ 2 & 1 & 3 \\ 1 & - 2 & - 1 \end{matrix}| = 3 (- 1 + 6) + 1 (- 2 - 3) + 2 (- 4 - 1) = 0 D_{1 =} |\begin{matrix} 3 & - 1 & 2 \\ 5 & 1 & 3 \\ 1 & - 2 & - 1 \end{matrix}| = 3 (- 1 + 6) + 1 (- 5 - 3) + 2 (- 10 - 1) = - 15 D_{2} = |\begin{matrix} 3 & 3 & 2 \\ 2 & 5 & 3 \\ 1 & 1 & - 1 \end{matrix}| = 3 (- 5 - 3) - 3 (- 2 - 3) + 2 (2 - 5) = - 15 D_{3} = |\begin{matrix} 3 & - 1 & 3 \\ 2 & 1 & 5 \\ 1 & - 2 & 1 \end{matrix}| = 3 (1 + 10) + 1 (2 - 5) + 3 (- 4 - 1) = - 15$

Here, D is zero, but D₁, D₂ and D₃ are non-zero. Thus, the system of linear equations is inconsistent.

Page No 5.84:

Question 25:

3x − y + 2z = 6
2x − y + z = 2
3x + 6y + 5z = 20.

Answer:

Given: 3x − y + 2z = 6
2x − y + z = 2
3x + 6y + 5z = 20

$D = |\begin{matrix} 3 & - 1 & 2 \\ 2 & - 1 & 1 \\ 3 & 6 & 5 \end{matrix}| 3 (- 5 - 6) + 1 (10 - 3) + 2 (12 + 3) = 4$

Since D is non-zero, the system of linear equations is consistent and has a unique solution.

$D_{1} = |\begin{matrix} 6 & - 1 & 2 \\ 2 & - 1 & 1 \\ 20 & 6 & 5 \end{matrix}| = 6 (- 5 - 6) + 1 (10 - 20) + 2 (12 + 20) = - 66 - 10 + 64 = - 12 D_{2} = |\begin{matrix} 3 & 6 & 2 \\ 2 & 2 & 1 \\ 3 & 20 & 5 \end{matrix}| = 3 (10 - 20) - 6 (10 - 3) + 2 (40 - 6) = - 30 - 42 + 68 = - 4 D_{3} = |\begin{matrix} 3 & - 1 & 6 \\ 2 & - 1 & 2 \\ 3 & 6 & 20 \end{matrix}| = 3 (- 20 - 12) + 1 (40 - 6) + 6 (12 + 3) = - 96 + 34 + 90 = 28 Now, x = \frac{D_{1}}{D} = \frac{- 12}{4} = - 3 y = \frac{D_{2}}{D} = \frac{- 4}{4} = - 1 z = \frac{D_{3}}{D} = \frac{28}{4} = 7 ∴ x = - 3, y = - 1 and z = 7$

Page No 5.85:

Question 26:

x − y + z = 3
2x + y − z = 2
− x − 2y + 2z = 1

Answer:

$Using the equations we get D = |\begin{matrix} 1 & - 1 & 1 \\ 2 & 1 & - 1 \\ - 1 & - 2 & 2 \end{matrix}| \Rightarrow 1 (2 - 2) + 1 (4 - 1) + 1 (- 4 + 1) = 0 D_{1} = |\begin{matrix} 3 & - 1 & 1 \\ 2 & 1 & - 1 \\ 1 & - 2 & 2 \end{matrix}| \Rightarrow 3 (2 - 2) + 1 (4 + 1) + 1 (- 4 - 1) = 0 D_{2} = |\begin{matrix} 1 & 3 & 1 \\ 2 & 2 & - 1 \\ - 1 & 1 & 2 \end{matrix}| \Rightarrow 1 (4 + 1) - 3 (4 - 1) + 1 (2 + 2) = 0 D_{3} = |\begin{matrix} 1 & - 1 & 3 \\ 2 & 1 & 2 \\ - 1 & - 2 & 1 \end{matrix}| \Rightarrow 1 (1 + 4) + 1 (2 + 2) + 3 (- 4 + 1) = 0$

Here,
$D = D_{1} = D_{2} = D_{3} = 0$
Thus, the system of linear equations has infinitely many solutions.

