Rd Sharma XII Vol 1 2018 for Class 12 Science Math Chapter 7 - Adjoint And Inverse Of Matrix

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Rd Sharma XII Vol 1 2018 Solutions for Class 12 Science Math Chapter 7 Adjoint And Inverse Of Matrix are provided here with simple step-by-step explanations. These solutions for Adjoint And Inverse Of Matrix are extremely popular among Class 12 Science students for Math Adjoint And Inverse Of Matrix Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2018 Book of Class 12 Science Math Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2018 Solutions. All Rd Sharma XII Vol 1 2018 Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

Page No 7.22:

Question 1:

Find the adjoint of each of the following matrices:

(i) $[\begin{matrix} - 3 & 5 \\ 2 & 4 \end{matrix}]$

(ii) $[\begin{matrix} a & b \\ c & d \end{matrix}]$

(iii) $[\begin{matrix} \cos α & \sin α \\ \sin α & \cos α \end{matrix}]$

(iv) $[\begin{matrix} 1 & \tan α / 2 \\ - \tan α / 2 & 1 \end{matrix}]$

Verify that (adj A) A = |A| I = A (adj A) for the above matrices.

Answer:

Given below are the squares matrices. Here, we will interchange the diagonal elements and change the signs of
the off-diagonal elements.

s. $(i) A = [\begin{matrix} - 3 & 5 \\ 2 & 4 \end{matrix}] adj A = [\begin{matrix} 4 & - 5 \\ - 2 & - 3 \end{matrix}] (adj A) A = [\begin{matrix} - 22 & 0 \\ 0 & - 22 \end{matrix}] |A| = - 22 |A| I = [\begin{matrix} - 22 & 0 \\ 0 & - 22 \end{matrix}] A (adj A) = [\begin{matrix} - 22 & 0 \\ 0 & - 22 \end{matrix}] ∴ (adj A) A = |A| I = A (adj A) Hence verified . (i i) B = [\begin{matrix} a & b \\ c & d \end{matrix}] adj B = [\begin{matrix} d & - b \\ - c & a \end{matrix}] (adj B) B = [\begin{matrix} a d - b c & 0 \\ 0 & - c b + a d \end{matrix}] |B| = a d - b c |B| I = [\begin{matrix} a d - b c & 0 \\ 0 & - c b + a d \end{matrix}] B (adj B) = [\begin{matrix} a d - b c & 0 \\ 0 & - c b + a d \end{matrix}] ∴ (adj B) B = |B| I = B (adj B) Hence verified . (i i i) C = [\begin{matrix} \cos α & \sin α \\ \sin α & \cos α \end{matrix}] adj C = [\begin{matrix} \cos α & - \sin α \\ - \sin α & \cos α \end{matrix}] (adj C) C = [\begin{matrix} \cos^{2} α - \sin^{2} α & 0 \\ 0 & \cos^{2} α - \sin^{2} α \end{matrix}] |C| = \cos^{2} α - \sin^{2} α |C| I = [\begin{matrix} \cos^{2} α - \sin^{2} α & 0 \\ 0 & \cos^{2} α - \sin^{2} α \end{matrix}] C (adj C) = [\begin{matrix} \cos^{2} α - \sin^{2} α & 0 \\ 0 & \cos^{2} α - \sin^{2} α \end{matrix}] ∴ (adj C) C = |C| I = C (adj C) Hence verified . (i v) D = [\begin{matrix} 1 & \tan \frac{α}{2} \\ - \tan \frac{α}{2} & 1 \end{matrix}] adj D = [\begin{matrix} 1 & - \tan \frac{α}{2} \\ \tan \frac{α}{2} & 1 \end{matrix}] (adj D) D = [\begin{matrix} 1 + \tan^{2} \frac{α}{2} & 0 \\ 0 & 1 + \tan^{2} \frac{α}{2} \end{matrix}] |D| = 1 + \tan^{2} \frac{α}{2} |D| I = [\begin{matrix} 1 + \tan^{2} \frac{α}{2} & 0 \\ 0 & 1 + \tan^{2} \frac{α}{2} \end{matrix}] D (adj D) = [\begin{matrix} 1 + \tan^{2} \frac{α}{2} & 0 \\ 0 & 1 + \tan^{2} \frac{α}{2} \end{matrix}] ∴ (adj D) D = |D| I = D (adj D) Hence verified .$

Page No 7.22:

Question 2:

Compute the adjoint of each of the following matrices:

(i) $[\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}]$

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

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

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

Verify that (adj A) A = |A| I = A (adj A) for the above matrices.

Answer:

$(i) A = [\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}] Now, C_{11} = |\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix}| = - 3, C_{12} = - |\begin{matrix} 2 & 2 \\ 2 & 1 \end{matrix}| = 2 and C_{13} = |\begin{matrix} 2 & 1 \\ 2 & 2 \end{matrix}| = 2 C_{21} = - |\begin{matrix} 2 & 2 \\ 2 & 1 \end{matrix}| = 2, C_{22} = |\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix}| = - 3 and C_{23} = - |\begin{matrix} 1 & 2 \\ 2 & 2 \end{matrix}| = 2 C_{31} = |\begin{matrix} 2 & 2 \\ 1 & 2 \end{matrix}| = 2, C_{32} = - |\begin{matrix} 1 & 2 \\ 2 & 2 \end{matrix}| = 2 and C_{33} = |\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix}| = - 3 ∴ adj A = {[\begin{matrix} - 3 & 2 & 2 \\ 2 & - 3 & 2 \\ 2 & 2 & - 3 \end{matrix}]}^{T} = [\begin{matrix} - 3 & 2 & 2 \\ 2 & - 3 & 2 \\ 2 & 2 & - 3 \end{matrix}] and (adj A) A = [\begin{matrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{matrix}] Now, |A| = 5 ∴ |A| I = [\begin{matrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{matrix}] and A (adj A) = [\begin{matrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{matrix}] Thus, (adj A) A = |A| I = A (adj A) (ii) B = [\begin{matrix} 1 & 2 & 5 \\ 2 & 3 & 1 \\ - 1 & 1 & 1 \end{matrix}] Now, C_{11} = |\begin{matrix} 3 & 1 \\ 1 & 1 \end{matrix}| = 2, C_{12} = - |\begin{matrix} 2 & 1 \\ - 1 & 1 \end{matrix}| = - 3 and C_{13} = |\begin{matrix} 2 & 3 \\ - 1 & 1 \end{matrix}| = 5 C_{21} = - |\begin{matrix} 2 & 5 \\ 1 & 1 \end{matrix}| = 3, C_{22} = |\begin{matrix} 1 & 5 \\ - 1 & 1 \end{matrix}| = 6 and C_{23} = - |\begin{matrix} 1 & 2 \\ - 1 & 1 \end{matrix}| = - 3 C_{31} = |\begin{matrix} 2 & 5 \\ 3 & 1 \end{matrix}| = - 13, C_{32} = - |\begin{matrix} 1 & 5 \\ 2 & 1 \end{matrix}| = 9 and C_{33} = |\begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix}| = - 1 ∴ adj B = {[\begin{matrix} 2 & - 3 & 5 \\ 3 & 6 & - 3 \\ - 13 & 9 & - 1 \end{matrix}]}^{T} = [\begin{matrix} 2 & 3 & - 13 \\ - 3 & 6 & 9 \\ 5 & - 3 & - 1 \end{matrix}] (adj B) B = [\begin{matrix} 21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21 \end{matrix}] and |B| = 21 ∴ |B| I = [\begin{matrix} 21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21 \end{matrix}] and B (adj B) = [\begin{matrix} 21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21 \end{matrix}] Thus, (adj A) A = |A| I = A (adj A) (iii) C = [\begin{matrix} 2 & - 1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & - 1 \end{matrix}] Now, C_{11} = |\begin{matrix} 2 & 5 \\ 4 & - 1 \end{matrix}| = - 22, C_{12} = - |\begin{matrix} 4 & 5 \\ 0 & - 1 \end{matrix}| = 4 and C_{13} = |\begin{matrix} 4 & 2 \\ 0 & 4 \end{matrix}| = 16 C_{21} = - |\begin{matrix} - 1 & 3 \\ 4 & - 1 \end{matrix}| = 11, C_{22} = |\begin{matrix} 2 & 3 \\ 0 & - 1 \end{matrix}| = - 2 and C_{23} = - |\begin{matrix} 2 & - 1 \\ 0 & 4 \end{matrix}| = - 8 C_{31} = |\begin{matrix} - 1 & 3 \\ 2 & 5 \end{matrix}| = - 11, C_{32} = - |\begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix}| = 2 and C_{33} = |\begin{matrix} 2 & - 1 \\ 4 & 2 \end{matrix}| = 8 ∴ adj C = {[\begin{matrix} - 22 & 4 & 16 \\ 11 & - 2 & - 8 \\ - 11 & 2 & 8 \end{matrix}]}^{T} = [\begin{matrix} - 22 & 11 & - 11 \\ 4 & - 2 & 2 \\ 16 & - 8 & 8 \end{matrix}] (adj C) C = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] and |C| = 0 ∴ |C| I = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] and C (adj C) = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] Thus, (adj A) A = |A| I = A (adj A) (iv) D = [\begin{matrix} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 1 & 1 & 3 \end{matrix}] Now, C_{11} = |\begin{matrix} 1 & 0 \\ 1 & 3 \end{matrix}| = 3, C_{12} = - |\begin{matrix} 5 & 0 \\ 1 & 3 \end{matrix}| = - 15 and C_{13} = |\begin{matrix} 5 & 1 \\ 1 & 1 \end{matrix}| = 4 C_{21} = - |\begin{matrix} 0 & - 1 \\ 1 & 3 \end{matrix}| = - 1, C_{22} = |\begin{matrix} 2 & - 1 \\ 1 & 3 \end{matrix}| = 7 and C_{23} = - |\begin{matrix} 2 & 0 \\ 1 & 1 \end{matrix}| = - 2 C_{31} = |\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}| = 1, C_{32} = - |\begin{matrix} 2 & - 1 \\ 5 & 0 \end{matrix}| = - 5 and C_{33} = |\begin{matrix} 2 & 0 \\ 5 & 1 \end{matrix}| = 2 ∴ adj D = {[\begin{matrix} 3 & - 15 & 4 \\ - 1 & 7 & - 2 \\ 1 & - 5 & 2 \end{matrix}]}^{T} = [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 7 & - 5 \\ 4 & - 2 & 2 \end{matrix}] (adj D) D = [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}] and |D| = 2 ∴ |D| I = [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}] and D (adj D) = [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}] Thus, (adj A) A = |A| I = A (adj A)$