Page No 5.85:

Question 27:

x + 2y = 5
3x + 6y = 15

Answer:

$Using the equations, we get D = |\begin{matrix} 1 & 2 \\ 3 & 6 \end{matrix}| = 6 - 6 = 0 D_{1 =} |\begin{matrix} 5 & 2 \\ 15 & 6 \end{matrix}| = 30 - 30 = 0 D_{2} = |\begin{matrix} 1 & 5 \\ 3 & 15 \end{matrix}| = 15 - 15 = 0$

$∴ D = D_{1} = D_{2}$

Hence, the system of linear equation has infinitely many solutions.

Page No 5.85:

Question 28:

x + y − z = 0
x − 2y + z = 0
3x + 6y − 5z = 0

Answer:

$Using the equations we get D = |\begin{matrix} 1 & 1 & - 1 \\ 1 & - 2 & 1 \\ 3 & 6 & - 5 \end{matrix}| \Rightarrow 1 (10 - 6) - 1 (- 5 - 3) - 1 (6 + 6) = 0 D_{1} = |\begin{matrix} 0 & 1 & - 1 \\ 0 & - 2 & 1 \\ 0 & 6 & - 5 \end{matrix}| \Rightarrow 0 (10 - 6) - 1 (0 - 0) - 1 (0 + 0) = 0 D_{2} = |\begin{matrix} 1 & 0 & - 1 \\ 1 & 0 & 1 \\ 3 & 0 & - 5 \end{matrix}| \Rightarrow 1 (0 - 0) - 0 (- 5 - 3) - 1 (0 - 0) = 0 D_{3} = |\begin{matrix} 1 & 1 & 0 \\ 1 & - 2 & 0 \\ 3 & 6 & 0 \end{matrix}| \Rightarrow 1 (0 - 0) - 1 (0 - 0) + 0 (6 + 6) = 0$

$∴ D = D_{1} = D_{2}$

Hence, the system of linear equations has infinitely many solutions.

Page No 5.85:

Question 29:

2x + y − 2z = 4
x − 2y + z = − 2
5x − 5y + z = − 2

Answer:

$Using the equations we get D = |\begin{matrix} 2 & 1 & - 2 \\ 1 & - 2 & 1 \\ 5 & - 5 & 1 \end{matrix}| \Rightarrow 2 (- 2 + 5) - 1 (1 - 5) - 2 (- 5 + 10) = 0 D_{1} = |\begin{matrix} 4 & 1 & - 2 \\ - 2 & - 2 & 1 \\ - 2 & - 5 & 1 \end{matrix}| \Rightarrow 4 (- 2 + 5) - 1 (- 2 + 2) - 2 (10 - 4) = 0 D_{2} = |\begin{matrix} 2 & 4 & - 2 \\ 1 & - 2 & 1 \\ 5 & - 2 & 1 \end{matrix}| \Rightarrow 2 (- 2 + 2) - 4 (1 - 5) - 2 (- 2 + 10) = 0 D_{3} = |\begin{matrix} 2 & 1 & 4 \\ 1 & - 2 & - 2 \\ 5 & - 5 & - 2 \end{matrix}| \Rightarrow 2 (4 - 10) - 1 (- 2 + 10) + 4 (- 5 + 10) = 0$

$∴ D = D_{1} = D_{2} = 0$

Hence, the system of linear equations has infinitely many solutions.

Page No 5.85:

Question 30:

x − y + 3z = 6
x + 3y − 3z = − 4
5x + 3y + 3z = 10

Answer:

$Using the equations, we get D = |\begin{matrix} 1 & - 1 & 3 \\ 1 & 3 & - 3 \\ 5 & 3 & 3 \end{matrix}| = 1 (9 + 9) + 1 (3 + 15) + 3 (3 - 15) = 18 + 18 - 36 = 0 D_{1} = |\begin{matrix} 6 & - 1 & 3 \\ - 4 & 3 & - 3 \\ 10 & 3 & 3 \end{matrix}| = 6 (9 + 9) + 1 (- 12 + 30) + 3 (- 12 - 30) = 108 + 18 - 126 = 0 D_{2} = |\begin{matrix} 1 & 6 & 3 \\ 1 & - 4 & - 3 \\ 5 & 10 & 3 \end{matrix}| = 1 (- 12 + 30) - 6 (3 + 15) + 3 (10 + 20) = 18 - 108 + 90 = 0 D_{3} = |\begin{matrix} 1 & - 1 & 6 \\ 1 & 3 & - 4 \\ 5 & 3 & 10 \end{matrix}| = 1 (30 + 12) + 1 (10 + 20) + 6 (3 - 15) = 42 + 30 - 72 = 0 ∴ D = D_{1} = D_{2} = D_{3} = 0$

Hence, the system of equations has infinitely many solutions.

Page No 5.85:

Question 31:

A salesman has the following record of sales during three months for three items A, B and C which have different rates of commission

Month	Sale of units		Total commission drawn (in Rs)
	A	B	C
Jan	90	100	20	800
Feb	130	50	40	900
March	60	100	30	850

Find out the rates of commission on items A, B and C by using determinant method.

Answer:

Let x, y and z be the rates of commission on items A, B and C respectively. Based on the given data, we get

$90 x + 100 y + 20 z = 800 130 x + 50 y + 40 z = 900 60 x + 100 y + 30 z = 850$

Dividing all the equations by 10 on both sides, we get

$9 x + 10 y + 2 z = 80 13 x + 5 y + 4 z = 90 6 x + 10 y + 3 z = 85$

$D = |\begin{matrix} 9 & 10 & 2 \\ 13 & 5 & 4 \\ 6 & 10 & 3 \end{matrix}| [Expressing the equation as a determinant] = 9 (15 - 40) - 10 (39 - 24) + 2 (130 - 30) = 9 (- 25) - 10 (15) + 2 (100) = - 175 D_{1} = |\begin{matrix} 80 & 10 & 2 \\ 90 & 5 & 4 \\ 85 & 10 & 3 \end{matrix}| = 80 (15 - 40) - 10 (270 - 340) + 2 (900 - 425) = 80 (- 25) - 10 (- 70) + 2 (475) = - 350 D_{2} = |\begin{matrix} 9 & 80 & 2 \\ 13 & 90 & 4 \\ 6 & 85 & 3 \end{matrix}| = 9 (270 - 340) - 80 (39 - 24) + 2 (1105 - 540) = 9 (- 70) - 80 (15) + 2 (565) = - 700 D_{3} = |\begin{matrix} 9 & 10 & 80 \\ 13 & 5 & 90 \\ 6 & 10 & 85 \end{matrix}| = 9 (425 - 900) - 10 (1105 - 540) + 80 (130 - 30) = 9 (- 475) - 10 (565) + 80 (100) = - 1925 Thus, x = \frac{D_{1}}{D} = \frac{- 350}{- 175} = 2 y = \frac{D_{2}}{D} = \frac{- 700}{- 175} = 4 z = \frac{D_{3}}{D} = \frac{- 1925}{- 175} = 11$

Therefore, the rates of commission on items A, B and C are 2, 4 and 11, respectively.

Page No 5.85:

Question 32:

An automobile company uses three types of steel S₁, S₂ and S₃ for producing three types of cars C₁, C₂ and C₃. Steel requirements (in tons) for each type of cars are given below :

	Cars C₁	C₂	C₃
Steel S₁	2	3	4
S₂	1	1	2
S₃	3	2	1

Using Cramer's rule, find the number of cars of each type which can be produced using 29, 13 and 16 tons of steel of three types respectively.