Page No 7.22:

Question 3:

For the matrix $A = [\begin{matrix} 1 & - 1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10 \end{matrix}]$ , show that A (adj A) = O.

Answer:

$A = [\begin{matrix} 1 & - 1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10 \end{matrix}] Now, C_{11} = |\begin{matrix} 3 & 0 \\ 2 & 10 \end{matrix}| = 30 C_{12} = - |\begin{matrix} 2 & 0 \\ 18 & 10 \end{matrix}| = - 20 C_{13} = |\begin{matrix} 2 & 3 \\ 18 & 2 \end{matrix}| = - 50 C_{21} = - |\begin{matrix} - 1 & 1 \\ 2 & 10 \end{matrix}| = 12 C_{22} = |\begin{matrix} 1 & 1 \\ 18 & 10 \end{matrix}| = - 8 C_{23} = - |\begin{matrix} 1 & - 1 \\ 18 & 2 \end{matrix}| = - 20 C_{31} = |\begin{matrix} - 1 & 1 \\ 3 & 0 \end{matrix}| = - 3 C_{32} = - |\begin{matrix} 1 & 1 \\ 2 & 0 \end{matrix}| = 2 C_{33} = |\begin{matrix} 1 & - 1 \\ 2 & 3 \end{matrix}| = 5 adj A = {[\begin{matrix} 30 & - 20 & - 50 \\ 12 & - 8 & - 20 \\ - 3 & 2 & 5 \end{matrix}]}^{T} = [\begin{matrix} 30 & 12 & - 3 \\ - 20 & - 8 & 2 \\ - 50 & - 20 & 5 \end{matrix}] ∴ A (adj A) = [\begin{matrix} 1 & - 1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10 \end{matrix}] [\begin{matrix} 30 & 12 & - 3 \\ - 20 & - 8 & 2 \\ - 50 & - 20 & 5 \end{matrix}] = [\begin{matrix} 30 + 20 - 50 & 12 + 18 - 20 & - 3 - 2 + 5 \\ 60 - 60 - 0 & 24 - 24 - 0 & - 6 + 6 + 0 \\ 540 - 40 - 500 & 216 - 16 - 200 & - 54 + 4 + 50 \end{matrix}] = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$

Page No 7.22:

Question 4:

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

Answer:

$A = [\begin{matrix} - 4 & - 3 & - 3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{matrix}] Now, C_{11} = |\begin{matrix} 0 & 1 \\ 4 & 3 \end{matrix}| = - 4, C_{12} = - |\begin{matrix} 1 & 1 \\ 4 & 3 \end{matrix}| = 1 and C_{13} = |\begin{matrix} 1 & 0 \\ 4 & 4 \end{matrix}| = 4 C_{21} = - |\begin{matrix} - 3 & - 3 \\ 4 & 3 \end{matrix}| = - 3, C_{22} = |\begin{matrix} - 4 & - 3 \\ 4 & 3 \end{matrix}| = 0 and C_{23} = - |\begin{matrix} - 4 & - 3 \\ 4 & 4 \end{matrix}| = 4 C_{31} = |\begin{matrix} - 3 & - 3 \\ 0 & 1 \end{matrix}| = - 3, C_{32} = - |\begin{matrix} - 4 & - 3 \\ 1 & 1 \end{matrix}| = 1 and C_{33} = |\begin{matrix} - 4 & - 3 \\ 1 & 0 \end{matrix}| = 3 ∴ adj A = {[\begin{matrix} - 4 & 1 & 4 \\ - 3 & 0 & 4 \\ - 3 & 1 & 3 \end{matrix}]}^{T} = [\begin{matrix} - 4 & - 3 & - 3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{matrix}] = A Hence proved .$

Page No 7.23:

Question 5:

If $A = [\begin{matrix} - 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1 \end{matrix}]$ , show that adj A = 3A^T.

Answer:

$A = [\begin{matrix} - 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1 \end{matrix}] Now, C_{11} = |\begin{matrix} 1 & - 2 \\ - 2 & 1 \end{matrix}| = - 3, C_{12} = - |\begin{matrix} 2 & - 2 \\ 2 & 1 \end{matrix}| = - 6 and C_{13} = |\begin{matrix} 2 & 1 \\ 2 & - 2 \end{matrix}| = - 6 C_{21} = - |\begin{matrix} - 2 & - 2 \\ - 2 & 1 \end{matrix}| = 6, C_{22} = |\begin{matrix} - 1 & - 2 \\ 2 & 1 \end{matrix}| = 3 and C_{23} = - |\begin{matrix} - 1 & - 2 \\ 2 & - 2 \end{matrix}| = - 6 C_{31} = |\begin{matrix} - 2 & - 2 \\ 1 & - 2 \end{matrix}| = 6, C_{32} = - |\begin{matrix} - 1 & - 2 \\ 2 & - 2 \end{matrix}| = - 6 and C_{33} = |\begin{matrix} - 1 & - 2 \\ 2 & 1 \end{matrix}| = 3 adj A = {[\begin{matrix} - 3 & - 6 & - 6 \\ 6 & 3 & - 6 \\ 6 & - 6 & 3 \end{matrix}]}^{T} = [\begin{matrix} - 3 & 6 & 6 \\ - 6 & 3 & - 6 \\ - 6 & - 6 & 3 \end{matrix}] A^{T} = [\begin{matrix} - 1 & 2 & 2 \\ - 2 & 1 & - 2 \\ - 2 & - 2 & 1 \end{matrix}] \Rightarrow 3 A^{T} = [\begin{matrix} - 3 & 6 & 6 \\ - 6 & 3 & - 6 \\ - 6 & - 6 & 3 \end{matrix}] \Rightarrow 3 A^{T} = adj A$

Page No 7.23:

Question 6:

Find A (adj A) for the matrix $A = [\begin{matrix} 1 & - 2 & 3 \\ 0 & 2 & - 1 \\ - 4 & 5 & 2 \end{matrix}] .$

Answer:

$A = [\begin{matrix} 1 & - 2 & 3 \\ 0 & 2 & - 1 \\ - 4 & 5 & 2 \end{matrix}] Now, C_{11} = |\begin{matrix} 2 & - 1 \\ 5 & 2 \end{matrix}| = 9, C_{12} = - |\begin{matrix} 0 & - 1 \\ - 4 & 2 \end{matrix}| = 4 and C_{13} = |\begin{matrix} 0 & 2 \\ - 4 & 5 \end{matrix}| = 8 C_{21} = - |\begin{matrix} - 2 & 3 \\ 5 & 2 \end{matrix}| = 19, C_{22} = |\begin{matrix} 1 & 3 \\ - 4 & 2 \end{matrix}| = 14 and C_{23} = - |\begin{matrix} 1 & - 2 \\ - 4 & 5 \end{matrix}| = 3 C_{31} = |\begin{matrix} - 2 & 3 \\ 2 & - 1 \end{matrix}| = - 4, C_{32} = - |\begin{matrix} 1 & 3 \\ 0 & - 1 \end{matrix}| = 1 and C_{33} = |\begin{matrix} 1 & - 2 \\ 0 & 2 \end{matrix}| = 2 a d j A = {[\begin{matrix} 9 & 4 & 8 \\ 19 & 14 & 3 \\ - 4 & 1 & 2 \end{matrix}]}^{T} = [\begin{matrix} 9 & 19 & - 4 \\ 4 & 14 & 1 \\ 8 & 3 & 2 \end{matrix}] ∴ A (a d j A) = [\begin{matrix} 25 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 25 \end{matrix}]$

Page No 7.23:

Question 7:

Find the inverse of each of the following matrices:

(i) $[\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]$

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

(iii) $[\begin{matrix} a & b \\ c & \frac{1 + b c}{a} \end{matrix}]$

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

Answer:

$(i) A = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] |A| = \cos^{2} θ + \sin^{2} θ = 1 \neq 0 A is a singular matrix; therefore, it is invertible . Let C_{i j} be a cofactor of a_{i j} in A . Now, C_{11} = \cos θ C_{12} = \sin θ C_{21} = - \sin θ C_{22} = \cos θ adj A = {[\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]}^{T} = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}] ∴ A^{- 1} = \frac{1}{|A|} adj A = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}] (i i) B = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] |B| = 0 - 1 = - 1 \neq 0 B is a singular matrix; therefore, it is invertible . Let C_{i j} be a cofactor of b_{i j} in B . Now, C_{11} = 0 C_{12} = - 1 C_{21} = - 1 C_{22} = 0 adj B = {[\begin{matrix} 0 & - 1 \\ - 1 & 0 \end{matrix}]}^{T} = [\begin{matrix} 0 & - 1 \\ - 1 & 0 \end{matrix}] ∴ B^{- 1} = \frac{1}{|B|} adj B = \frac{1}{- 1} [\begin{matrix} 0 & - 1 \\ - 1 & 0 \end{matrix}] = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] (i i i) C = [\begin{matrix} a & b \\ c & \frac{1 + b c}{a} \end{matrix}] |C| = 1 + b c - b c = 1 \neq 0 C is a singular matrix; therefore, it is invertible . Let C_{i j} be a cofactor of c_{i j} in C . Now, C_{11} = \frac{1 + b c}{a} C_{12} = - c C_{21} = - b C_{22} = a adj C = {[\begin{matrix} \frac{1 + b c}{a} & - c \\ - b & a \end{matrix}]}^{T} = [\begin{matrix} \frac{1 + b c}{a} & - b \\ - c & a \end{matrix}] ∴ C^{- 1} = \frac{1}{|C|} adj C = [\begin{matrix} \frac{1 + b c}{a} & - b \\ - c & a \end{matrix}] (i v) D = [\begin{matrix} 2 & 5 \\ - 3 & 1 \end{matrix}] |D| = 2 + 15 = 17 \neq 0 D is a singular matrix; therefore, it is invertible . Let C_{i j} be a cofactor of d_{i j} in D . Now, C_{11} = 1 C_{12} = 3 C_{21} = - 5 C_{22} = 2 adj D = {[\begin{matrix} 1 & 3 \\ - 5 & 2 \end{matrix}]}^{T} = [\begin{matrix} 1 & - 5 \\ 3 & 2 \end{matrix}] ∴ D^{- 1} = \frac{1}{|D|} adj D = \frac{1}{17} [\begin{matrix} 1 & - 5 \\ 3 & 2 \end{matrix}]$

Page No 7.23:

Question 8:

Find the inverse of each of the following matrices.

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

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

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

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

(v) $[\begin{matrix} 0 & 1 & - 1 \\ 4 & - 3 & 4 \\ 3 & - 3 & 4 \end{matrix}]$

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

(vii) $[\begin{matrix} 1 & 0 & 0 \\ 0 & \cos α & \sin α \\ 0 & \sin α & - \cos α \end{matrix}]$

Answer:

$(i) A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{matrix}] Now, C_{11} = |\begin{matrix} 3 & 1 \\ 1 & 2 \end{matrix}| = 5, C_{12} = - |\begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix}| = - 1 and C_{13} = |\begin{matrix} 2 & 3 \\ 3 & 1 \end{matrix}| = - 7 C_{21} = - |\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}| = - 1, C_{22} = |\begin{matrix} 1 & 3 \\ 3 & 2 \end{matrix}| = - 7 and C_{23} = - |\begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix}| = 5 C_{31} = |\begin{matrix} 2 & 3 \\ 3 & 1 \end{matrix}| = - 7, C_{32} = - |\begin{matrix} 1 & 3 \\ 2 & 1 \end{matrix}| = 5 and C_{33} = |\begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix}| = - 1 adj A = {[\begin{matrix} 5 & - 1 & - 7 \\ - 1 & - 7 & 5 \\ - 7 & 5 & - 1 \end{matrix}]}^{T} = [\begin{matrix} 5 & - 1 & - 7 \\ - 1 & - 7 & 5 \\ - 7 & 5 & - 1 \end{matrix}] and |A| = - 18 ∴ A^{- 1} = - \frac{1}{18} [\begin{matrix} 5 & - 1 & - 7 \\ - 1 & - 7 & 5 \\ - 7 & 5 & - 1 \end{matrix}] (ii) B = [\begin{matrix} 1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1 \end{matrix}] Now, C_{11} = |\begin{matrix} - 1 & - 1 \\ 3 & - 1 \end{matrix}| = 4, C_{12} = - |\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}| = - 1 and C_{13} = |\begin{matrix} 1 & - 1 \\ 2 & 3 \end{matrix}| = 5 C_{21} = - |\begin{matrix} 2 & 5 \\ 3 & - 1 \end{matrix}| = 17, C_{22} = |\begin{matrix} 1 & 5 \\ 2 & - 1 \end{matrix}| = - 11 and C_{23} = - |\begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix}| = 1 C_{31} = |\begin{matrix} 2 & 5 \\ - 1 & - 1 \end{matrix}| = 3, C_{32} = - |\begin{matrix} 1 & 5 \\ 1 & - 1 \end{matrix}| = 6 and C_{33} = |\begin{matrix} 1 & 2 \\ 1 & - 1 \end{matrix}| = - 3 adj B = {[\begin{matrix} 4 & - 1 & 5 \\ 17 & - 11 & 1 \\ 3 & 6 & - 3 \end{matrix}]}^{T} = [\begin{matrix} 4 & 17 & 3 \\ - 1 & - 11 & 6 \\ 5 & 1 & - 3 \end{matrix}] and |B| = 27 ∴ B^{- 1} = \frac{1}{27} [\begin{matrix} 4 & 17 & 3 \\ - 1 & - 11 & 6 \\ 5 & 1 & - 3 \end{matrix}] (iii) C = [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}] Now, C_{11} = |\begin{matrix} 2 & - 1 \\ - 1 & 2 \end{matrix}| = 3, C_{12} = - |\begin{matrix} - 1 & - 1 \\ 1 & 2 \end{matrix}| = 1 and C_{13} = |\begin{matrix} - 1 & 2 \\ 1 & - 1 \end{matrix}| = - 1 C_{21} = - |\begin{matrix} - 1 & 1 \\ - 1 & 2 \end{matrix}| = 1, C_{22} = |\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix}| = 3 and C_{23} = - |\begin{matrix} 2 & - 1 \\ 1 & - 1 \end{matrix}| = 1 C_{31} = |\begin{matrix} - 1 & 1 \\ 2 & - 1 \end{matrix}| = - 1, C_{32} = - |\begin{matrix} 2 & 1 \\ - 1 & - 1 \end{matrix}| = 1 and C_{33} = |\begin{matrix} 2 & - 1 \\ - 1 & 2 \end{matrix}| = 3 adj C = {[\begin{matrix} 3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{matrix}]}^{T} = [\begin{matrix} 3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{matrix}] and |C| = 4 ∴ C^{- 1} = \frac{1}{4} [\begin{matrix} 3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{matrix}] (iv) D = [\begin{matrix} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix}] Now, C_{11} = |\begin{matrix} 1 & 0 \\ 1 & 3 \end{matrix}| = 3, C_{12} = - |\begin{matrix} 5 & 0 \\ 0 & 3 \end{matrix}| = - 15 and C_{13} = |\begin{matrix} 5 & 1 \\ 0 & 1 \end{matrix}| = 5 C_{21} = - |\begin{matrix} 0 & - 1 \\ 1 & 3 \end{matrix}| = - 1, C_{22} = |\begin{matrix} 2 & - 1 \\ 0 & 3 \end{matrix}| = 6 and C_{23} = - |\begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix}| = - 2 C_{31} = |\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}| = 1, C_{32} = - |\begin{matrix} 2 & - 1 \\ 5 & 0 \end{matrix}| = - 5 and C_{33} = |\begin{matrix} 2 & 0 \\ 5 & 1 \end{matrix}| = 2 adj D = {[\begin{matrix} 3 & - 15 & 5 \\ - 1 & 6 & - 2 \\ 1 & - 5 & 2 \end{matrix}]}^{T} = [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}] and |D| = 1 ∴ D^{- 1} = [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}] (v) E = [\begin{matrix} 0 & 1 & - 1 \\ 4 & - 3 & 4 \\ 3 & - 3 & 4 \end{matrix}] Now, C_{11} = |\begin{matrix} - 3 & 4 \\ - 3 & 4 \end{matrix}| = 0, C_{12} = - |\begin{matrix} 4 & 4 \\ 3 & 4 \end{matrix}| = - 4 and C_{13} = |\begin{matrix} 4 & - 3 \\ 3 & - 3 \end{matrix}| = - 3 C_{21} = - |\begin{matrix} 1 & - 1 \\ - 3 & 4 \end{matrix}| = - 1, C_{22} = |\begin{matrix} 0 & - 1 \\ 3 & 4 \end{matrix}| = 3 and C_{23} = - |\begin{matrix} 0 & 1 \\ 3 & - 3 \end{matrix}| = 3 C_{31} = |\begin{matrix} 1 & - 1 \\ - 3 & 4 \end{matrix}| = 1, C_{32} = - |\begin{matrix} 0 & - 1 \\ 4 & 4 \end{matrix}| = - 4 and C_{33} = |\begin{matrix} 0 & 1 \\ 4 & - 3 \end{matrix}| = - 4 adj E = {[\begin{matrix} 0 & - 4 & - 3 \\ - 1 & 3 & 3 \\ 1 & - 4 & - 4 \end{matrix}]}^{T} = [\begin{matrix} 0 & - 1 & 1 \\ - 4 & 3 & - 4 \\ - 3 & 3 & - 4 \end{matrix}] and |E| = - 1 ∴ E^{- 1} = - 1 [\begin{matrix} 0 & - 1 & 1 \\ - 4 & 3 & - 4 \\ - 3 & 3 & - 4 \end{matrix}] = [\begin{matrix} 0 & 1 & - 1 \\ 4 & - 3 & 4 \\ 3 & - 3 & 4 \end{matrix}] (vi) F = [\begin{matrix} 0 & 0 & - 1 \\ 3 & 4 & 5 \\ - 2 & - 4 & - 7 \end{matrix}] Now, C_{11} = |\begin{matrix} 4 & 5 \\ - 4 & - 7 \end{matrix}| = - 8, C_{12} = - |\begin{matrix} 3 & 5 \\ - 2 & - 7 \end{matrix}| = 11 and C_{13} = |\begin{matrix} 3 & 4 \\ - 2 & - 4 \end{matrix}| = - 4 C_{21} = - |\begin{matrix} 0 & - 1 \\ - 4 & - 7 \end{matrix}| = 4, C_{22} = |\begin{matrix} 0 & - 1 \\ - 2 & - 7 \end{matrix}| = - 2 and C_{23} = - |\begin{matrix} 0 & 0 \\ - 2 & - 4 \end{matrix}| = 0 C_{31} = |\begin{matrix} 0 & - 1 \\ 4 & 5 \end{matrix}| = 4, C_{32} = - |\begin{matrix} 0 & - 1 \\ 3 & 5 \end{matrix}| = - 3 and C_{33} = |\begin{matrix} 0 & 0 \\ 3 & 4 \end{matrix}| = 0 adj F = {[\begin{matrix} - 8 & 11 & - 4 \\ 4 & - 2 & 0 \\ 4 & - 3 & 0 \end{matrix}]}^{T} = [\begin{matrix} - 8 & 4 & 4 \\ 11 & - 2 & - 3 \\ - 4 & 0 & 0 \end{matrix}] and |F| = 4 ∴ F^{- 1} = \frac{1}{4} [\begin{matrix} - 8 & 4 & 4 \\ 11 & - 2 & - 3 \\ - 4 & 0 & 0 \end{matrix}] (vii) G = [\begin{matrix} 1 & 0 & 0 \\ 0 & \cos α & \sin α \\ 0 & \sin α & - \cos α \end{matrix}] Now, C_{11} = |\begin{matrix} \cos α & \sin α \\ \sin α & - \cos α \end{matrix}| = - 1, C_{12} = - |\begin{matrix} 0 & \sin α \\ 0 & - \cos α \end{matrix}| = 0 and C_{13} = |\begin{matrix} 0 & \cos α \\ 0 & \sin α \end{matrix}| = 0 C_{21} = - |\begin{matrix} 0 & 0 \\ \sin α & - \cos α \end{matrix}| = 0, C_{22} = |\begin{matrix} 1 & 0 \\ 0 & - \cos α \end{matrix}| = - \cos α and C_{23} = - |\begin{matrix} 1 & 0 \\ 0 & \sin α \end{matrix}| = - \sin α C_{31} = |\begin{matrix} 0 & 0 \\ \cos α & \sin α \end{matrix}| = 0, C_{32} = - |\begin{matrix} 1 & 0 \\ 0 & \sin α \end{matrix}| = - \sin α and C_{33} = |\begin{matrix} 1 & 0 \\ 0 & \cos α \end{matrix}| = \cos α adj F = {[\begin{matrix} - 1 & 0 & 0 \\ 0 & - \cos α & - \sin α \\ 0 & - \sin α & \cos α \end{matrix}]}^{T} = [\begin{matrix} - 1 & 0 & 0 \\ 0 & - \cos α & - \sin α \\ 0 & - \sin α & \cos α \end{matrix}] and |F| = - 1 ∴ F^{- 1} = - 1 [\begin{matrix} - 1 & 0 & 0 \\ 0 & - \cos α & - \sin α \\ 0 & - \sin α & \cos α \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & \cos α & \sin α \\ 0 & \sin α & - \cos α \end{matrix}]$