Answer:

$Let x, y and z denote the number of cars that can be produced of each type . Then, 2 x + 3 y + 4 z = 29 x + y + 2 z = 13 3 x + 2 y + z = 16 Using Cramer' s rule, we get D = |\begin{matrix} 2 & 3 & 4 \\ 1 & 1 & 2 \\ 3 & 2 & 1 \end{matrix}| = 2 (1 - 4) - 3 (1 - 6) + 4 (2 - 3) = - 6 + 15 - 4 = 5 D_{1} = |\begin{matrix} 29 & 3 & 4 \\ 13 & 1 & 2 \\ 16 & 2 & 1 \end{matrix}| = 29 (1 - 4) - 3 (13 - 32) + 4 (26 - 16) = - 87 + 57 + 40 = 10 D_{2} = |\begin{matrix} 2 & 29 & 4 \\ 1 & 13 & 2 \\ 3 & 16 & 1 \end{matrix}| = 2 (13 - 32) - 29 (1 - 6) + 4 (16 - 39) = - 38 + 145 - 92 = 15 D_{3} = |\begin{matrix} 2 & 3 & 29 \\ 1 & 1 & 13 \\ 3 & 2 & 16 \end{matrix}| = 2 (16 - 26) - 3 (16 - 39) + 29 (2 - 3) = - 20 + 69 - 29 = 20 Thus, x = \frac{D_{1}}{D} = \frac{10}{5} = 2 y = \frac{D_{2}}{D} = \frac{15}{5} = 3 z = \frac{D_{3}}{D} = \frac{20}{5} = 4$

Therefore, 2 C₁ cars, 3 C₂ cars and 4 C₃ cars can be produced using the three types of steel.

Page No 5.89:

Question 1:

Solve each of the following system of homogeneous linear equations.
x + y − 2z = 0
2x + y − 3z = 0
5x + 4y − 9z = 0

Answer:

Given: x + y − 2z = 0
2x + y − 3z = 0
5x + 4y − 9z = 0

$D = |\begin{matrix} 1 & 1 & - 2 \\ 2 & 1 & - 3 \\ 5 & 4 & - 9 \end{matrix}| = 1 (- 9 + 12) - 1 (- 18 + 15) - 2 (8 - 5) = 0 So, the system has infinitely many solutions . Putting z = k in the first two equations, we get x + y = 2 k 2 x + y = 3 k Using Cramer' s rule, we get x = \frac{D_{1}}{D} = \frac{|\begin{matrix} 2 k & 1 \\ 3 k & 1 \end{matrix}|}{|\begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix}|} = \frac{- k}{- 1} = k y = \frac{D_{2}}{D} = \frac{|\begin{matrix} 1 & 2 k \\ 2 & 3 k \end{matrix}|}{|\begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix}|} = \frac{- k}{- 1} = k z = k Clearly, these values satisfy the third equation . Thus, x = y = z = k [k \in R]$

Page No 5.89:

Question 2:

Solve each of the following system of homogeneous linear equations.
2x + 3y + 4z = 0
x + y + z = 0
2x − y + 3z = 0

Answer:

$D = |\begin{matrix} 2 & 3 & 4 \\ 1 & 1 & 1 \\ 2 & - 1 & 3 \end{matrix}| = 2 (3 + 1) - 3 (3 - 2) + 4 (- 1 - 2) = 8 - 3 - 12 = - 7 So, the given system of equations has only the trivial solution i . e . x = 0, y = 0, z = 0$

Page No 5.89:

Question 3:

Solve each of the following system of homogeneous linear equations.
3x + y + z = 0
x − 4y + 3z = 0
2x + 5y − 2z = 0