Page No 7.23:

Question 9:

Find the inverse of each of the following matrices and verify that $A^{- 1} A = I_{3}$ .

(i) $[\begin{matrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{matrix}]$

(ii) $[\begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix}]$

Answer:

$(i) A = [\begin{matrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{matrix}] Now, C_{11} = |\begin{matrix} 4 & 3 \\ 3 & 4 \end{matrix}| = 7, C_{12} = - |\begin{matrix} 1 & 3 \\ 1 & 4 \end{matrix}| = - 1 and C_{13} = |\begin{matrix} 1 & 4 \\ 1 & 3 \end{matrix}| = - 1 C_{21} = - |\begin{matrix} 3 & 3 \\ 3 & 4 \end{matrix}| = - 3, C_{22} = |\begin{matrix} 1 & 3 \\ 1 & 4 \end{matrix}| = 1 and C_{23} = - |\begin{matrix} 1 & 3 \\ 1 & 3 \end{matrix}| = 0 C_{31} = |\begin{matrix} 3 & 3 \\ 4 & 3 \end{matrix}| = - 3, C_{32} = - |\begin{matrix} 1 & 3 \\ 1 & 3 \end{matrix}| = 0 and C_{33} = |\begin{matrix} 1 & 3 \\ 1 & 4 \end{matrix}| = 1 adj A = {[\begin{matrix} 7 & - 1 & - 1 \\ - 3 & 1 & 0 \\ - 3 & 0 & 1 \end{matrix}]}^{T} = [\begin{matrix} 7 & - 3 & - 3 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{matrix}] and |A| = 1 A^{- 1} = [\begin{matrix} 7 & - 3 & - 3 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{matrix}] Now, A^{- 1} A = [\begin{matrix} 7 & - 3 & - 3 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = I_{3} (ii) B = [\begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix}] Now, C_{11} = |\begin{matrix} 4 & 1 \\ 7 & 2 \end{matrix}| = 1, C_{12} = - |\begin{matrix} 3 & 1 \\ 3 & 2 \end{matrix}| = - 3 and C_{13} = |\begin{matrix} 3 & 4 \\ 3 & 7 \end{matrix}| = 9 C_{21} = - |\begin{matrix} 3 & 1 \\ 7 & 2 \end{matrix}| = 1, C_{22} = |\begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix}| = 1 and C_{23} = - |\begin{matrix} 2 & 3 \\ 3 & 7 \end{matrix}| = - 5 C_{31} = |\begin{matrix} 3 & 1 \\ 4 & 1 \end{matrix}| = - 1, C_{32} = - |\begin{matrix} 2 & 1 \\ 3 & 1 \end{matrix}| = 1 and C_{33} = |\begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix}| = - 1 adj B = {[\begin{matrix} 1 & - 3 & 9 \\ 1 & 1 & - 5 \\ - 1 & 1 & - 1 \end{matrix}]}^{T} = [\begin{matrix} 1 & 1 & - 1 \\ - 3 & 1 & 1 \\ 9 & - 5 & - 1 \end{matrix}] and |B| = 2 B^{- 1} = \frac{1}{2} [\begin{matrix} 1 & 1 & - 1 \\ - 3 & 1 & 1 \\ 9 & - 5 & - 1 \end{matrix}] Now, B^{- 1} B = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = I_{3}$

Page No 7.23:

Question 10:

For the following pairs of matrices verity that ${(A B)}^{- 1} = B^{- 1} A^{- 1} :$

(i) $A = [\begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix}] and B [\begin{matrix} 4 & 6 \\ 3 & 2 \end{matrix}]$

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

Answer:

$(i) We have, A = [\begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix}] and B = [\begin{matrix} 4 & 6 \\ 3 & 2 \end{matrix}] ∴ A B = [\begin{matrix} 18 & 22 \\ 43 & 52 \end{matrix}] |A B| = - 10 Since, |A B| \neq 0 Hence, A B is invertible . Let C_{i j} be the cofactor of a_{i n} in A B = [a_{i j}] C_{11} = 52, C_{12} = - 43, C_{21} = - 22 and C_{22} = 18 adj (A B) = {[\begin{matrix} 52 & - 43 \\ - 22 & 18 \end{matrix}]}^{T} = [\begin{matrix} 52 & - 22 \\ - 43 & 18 \end{matrix}] ∴ {(A B)}^{- 1} = - \frac{1}{10} [\begin{matrix} 52 & - 22 \\ - 43 & 18 \end{matrix}] . . . (1) Now, B = [\begin{matrix} 4 & 6 \\ 3 & 2 \end{matrix}] |B| = - 10 Since, |B| \neq 0 Hence, B is invertible . Let C_{i j} be the cofactor of a_{i n} in B = [a_{i j}] C_{11} = 2, C_{12} = - 3, C_{21} = - 6 and C_{22} = 4 adj B = {[\begin{matrix} 2 & - 3 \\ - 6 & 4 \end{matrix}]}^{T} = [\begin{matrix} 2 & - 6 \\ - 3 & 4 \end{matrix}] ∴ B^{- 1} = - \frac{1}{10} [\begin{matrix} 2 & - 6 \\ - 3 & 4 \end{matrix}] |A| = 1 Since, |A| \neq 0 Hence, A is invertible . Let C_{i j} be the cofactor of a_{i n} in A = [a_{i j}] C_{11} = 5, C_{12} = - 7, C_{21} = - 2 and C_{22} = 3 adj A = {[\begin{matrix} 5 & - 7 \\ - 2 & 3 \end{matrix}]}^{T} = [\begin{matrix} 5 & - 2 \\ - 7 & 3 \end{matrix}] ∴ A^{- 1} = [\begin{matrix} 5 & - 2 \\ - 7 & 3 \end{matrix}] Now, B^{- 1} A^{- 1} = - \frac{1}{10} [\begin{matrix} 52 & - 22 \\ - 43 & 18 \end{matrix}] . . . (2) From eq . (1) and (2), we have {(A B)}^{- 1} = B^{- 1} A^{- 1} Hence verified . (ii) We have, A = [\begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix}] and B = [\begin{matrix} 4 & 5 \\ 3 & 4 \end{matrix}] ∴ A B = [\begin{matrix} 11 & 14 \\ 29 & 37 \end{matrix}] Now, |A B| = 1 Since, |A B| \neq 0 Hence, A B is invertible . Let C_{i j} be the cofactor of a_{i n} in A B = [a_{i j}] C_{11} = 37, C_{12} = - 29, C_{21} = - 14 and C_{22} = 11 adj (A B) = [\begin{matrix} 37 & - 14 \\ - 29 & 11 \end{matrix}] ∴ {(A B)}^{- 1} = [\begin{matrix} 37 & - 14 \\ - 29 & 11 \end{matrix}] . . . (1) |B| = 1 Since, |B| \neq 0 Hence, B is invertible . Let C_{i j} be the cofactor of a_{i n} in B = [a_{i j}] C_{11} = 4, C_{12} = - 3, C_{21} = - 5 and C_{22} = 4 adj B = [\begin{matrix} 4 & - 5 \\ - 3 & 4 \end{matrix}] ∴ B^{- 1} = [\begin{matrix} 4 & - 5 \\ - 3 & 4 \end{matrix}] |A| = 1 Since, |A| \neq 0 Hence, A is invertible . Let C_{i j} be the cofactor of a_{i n} in A = [a_{i j}] C_{11} = 3, C_{12} = - 5, C_{21} = - 1 and C_{22} = 2 adj A = [\begin{matrix} 3 & - 1 \\ - 5 & 2 \end{matrix}] ∴ A^{- 1} = [\begin{matrix} 3 & - 1 \\ - 5 & 2 \end{matrix}] Now, B^{- 1} A^{- 1} = [\begin{matrix} 37 & - 14 \\ - 29 & 11 \end{matrix}] . . . (2) From eq . (1) and (2), we have {(A B)}^{- 1} = B^{- 1} A^{- 1} Hence verified .$
(AB)−1=B−1 A−1 (AB)−1=B−1 A−1(AB)−1=B−1 A−1

Page No 7.23:

Question 11:

Let $A = [\begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix}] and B = [\begin{matrix} 6 & 7 \\ 8 & 9 \end{matrix}] . Find {(A B)}^{- 1}$

Answer:

$Given : A = [\begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix}] B = [\begin{matrix} 6 & 7 \\ 8 & 9 \end{matrix}] A B = [\begin{matrix} 34 & 39 \\ 82 & 94 \end{matrix}] Now, |A B| = - 2 Since, |A B| \neq 0 Hence, A B is invertible . Let C_{i j} be the cofactor of a_{i n} in A B = [a_{i j}] C_{11} = 94, C_{12} = - 82, C_{21} = - 39 and C_{22} = 34 adj (A B) = {[\begin{matrix} 94 & - 82 \\ - 39 & 34 \end{matrix}]}^{T} = [\begin{matrix} 94 & - 39 \\ - 82 & 34 \end{matrix}] ∴ {(A B)}^{- 1} = - \frac{1}{2} [\begin{matrix} 94 & - 39 \\ - 82 & 34 \end{matrix}] = [\begin{matrix} - 47 & \frac{39}{2} \\ 41 & - 17 \end{matrix}]$

Page No 7.23:

Question 12:

Given $A = [\begin{matrix} 2 & - 3 \\ - 4 & 7 \end{matrix}]$ , compute A⁻¹ and show that $2 A^{- 1} = 9 I - A .$

Answer:

$We have, A = [\begin{matrix} 2 & - 3 \\ - 4 & 7 \end{matrix}] Now, adj (A) = [\begin{matrix} 7 & 3 \\ 4 & 2 \end{matrix}] and |A| = 2 ∴ A^{- 1} = \frac{1}{2} [\begin{matrix} 7 & 3 \\ 4 & 2 \end{matrix}] Now, 2 A^{- 1} = 9 I - A LHS = 2 A^{- 1} = [\begin{matrix} 7 & 3 \\ 4 & 2 \end{matrix}] RHS = 9 I - A = 9 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] - [\begin{matrix} 2 & - 3 \\ - 4 & 7 \end{matrix}] = [\begin{matrix} 7 & 3 \\ 4 & 2 \end{matrix}] = LHS Hence proved .$

Page No 7.23:

Question 13:

If $A = [\begin{matrix} 4 & 5 \\ 2 & 1 \end{matrix}]$ , then show that $A - 3 I = 2 (I + 3 A^{- 1}) .$

Answer:

$We have, A = [\begin{matrix} 4 & 5 \\ 2 & 1 \end{matrix}] Now, adj (A) = [\begin{matrix} 1 & - 5 \\ - 2 & 4 \end{matrix}] and |A| = - 6 ∴ A^{- 1} = - \frac{1}{6} [\begin{matrix} 1 & - 5 \\ - 2 & 4 \end{matrix}] Now, A - 3 I = I + 3 A^{- 1} LHS = A - 3 I = [\begin{matrix} 4 & 5 \\ 2 & 1 \end{matrix}] - 3 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 5 \\ 2 & - 2 \end{matrix}] RHS = 2 (I + 3 A^{- 1}) = 2 \{[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] - 3 \times \frac{1}{6} [\begin{matrix} 1 & - 5 \\ - 2 & 4 \end{matrix}]\} = 2 [\begin{matrix} 0.5 & 2.5 \\ 1 & - 1 \end{matrix}] = [\begin{matrix} 1 & 5 \\ 2 & - 2 \end{matrix}] = LHS Hence proved .$

Page No 7.23:

Question 14:

Find the inverse of the matrix $A = [\begin{matrix} a & b \\ c & \frac{1 + b c}{a} \end{matrix}]$ and show that $a A^{- 1} = (a^{2} + b c + 1) I - a A .$

Answer:

$We have, A = [\begin{matrix} a & b \\ c & \frac{1 + b c}{a} \end{matrix}] So, adj (A) = [\begin{matrix} \frac{1 + b c}{a} & - b \\ - c & a \end{matrix}] and |A| = 1 ∴ A^{- 1} = [\begin{matrix} \frac{1 + b c}{a} & - b \\ - c & a \end{matrix}] Now, a A^{- 1} = (a^{2} + b c + 1) I - a A LHS = a A^{- 1} = a [\begin{matrix} \frac{1 + b c}{a} & - b \\ - c & a \end{matrix}] = [\begin{matrix} 1 + b c & - b a \\ - c a & a^{2} \end{matrix}] RHS = (a^{2} + b c + 1) I - a A = a^{2} + b c + 1 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] - [\begin{matrix} a^{2} & b a \\ c a & 1 + b c \end{matrix}] = [\begin{matrix} a^{2} + b c + 1 & 0 \\ 0 & a^{2} + b c + 1 \end{matrix}] - [\begin{matrix} a^{2} & b a \\ c a & 1 + b c \end{matrix}] = [\begin{matrix} 1 + b c & - b a \\ - c a & a^{2} \end{matrix}] = LHS Hence proved .$

Page No 7.23:

Question 15:

Given $A = [\begin{matrix} 5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1 \end{matrix}], B^{- 1} = [\begin{matrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{matrix}]$ . Compute (AB)⁻¹.