Answer:

Given: 3x + y + z = 0
x − 4y + 3z = 0
2x + 5y − 2z = 0

$D = |\begin{matrix} 3 & 1 & 1 \\ 1 & - 4 & 3 \\ 2 & 5 & - 2 \end{matrix}| = 0 The system has infinitely many solutions . Putting z = k in the first two equations, we get 3 x + y = - k x - 4 y = - 3 k Solving these equations by Cramer' s rule, we get x = \frac{D_{1}}{D} = \frac{|\begin{matrix} - k & 1 \\ - 3 k & - 4 \end{matrix}|}{|\begin{matrix} 3 & 1 \\ 1 & - 4 \end{matrix}|} = - \frac{7 k}{13} y = \frac{D_{2}}{D} = \frac{|\begin{matrix} 3 & - k \\ 1 & - 3 k \end{matrix}|}{|\begin{matrix} 3 & 1 \\ 1 & - 4 \end{matrix}|} = \frac{8 k}{13} z = k \Rightarrow x = - \frac{7 k}{13}, y = \frac{8 k}{13} and z = k or x = - 7 k, y = 8 k and z = 13 k Clearly, these values satisfy the third equation . Thus, x = - 7 k y = 8 k [k \in R] z = 13 k$

Page No 5.89:

Question 4:

Find the real values of $λ$ for which the following system of linear equations has non-trivial solutions. Also, find the non-trivial solutions

$2 λ x - 2 y + 3 z = 0 x + λ y + 2 z = 0 2 x + λ z = 0$

Answer:

$The given system of equations can be written as 2 λ x - 2 y + 3 z = 0 x + λ y + 2 z = 0 2 x + 0 y + λ z = 0 The given system of equations will have non - trivial solution s if D = 0 . \Rightarrow |\begin{matrix} 2 λ & - 2 & 3 \\ 1 & λ & 2 \\ 2 & 0 & λ \end{matrix}| = 0 \Rightarrow 2 λ (λ^{2}) + 2 (λ - 4) + 3 (- 2 λ) = 0 \Rightarrow 2 λ^{3} - 4 λ - 8 = 0 \Rightarrow λ = 2 So, the given system of equations will have non - trivial solutions if λ = 2 . Now, we shall find solutions for λ = 2 . Replacing z by k in the first two equations, we get 2 λ x - 2 y = - 3 k x + λ y = - 2 k Solving these by Cramer' s rule, we get x = \frac{|\begin{matrix} - 3 k & - 2 \\ - 2 k & λ \end{matrix}|}{|\begin{matrix} 2 λ & - 2 \\ 1 & λ \end{matrix}|} = \frac{- 3 k λ - 4 k}{2 λ^{2} + 2} = \frac{- 3 k (2) - 4 k}{2 (2)^{2} + 2} = \frac{- 6 k - 4 k}{10} = - k y = \frac{|\begin{matrix} 2 λ & - 3 k \\ 1 & - 2 k \end{matrix}|}{|\begin{matrix} 2 λ & - 2 \\ 1 & λ \end{matrix}|} = \frac{- 4 k λ + 3 k}{2 λ^{2} + 2} = \frac{- 4 k (2) + 3 k}{2 (2)^{2} + 2} = \frac{- 5 k}{10} = \frac{- k}{2} Substituting these values of x and y in the third equation, we get LHS = 2 (- k) + 0 (- \frac{k}{2}) + 2 (k) = 0 = RHS Thus, λ = 2, x = - k, y = - \frac{k}{2} and z = k [k \in R]$

Page No 5.89:

Question 5:

If a, b, c are non-zero real numbers and if the system of equations
(a − 1) x = y + z
(b − 1) y = z + x
(c − 1) z = x + y
has a non-trivial solution, then prove that ab + bc + ca = abc.