Answer:

We have,
$A = [\begin{matrix} 5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1 \end{matrix}] B^{- 1} = [\begin{matrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{matrix}] We know (A B)^{- 1} = B^{- 1} A^{- 1} For matrix A, C_{11} = |\begin{matrix} 3 & 2 \\ 2 & 1 \end{matrix}| = - 1, C_{12} = - |\begin{matrix} 2 & 2 \\ 1 & 1 \end{matrix}| = 0 and C_{13} = |\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}| = 1 C_{21} = - |\begin{matrix} 0 & 4 \\ 2 & 1 \end{matrix}| = 8, C_{22} = |\begin{matrix} 5 & 4 \\ 1 & 1 \end{matrix}| = 1 and C_{23} = - |\begin{matrix} 5 & 0 \\ 1 & 2 \end{matrix}| = - 10 C_{31} = |\begin{matrix} 0 & 4 \\ 3 & 2 \end{matrix}| = - 12, C_{32} = - |\begin{matrix} 5 & 4 \\ 2 & 2 \end{matrix}| = - 2 and C_{33} = |\begin{matrix} 5 & 0 \\ 2 & 3 \end{matrix}| = 15 Now, adj (A) = {[\begin{matrix} - 1 & 0 & 1 \\ 8 & 1 & - 10 \\ - 12 & - 2 & 15 \end{matrix}]}^{T} = [\begin{matrix} - 1 & 8 & - 12 \\ 0 & 1 & - 2 \\ 1 & - 10 & 15 \end{matrix}] and |A| = - 1 ∴ A^{- 1} = - [\begin{matrix} - 1 & 8 & - 12 \\ 0 & 1 & - 2 \\ 1 & - 10 & 15 \end{matrix}] = [\begin{matrix} 1 & - 8 & 12 \\ 0 & - 1 & 2 \\ - 1 & 10 & - 15 \end{matrix}] So, B^{- 1} A^{- 1} = [\begin{matrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{matrix}] [\begin{matrix} 1 & - 8 & 12 \\ 0 & - 1 & 2 \\ - 1 & 10 & - 15 \end{matrix}] = [\begin{matrix} - 2 & 19 & - 27 \\ - 2 & 18 & - 25 \\ - 3 & 29 & - 42 \end{matrix}]$

Page No 7.23:

Question 16:

Let $F (α) = [\begin{matrix} \cos α & - \sin α & 0 \\ \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}] and G (β) = [\begin{matrix} \cos β & 0 & \sin β \\ 0 & 1 & 0 \\ - \sin β & 0 & \cos β \end{matrix}]$

Show that
(i) ${[F (α)]}^{- 1} = F (- α)$
(ii) ${[G (β)]}^{- 1} = G (- β)$
(iii) ${[F (α) G (β)]}^{- 1} = G (- β) F (- α)$ .

Answer:

$(i) F (α) = [\begin{matrix} \cos α & - \sin α & 0 \\ \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow F (- α) = [\begin{matrix} \cos (- α) & - \sin (- α) & 0 \\ \sin (- α) & \cos (- α) & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \cos α & \sin α & 0 \\ - \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}] Now, C_{11} = |\begin{matrix} \cos α & 0 \\ 0 & 1 \end{matrix}| = \cos α, C_{12} = - |\begin{matrix} \sin α & 0 \\ 0 & 1 \end{matrix}| = - \sin α and C_{13} = |\begin{matrix} \sin α & \cos α \\ 0 & 0 \end{matrix}| = 0 C_{21} = - |\begin{matrix} - \sin α & 0 \\ 0 & 1 \end{matrix}| = \sin α, C_{22} = |\begin{matrix} \cos α & 0 \\ 0 & 1 \end{matrix}| = \cos α and C_{23} = - |\begin{matrix} \cos α & - \sin α \\ 0 & 0 \end{matrix}| = 0 C_{31} = |\begin{matrix} - \sin α & 0 \\ \cos α & 0 \end{matrix}| = 0, C_{32} = - |\begin{matrix} \cos α & 0 \\ \sin α & 0 \end{matrix}| = 0 and C_{33} = |\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}| = 1 \Rightarrow adj \{F (α)\} = {[\begin{matrix} \cos α & - \sin α & 0 \\ \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}]}^{T} = [\begin{matrix} \cos α & \sin α & 0 \\ - \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow |F (α)| = 1 ∴ {[F (α)]}^{- 1} = [\begin{matrix} \cos α & \sin α & 0 \\ - \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}] . . . (1) \Rightarrow {[F (α)]}^{- 1} = F (- α) (ii) G (β) = [\begin{matrix} \cos β & 0 & \sin β \\ 0 & 1 & 0 \\ - \sin β & 0 & \cos β \end{matrix}] \Rightarrow G (- β) = [\begin{matrix} \cos (- β) & 0 & \sin (- β) \\ 0 & 1 & 0 \\ - \sin (- β) & 0 & \cos (- β) \end{matrix}] = [\begin{matrix} \cos β & 0 & - \sin β \\ 0 & 1 & 0 \\ \sin β & 0 & \cos β \end{matrix}] Now, C_{11} = |\begin{matrix} 1 & 0 \\ 0 & \cos β \end{matrix}| = \cos β, C_{12} = - |\begin{matrix} 0 & 0 \\ - \sin β & \cos β \end{matrix}| = 0 and C_{13} = |\begin{matrix} 0 & 1 \\ - \sin β & 0 \end{matrix}| = \sin β C_{21} = - |\begin{matrix} 0 & \sin β \\ 0 & \cos β \end{matrix}| = 0, C_{22} = |\begin{matrix} \cos β & \sin β \\ - \sin β & \cos β \end{matrix}| = 1 and C_{23} = - |\begin{matrix} \cos β & 0 \\ - \sin β & 0 \end{matrix}| = 0 C_{31} = |\begin{matrix} 0 & \sin β \\ 1 & 0 \end{matrix}| = - \sin β, C_{32} = - |\begin{matrix} \cos β & \sin β \\ 0 & 0 \end{matrix}| = 0 and C_{33} = |\begin{matrix} \cos β & 0 \\ 0 & 1 \end{matrix}| = \cos β adj \{G (β)\} = {[\begin{matrix} \cos β & 0 & \sin β \\ 0 & 1 & 0 \\ - \sin β & 0 & \cos β \end{matrix}]}^{T} = [\begin{matrix} \cos β & 0 & - \sin β \\ 0 & 1 & 0 \\ \sin β & 0 & \cos β \end{matrix}] |G (β)| = 1 ∴ G (β)^{- 1} = [\begin{matrix} \cos β & 0 & - \sin β \\ 0 & 1 & 0 \\ \sin β & 0 & \cos β \end{matrix}] . . . (2) \Rightarrow G (β)^{- 1} = = G (- β) (iii) F (α) = [\begin{matrix} \cos α & - \sin α & 0 \\ \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow F (- α) = [\begin{matrix} \cos (- α) & - \sin (- α) & 0 \\ \sin (- α) & \cos (- α) & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \cos α & \sin α & 0 \\ - \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}] . . . (3) G (β) = [\begin{matrix} \cos β & 0 & \sin β \\ 0 & 1 & 0 \\ - \sin β & 0 & \cos β \end{matrix}] \Rightarrow G (- β) = [\begin{matrix} \cos (- β) & 0 & \sin (- β) \\ 0 & 1 & 0 \\ - \sin (- β) & 0 & \cos (- β) \end{matrix}] = [\begin{matrix} \cos β & 0 & - \sin β \\ 0 & 1 & 0 \\ \sin β & 0 & \cos β \end{matrix}] . . . (4) {[F (α) G (β)]}^{- 1} = {[G (β)]}^{- 1} {[F (α)]}^{- 1} = [\begin{matrix} \cos β & 0 & - \sin β \\ 0 & 1 & 0 \\ \sin β & 0 & \cos β \end{matrix}] [\begin{matrix} \cos α & \sin α & 0 \\ - \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}] [Using eqn (1) and (2)] = G (- β) F (- α) [Using eqn (3) and (4)]$

Page No 7.23:

Question 17:

If $A = [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}]$ , verify that $A^{2} - 4 A + I = O, where I = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] and O = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$ . Hence, find A⁻¹.

Answer:

$A = [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}] ∴ A^{2} = [\begin{matrix} 7 & 12 \\ 4 & 7 \end{matrix}] and A^{2} - 4 A + I = [\begin{matrix} 7 & 12 \\ 4 & 7 \end{matrix}] - [\begin{matrix} 8 & 12 \\ 4 & 8 \end{matrix}] + [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow A^{2} - 4 A + I = [\begin{matrix} 7 - 8 + 1 & 12 - 12 + 0 \\ 4 - 4 + 0 & 7 - 8 + 1 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = O \Rightarrow A^{2} - 4 A + I = 0 \Rightarrow A^{2} - 4 A = - I \Rightarrow A^{- 1} A^{2} - 4 A A^{- 1} = - I A^{- 1} [Pre - multiplying both sides by A^{- 1}] \Rightarrow A - 4 I = - A^{- 1} \Rightarrow A^{- 1} = 4 I - A \Rightarrow A^{- 1} = \{[\begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix}] - [\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}]\} = [\begin{matrix} 2 & - 3 \\ - 1 & 2 \end{matrix}]$

Page No 7.24:

Question 18:

Show that $A = [\begin{matrix} - 8 & 5 \\ 2 & 4 \end{matrix}]$ satisfies the equation $A^{2} + 4 A - 42 I = O$ . Hence, find A⁻¹.