Answer:

The three equations can be expressed as

$(a - 1) x - y - z = 0 - x + (b - 1) y - z = 0 - x - y + (c - 1) z = 0$

Expressing this as a determinant, we get

$∆ = |\begin{matrix} (a - 1) & - 1 & - 1 \\ - 1 & (b - 1) & - 1 \\ - 1 & - 1 & (c - 1) \end{matrix}|$

If the matrix has a non-trivial solution, then

$|\begin{matrix} (a - 1) & - 1 & - 1 \\ - 1 & (b - 1) & - 1 \\ - 1 & - 1 & (c - 1) \end{matrix}| = 0$

$\Rightarrow (a - 1) [(b - 1) (c - 1) - 1] + 1 [- (c - 1) - 1] - 1 [1 + b - 1] = 0 \Rightarrow (a - 1) [b c - c - b + 1 - 1] + 1 [- c + 1 - 1] - 1 [b] = 0 \Rightarrow (a - 1) [b c - b - c] - c - b = 0 \Rightarrow a b c - a b - a c - b c + b + c - b - c = 0 \Rightarrow a b + a c + b c = a b c$

Hence proved.

Page No 5.90:

Question 1:

If A and B are square matrices of order 2, then det (A + B) = 0 is possible only when
(a) det (A) = 0 or det (B) = 0
(b) det (A) + det (B) = 0
(c) det (A) = 0 and det (B) = 0
(d) A + B = O

Answer:

(d) A + B = O

$Let A = [a_{i j}] and B = [b_{i j}] be a square matrix of order 2 . As their orders are same, A + B is defined as A + B = [a_{i j} + b_{i j}] \Rightarrow |A + B| = |a_{i j} + b_{i j}| Now, |A + B| = 0 \Rightarrow |a_{i j} + b_{i j}| = 0 \Rightarrow [a_{i j} + b_{i j}] = 0 [each corrsponding term is 0] \Rightarrow A + B = 0$

Page No 5.90:

Question 2:

Which of the following is not correct?
(a) $| A | = | A^{T} |, where A = {[a_{i j}]}_{3 \times 3}$

(b) $| k A | = | k^{3} |, where A = {[a_{i j}]}_{3 \times 3}$

(c) If A is a skew-symmetric matrix of odd order, then |A| = 0

(d) $|\begin{matrix} a + b & c + d \\ e + f & g + h \end{matrix}| = |\begin{matrix} a & c \\ e & g \end{matrix}| + |\begin{matrix} b & d \\ f & h \end{matrix}|$

Answer:

(d) $|\begin{matrix} a + b & c + d \\ e + f & g + h \end{matrix}| = |\begin{matrix} a & c \\ e & g \end{matrix}| + |\begin{matrix} b & d \\ f & h \end{matrix}|$

$|a + b c + d e + f g + h| = |a + b c e + f g| + |a + b d e + f h| = |a c e g| + |b c f g| + |a d e h| + |b d f h|$

Page No 5.90:

Question 3:

If $A = |\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}|$ and C_ij is cofactor of a_ij in A, then value of |A| is given by

(a) a₁₁ C₃₁ + a₁₂C₃₂ + a₁₃ C₃₃
(b) a₁₁ C₁₁ + a₁₂ C₂₁ + a₁₃ C₃₁
(c) a₂₁ C₁₁ + a₂₂ C₁₂ + a₂₃ C₁₃
(d) a₁₁ C₁₁ + a₂₁ C₂₁ + a₁₃ C₃₁

Answer:

(d) a₁₁ C₁₁ + a₂₁ C₂₁ + a₁₃ C₃₁

Properties of determinants state that if A is a square matrix of the order n, then Det (A) is the sum of products of elements of a row (or a column) with the corresponding cofactor of that element.