Answer:

$A = [\begin{matrix} - 8 & 5 \\ 2 & 4 \end{matrix}] ∴ A^{2} = [\begin{matrix} 74 & - 20 \\ - 8 & 26 \end{matrix}] and A^{2} + 4 A - 42 I = [\begin{matrix} 74 & - 20 \\ - 8 & 26 \end{matrix}] + [\begin{matrix} - 32 & 20 \\ 8 & 16 \end{matrix}] - [\begin{matrix} 42 & 0 \\ 0 & 42 \end{matrix}] \Rightarrow A^{2} + 4 A - 42 I = [\begin{matrix} 74 - 32 - 42 & - 20 + 20 - 0 \\ - 8 + 8 - 0 & 26 + 16 - 42 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = O Now, A^{2} + 4 A - 42 I = 0 \Rightarrow A^{2} + 4 A = 42 I \Rightarrow A^{- 1} A^{2} + 4 A^{- 1} A = 42 I A^{- 1} [Pre - multiplying both sides by A^{- 1}] \Rightarrow A + 4 I = 42 A^{- 1} \Rightarrow A^{- 1} = \frac{1}{42} (A + 4 I) \Rightarrow A^{- 1} = \frac{1}{42} \{[\begin{matrix} - 8 & 5 \\ 2 & 4 \end{matrix}] + [\begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix}]\} = \frac{1}{42} [\begin{matrix} - 4 & 5 \\ 2 & 8 \end{matrix}]$

Page No 7.24:

Question 19:

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

Answer:

$A = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] = [\begin{matrix} 9 - 1 & 3 + 2 \\ - 3 - 2 & - 1 + 4 \end{matrix}] = [\begin{matrix} 8 & 5 \\ - 5 & 3 \end{matrix}] and A^{2} - 5 A + 7 I = [\begin{matrix} 8 & 5 \\ - 5 & 3 \end{matrix}] - [\begin{matrix} 15 & 5 \\ - 5 & 10 \end{matrix}] + [\begin{matrix} 7 & 0 \\ 0 & 7 \end{matrix}] \Rightarrow A^{2} - 5 A + 7 I = [\begin{matrix} 8 - 15 + 7 & 5 - 5 + 0 \\ - 5 + 5 + 0 & 3 - 10 + 7 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = O Now, A^{2} - 5 A + 7 I = 0 \Rightarrow A^{2} - 5 A = - 7 I \Rightarrow A^{- 1} A^{2} - 5 A^{- 1} A = - 7 A^{- 1} I [Pre - multiplying both sides by A^{- 1}] \Rightarrow A - 5 I = - 7 A^{- 1} \Rightarrow A^{- 1} = - \frac{1}{7} (A - 5 I) \Rightarrow A^{- 1} = - \frac{1}{7} \{[\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] - [\begin{matrix} 5 & 0 \\ 0 & 5 \end{matrix}]\} = \frac{1}{7} [\begin{matrix} 2 & - 1 \\ 1 & 3 \end{matrix}]$

Page No 7.24:

Question 20:

If $A = [\begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix}]$ , find x and y such that $A^{2} = x A + y I = O$ . Hence, evaluate A⁻¹.

Answer:

$A = [\begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix}] ∴ A^{2} = [\begin{matrix} 22 & 27 \\ 18 & 31 \end{matrix}] Now, A^{2} - x A + y I = O \Rightarrow [\begin{matrix} 22 & 27 \\ 18 & 31 \end{matrix}] - [\begin{matrix} 4 x & 3 x \\ 2 x & 5 x \end{matrix}] + [\begin{matrix} y & 0 \\ 0 & y \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \Rightarrow [\begin{matrix} 22 - 4 x - y & 27 - 3 x \\ 18 - 2 x & 31 - 5 x - y \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] Thus, we have 22 - 4 x + y = 0, 27 - 3 x = 0, 18 - 2 x = 0 and 31 - 5 x + y = 0 \Rightarrow - 3 x = - 27 \Rightarrow x = 9 On putting x = 9 in 22 - 4 x + y = 0, we get 22 - 36 + y = 0 \Rightarrow - 14 = - y \Rightarrow y = 14 Now, A^{2} - 9 A + 14 I = 0 \Rightarrow A^{2} - 9 A = - 14 I \Rightarrow A^{- 1} A^{2} - 9 A A^{- 1} = - 14 I A^{- 1} [Pre - multiplying both sides by A^{- 1}] \Rightarrow A - 9 I = - 14 A^{- 1} \Rightarrow A^{- 1} = - \frac{1}{14} (A - 9 I) \Rightarrow A^{- 1} = - \frac{1}{14} \{[\begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix}] - [\begin{matrix} 9 & 0 \\ 0 & 9 \end{matrix}]\} = \frac{1}{14} [\begin{matrix} 5 & - 3 \\ - 2 & 4 \end{matrix}]$

Page No 7.24:

Question 21:

If $A = [\begin{matrix} 3 & - 2 \\ 4 & - 2 \end{matrix}]$ , find the value of $λ$ so that $A^{2} = λ A - 2 I$ . Hence, find A⁻¹.

Answer:

$A = [3 - 2 4 - 2] ∴ A^{2} = [1 - 2 4 - 4] Given : A^{2} = λ A - 2 I . . . (1) \Rightarrow [1 - 2 4 - 4] = λ [3 - 2 4 - 2] - 2 [1 0 0 1] \Rightarrow [1 - 2 4 - 4] = [3 λ - 2 λ 4 λ - 2 λ] - [2 0 0 2] \Rightarrow [1 - 2 4 - 4] = [3 λ - 2 - 2 λ 4 λ - 2 λ - 2] On equating corresponding terms, we get - 2 λ = - 2 \Rightarrow λ = 1 On substituting λ = 1 in (1), we get A^{2} = A - 2 I \Rightarrow A^{2} - A = - 2 I \Rightarrow A - A^{2} = 2 I \Rightarrow A^{- 1} (A - A^{2}) = A^{- 1} \times 2 I (Pre - multiplying both sides with A^{- 1}) \Rightarrow I - A = 2 A^{- 1} 2 A^{- 1} = [1 0 0 1] - [3 - 2 4 - 2] = [1 - 3 0 + 2 0 - 4 1 + 2] \Rightarrow A^{- 1} = \frac{1}{2} [- 2 2 - 4 3]$

Page No 7.24:

Question 22:

Show that $A = [\begin{matrix} 5 & 3 \\ - 1 & - 2 \end{matrix}]$ satisfies the equation $x^{2} - 3 x - 7 = 0$ . Thus, find A⁻¹.

Answer:

$A = [5 3 - 1 - 2] A^{2} = [22 9 - 3 1] If I_{2} is the identity matrix of order 2, then A^{2} - 3 A - 7 I_{2} = [22 9 - 3 1] - 3 [5 3 - 1 - 2] - 7 [1 0 0 1] \Rightarrow A^{2} - 3 A - 7 I_{2} = [22 - 15 - 7 9 - 9 - 0 - 3 + 3 + 0 1 + 6 - 7] = [0 0 0 0] = 0 \Rightarrow A^{2} - 3 A - 7 I_{2} = 0 Thus, A satisfies x^{2} - 3 x - 7 = 0 . Now, A^{2} - 3 A - 7 I_{2} = 0 \Rightarrow A^{2} - 3 A = 7 I_{2} \Rightarrow A^{- 1} (A^{2} - 3 A) = A^{- 1} \times 7 I_{2} [Pre - multiplying both sides by A^{- 1}] \Rightarrow A - 3 I_{2} = 7 A^{- 1} \Rightarrow [5 3 - 1 - 2] - 3 [1 0 0 1] = 7 A^{- 1} \Rightarrow A^{- 1} = \frac{1}{7} [5 - 3 3 - 0 - 1 - 0 - 2 - 3] = \frac{1}{7} [2 3 - 1 - 5]$

Page No 7.24:

Question 23:

Show that $A = [\begin{matrix} 6 & 5 \\ 7 & 6 \end{matrix}]$ satisfies the equation $x^{2} - 12 x + 1 = O$ . Thus, find A⁻¹.

Answer:

$A = [6 5 7 6] ∴ A^{2} = [71 60 84 71] If I_{2} is the identity matrix of order 2, then A^{2} - 12 A + I_{2} = [71 60 84 71] - 12 [6 5 7 6] + [1 0 0 1] \Rightarrow A^{2} - 12 A + I_{2} = [71 - 72 + 1 60 - 60 + 0 84 - 84 + 0 71 - 72 + 1] \Rightarrow A^{2} - 12 A + I_{2} = 0 Thus, A satisfies x^{2} - 12 x + 1 = 0 . Now, A^{2} - 12 A + I_{2} = 0 \Rightarrow I_{2} = 12 A - A^{2} \Rightarrow A^{- 1} I_{2} = A^{- 1} (12 A - A^{2}) [Pre - multiplying both sides by A^{- 1}] \Rightarrow A^{- 1} = 12 I_{2} - A \Rightarrow A^{- 1} = 12 [1 0 0 1] - [6 5 7 6] \Rightarrow A^{- 1} = [12 - 6 0 - 5 0 - 7 12 - 6] \Rightarrow A^{- 1} = [6 - 5 - 7 6]$

Page No 7.24:

Question 24:

For the matrix $A = [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}]$ . Show that $A^{- 3} - 6 A^{2} + 5 A + 11 I_{3} = O$ . Hence, find A⁻¹.