$|A| = a_{11} C_{11} + a_{21} C_{21} + a_{31'} C_{31} [Calculating along C_{1}]$

Page No 5.90:

Question 4:

Which of the following is not correct in a given determinant of A, where A = [a_ij]_3×3.
(a) Order of minor is less than order of the det (A)
(b) Minor of an element can never be equal to cofactor of the same element
(c) Value of determinant is obtained by multiplying elements of a row or column by corresponding cofactors
(d) Order of minors and cofactors of elements of A is same

Answer:

(b) Minor of an element can never be equal to the cofactor of the same element.

$C_{i j} = {(- 1)}^{i + j} M_{i j} So, for even values of i + j, C_{i j} = M_{i j} .$

Page No 5.90:

Question 5:

Let $|\begin{matrix} x & 2 & x \\ x^{2} & x & 6 \\ x & x & 6 \end{matrix}| = a x^{4} + b x^{3} + c x^{2} + d x + e$

Then, the value of $5 a + 4 b + 3 c + 2 d + e$ is equal to
(a) 0
(b) − 16
(c) 16
(d) none of these

Answer:

(d) none of these

$∆ = |x 2 x x^{2} x 6 x x 6| = x |x 6 x 6| - x^{2} |2 x x 6| + x |2 x x 6| [Expanding along C_{1}] = 0 - x^{2} (12 - x^{2}) + x (12 - x^{2}) = x^{4} - 12 x^{2} + 12 x - x^{3} ∆ = a x^{4} + b x^{3} + c x^{2} + d x + e [Given] \Rightarrow x^{4} - 12 x^{2} + 12 x - x^{3} = a x^{4} + b x^{3} + c x^{2} + d x + e \Rightarrow a = 1, b = - 1, c = - 12, d = 12, e = 0 Thus, 5 a + 4 b + 3 c + 2 d + e = 5 - 4 - 36 + 24 + 0 = - 11$

Page No 5.90:

Question 6:

The value of the determinant

$|\begin{matrix} a^{2} & a & 1 \\ \cos n x & \cos (n + 1) x & \cos (n + 2) x \\ \sin n x & \sin (n + 1) x & \sin (n + 2) x \end{matrix}| is independent of$

(a) n
(b) a
(c) x
(d) none of these

Answer:

(a) n

$Let A = nx, B = (n + 1) x, C = (n + 2) x \Rightarrow C - B = x, B - A = x, C - A = 2 x Thus, the given determinant is |a^{2} a 1 \cos A \cos B \cos C \sin A \sin B \sin C| = a^{2} (\cos B \sin C - \cos C \sin B) - a \times (\cos A \sin C - \cos C \sin A) + 1 \times (\cos A \sin B - \sin A \cos B) = a^{2} \sin (C - B) - a \sin (C - A) + \sin (B - A) = a^{2} \sin x - a \sin 2 x + \sin x [Independent of n]$

Page No 5.90:

Question 7:

If $∆_{1} = |\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix}|, ∆_{2} = |\begin{matrix} 1 & b c & a \\ 1 & c a & b \\ 1 & a b & c \end{matrix}|, then$

(a) $∆_{1} + ∆_{2} = 0$
(b) $∆_{1} + 2 ∆_{2} = 0$
(c) $∆_{1} = ∆_{2}$
(d) none of these

Answer:

(a) $∆_{1} + ∆_{2} = 0$

$Δ_{2 =} |1 b c a 1 c a b 1 a b c| = \frac{1}{a b c} |a a b c a^{2} b b c a b^{2} c c a b c^{2}| [R_{1}, R_{2}, R_{3} are multiplied by a, b and c respectively, therefore we divide by abc] = \frac{a b c}{a b c} |a 1 a^{2} b 1 b^{2} c 1 c^{2}| [Taking a b c common from C_{2}] = - |1 a a^{2} 1 b b^{2} 1 c c^{2}| C_{1} \leftrightarrow C_{2} We know that the value of a determinant remains unchanged if its rows and columns are interchanged . So, ∆_{2} = - |1 1 1 a b c a^{2} b^{2} c^{2}| = - Δ_{1} \Rightarrow Δ_{1} + Δ_{2} = 0$

Page No 5.90:

Question 8:

If $D_{k} = |\begin{matrix} 1 & n & n \\ 2 k & n^{2} + n + 2 & n^{2} + n \\ 2 k - 1 & n^{2} & n^{2} + n + 2 \end{matrix}| and \sum_{k = 1}^{n} D_{k} = 48$ , then n equals

(a) 4
(b) 6
(c) 8
(d) none of these

Answer:

(a) 4

$D_{k} = |1 n n 2 k n^{2} + n + 2 n^{2} + n 2 k - 1 n^{2} n^{2} + n + 2| = |1 n n 1 n + 2 - 2 2 k - 1 n^{2} n^{2} + n + 2| [Applying R_{2} \to R_{2} - R_{3}] = |1 n n 0 2 - 2 - n 2 k - 1 n^{2} n^{2} + n + 2| [Applying R_{2} \to R_{2} - R_{1}] \sum_{k = 1}^{n} D_{k} = |1 n n 0 2 - 2 - n 1 n^{2} n^{2} + n + 2| + |1 n n 0 2 - 2 - n 3 n^{2} n^{2} + n + 2| + . . . + |1 n n 0 2 - 2 - n n n^{2} n^{2} + n + 2| \sum_{k = 1}^{n} D_{k} = 1 (2 (n^{2} + n + 2) + (2 + n) n^{2}) + 1 (n (- 2 - n) - 2 n) + 1 (2 (n^{2} + n + 2) + (2 + n) n^{2}) + 2 (n (- 2 - n) - 2 n) + . . . + 1 (2 (n^{2} + n + 2) + (2 + n) n^{2}) + n (n (- 2 - n) - 2 n) \sum_{k = 1}^{n} D_{k} = n (2 (n^{2} + n + 2) + (2 + n) n^{2}) + (n (- 2 - n) - 2 n) (1 + 3 + 5 + 7 + . . . + n) \sum_{k = 1}^{n} D_{k} = n (2 (n^{2} + n + 2) + (2 + n) n^{2}) + (n (- 2 - n) - 2 n) (n^{2}) \sum_{k = 1}^{n} D_{k} = 2 n^{2} + 4 n \Rightarrow 2 n^{2} + 4 n = 48 \Rightarrow (n - 6) (n - 4) = 0 \Rightarrow n = 4$

Page No 5.90:

Question 9:

Let $|\begin{matrix} x^{2} + 3 x & x - 1 & x + 3 \\ x + 1 & - 2 x & x - 4 \\ x - 3 & x + 4 & 3 x \end{matrix}| = a x^{4} + b x^{3} + c x^{2} + d x + e$

be an identity in x, where a, b, c, d, e are independent of x. Then the value of e is
(a) 4
(b) 0
(c) 1
(d) none of these

Answer:

(b) 0

$Let Δ = |x^{2} + 3 x x - 1 x + 3 x + 1 - 2 x x - 4 x - 3 x + 4 3 x| = (x^{2} + 3 x) |- 2 x x - 4 x + 4 3 x| - (x - 1) |x + 1 x - 4 x - 3 3 x| + (x + 3) |x + 1 - 2 x x - 3 x + 4| = (x^{2} + 3 x) (- 6 x - x^{2} + 16) - (x - 1) (3 x^{2} + 3 x - x^{2} + 7 x - 12) + (x + 3) (x^{2} + 5 x + 4 + 2 x^{2} - 6 x) = - 7 x^{4} + 16 x^{2} + 48 x + 21 x^{3} + 8 x^{2} - 22 x - 2 x^{3} - 12 + 8 x^{2} + x + 3 x^{3} + 12 = - 7 x^{4} + 22 x^{3} + 32 x^{2} + 27 x + 0 But x is a root of a x^{4} + b x^{3} + c x^{2} + d x + e . \Rightarrow e = 0$