Answer:

$A = [1 1 1 1 2 - 3 2 - 1 3] \Rightarrow |A| = |1 1 1 1 2 - 3 2 - 1 3| = (1 \times 3) - (1 \times 9) + (1 \times - 5) = 3 - 9 - 5 = - 11 Since, |A| \neq 0 Hence, A^{- 1} exists . Now, A^{2} = [1 1 1 1 2 - 3 2 - 1 3] [1 1 1 1 2 - 3 2 - 1 3] = [1 + 1 + 2 1 + 2 - 1 1 - 3 + 3 1 + 2 - 6 1 + 4 + 3 1 - 6 - 9 2 - 1 + 6 2 - 2 - 3 2 + 3 + 9] = [4 2 1 - 3 8 - 14 7 - 3 14] and A^{3} = A^{2} . A = [4 2 1 - 3 8 - 14 7 - 3 14] [1 1 1 1 2 - 3 2 - 1 3] = [4 + 2 + 2 4 + 4 - 1 4 - 6 + 3 - 3 + 8 - 28 - 3 + 16 + 14 - 3 - 24 - 42 7 - 3 + 28 7 - 6 - 14 7 + 9 + 42] = [8 7 1 - 23 27 - 69 32 - 13 58] Now, A^{3} - 6 A^{2} + 5 A + 11 I_{3} = [8 7 1 - 23 27 - 69 32 - 13 58] - 6 [4 2 1 - 3 8 - 14 7 - 3 14] + 5 [1 1 1 1 2 - 3 2 - 1 3] + 11 [1 0 0 0 1 0 0 0 1] = [8 - 24 + 5 + 11 7 - 12 + 5 + 0 1 - 6 + 5 + 0 - 23 + 18 + 5 + 0 27 - 48 + 10 + 11 - 69 + 84 - 15 + 0 32 - 42 + 10 + 0 - 13 + 18 - 5 + 0 58 - 84 + 15 + 11] = [0 0 0 0 0 0 0 0 0] = O (Null matrix) Again, A^{3} - 6 A^{2} + 5 A + 11 I_{3} = O \Rightarrow A^{- 1} \times (A^{3} - 6 A^{2} + 5 A + 11 I_{3}) = A^{- 1} \times O (Pre - multiplying both sides because A^{- 1} exists) \Rightarrow (A^{2} - 6 A + 5 I_{3} + 11 A^{- 1}) = 0 \Rightarrow [4 2 1 - 3 8 - 14 7 - 3 14] - 6 [1 1 1 1 2 - 3 2 - 1 3] + 5 [1 0 0 0 1 0 0 0 1] = - 11 A^{- 1} \Rightarrow [4 - 6 + 5 2 - 6 + 0 1 - 6 + 0 - 3 - 6 + 0 8 - 12 + 5 - 14 + 18 + 0 7 - 12 + 0 - 3 + 6 + 0 14 - 18 + 5] = - 11 A^{- 1} \Rightarrow A^{- 1} = - \frac{1}{11} [3 - 4 - 5 - 9 1 4 - 5 3 1]$

Page No 7.24:

Question 25:

Show that the matrix, $A = [\begin{matrix} 1 & 0 & - 2 \\ - 2 & - 1 & 2 \\ 3 & 4 & 1 \end{matrix}]$ satisfies the equation, $A^{3} - A^{2} - 3 A - I_{3} = O$ . Hence, find A⁻¹.

Answer:

$We have, A = [1 0 - 2 - 2 - 1 2 3 4 1] \Rightarrow |A| = |1 0 - 2 - 2 - 1 2 3 4 1| = 1 (- 9) + 0 - 2 (- 8) = - 9 + 16 = 7 Since, |A| \neq 0 Hence, A^{- 1} exists . Now, A^{2} = [1 0 - 2 - 2 - 1 2 3 4 1] [1 0 - 2 - 2 - 1 2 3 4 1] = [1 + 0 - 6 0 + 0 - 8 - 2 + 0 - 2 - 2 + 2 + 6 0 + 1 + 8 4 - 2 + 2 3 - 8 + 3 0 - 4 + 4 - 6 + 8 + 1] = [- 5 - 8 - 4 6 9 4 - 2 0 3] A^{3} = A^{2} . A = [- 5 - 8 - 4 6 9 4 - 2 0 3] [1 0 - 2 - 2 - 1 2 3 4 1] = [- 5 + 16 - 12 0 + 8 - 16 10 - 16 - 4 6 - 18 + 12 0 - 9 + 16 - 12 + 18 + 4 - 2 + 0 + 9 0 + 0 + 12 4 + 0 + 3] = [- 1 - 8 - 10 0 7 10 7 12 7] Now, A^{3} - A^{2} - 3 A - I_{3} = [- 1 - 8 - 10 0 7 10 7 12 7] - [- 5 - 8 - 4 6 9 4 - 2 0 3] - 3 [1 0 - 2 - 2 - 1 2 3 4 1] - [1 0 0 0 1 0 0 0 1] = [- 1 + 5 - 3 - 1 - 8 + 8 + 0 + 0 - 10 + 4 + 6 - 0 0 - 6 + 6 - 0 7 - 9 + 3 - 1 10 - 4 - 6 - 0 7 + 2 - 9 - 0 12 + 0 - 12 - 0 7 - 3 - 3 - 1] = [0 0 0 0 0 0 0 0 0] = O Hence proved . Now, A^{3} - A^{2} - 3 A - I_{3} = O (Null matrix) \Rightarrow A^{- 1} (A^{3} - A^{2} - 3 A - I_{3}) = A^{- 1} O (Pre - multiplying by A^{- 1}) \Rightarrow A^{2} - A^{1} - 3 I_{3} = A^{- 1} \Rightarrow [- 5 - 8 - 4 6 9 4 - 2 0 3] - [1 0 - 2 - 2 - 1 2 3 4 1] - 3 [1 0 0 0 1 0 0 0 1] = A^{- 1} \Rightarrow [- 5 - 1 - 3 - 8 - 0 - 0 - 4 + 2 + 0 6 + 2 + 0 9 + 1 - 3 4 - 2 - 2 - 3 - 0 0 - 4 - 0 3 - 1 - 3] = [- 9 - 8 - 2 8 7 2 - 5 - 4 - 1] = A^{- 1} \Rightarrow A^{- 1} = [- 9 - 8 - 2 8 7 2 - 5 - 4 - 1]$

Page No 7.24:

Question 26:

If $A = [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}]$ . Verify that $A^{3} - 6 A^{2} + 9 A - 4 I = O$ and hence find A⁻¹.

Answer:

$A = [2 - 1 1 - 1 2 - 1 1 - 1 2] \Rightarrow |A| = |2 - 1 1 - 1 2 - 1 1 - 1 2| = 2 \times (4 - 1) + 1 (- 2 + 1) + 1 (1 - 2) = 6 - 1 - 1 = 4 Since, |A| \neq 0 Hence, A^{- 1} exists . Now, A^{2} = [2 - 1 1 - 1 2 - 1 1 - 1 2] [2 - 1 1 - 1 2 - 1 1 - 1 2] = [4 + 1 + 1 - 2 - 2 - 1 2 + 1 + 2 - 2 - 2 - 1 1 + 4 + 1 - 1 - 2 - 2 2 + 1 + 2 - 1 - 2 - 2 1 + 1 + 4] = [6 - 5 5 - 5 6 - 5 5 - 5 6] A^{3} = A^{2} . A = [6 - 5 5 - 5 6 - 5 5 - 5 6] [2 - 1 1 - 1 2 - 1 1 - 1 2] = [12 + 5 + 5 - 6 - 10 - 5 6 + 5 + 10 - 10 - 6 - 5 5 + 12 + 5 - 5 - 6 - 10 10 + 5 + 6 - 5 - 10 - 6 5 + 5 + 12] = [22 - 21 21 - 21 22 - 21 21 - 21 22] Now, A^{3} - 6 A^{2} + 9 A - 4 I = [22 - 21 21 - 21 22 - 21 21 - 21 22] - 6 [6 - 5 5 - 5 6 - 5 5 - 5 6] + 9 [2 - 1 1 - 1 2 - 1 1 - 1 2] - 4 [1 0 0 0 1 0 0 0 1] = [22 - 36 + 18 - 4 - 21 + 30 - 9 - 0 21 - 30 + 9 - 0 - 21 + 30 - 9 - 0 22 - 36 + 18 - 4 - 21 + 30 - 9 - 0 21 - 30 + 9 - 0 - 21 + 30 - 9 - 0 22 - 36 + 18 - 4] = [0 0 0 0 0 0 0 0 0] = O [Null matrix] Hence proved . Now, A^{3} - 6 A^{2} + 9 A - 4 I = O \Rightarrow A^{- 1} \times (A^{3} - 6 A^{2} + 9 A - 4 I) = A^{- 1} \times O \Rightarrow A^{2} - 6 A + 9 I - 4 A^{- 1} = O \Rightarrow 4 A^{- 1} = A^{2} - 6 A + 9 I \Rightarrow 4 A^{- 1} = [6 - 5 5 - 5 6 - 5 5 - 5 6] - 6 [2 - 1 1 - 1 2 - 1 1 - 1 2] + 9 [1 0 0 0 1 0 0 0 1] \Rightarrow 4 A^{- 1} = [6 - 12 + 9 - 5 + 6 + 0 5 - 6 + 0 - 5 + 6 + 0 6 - 12 + 9 - 5 + 6 + 0 5 - 6 + 0 - 5 + 6 + 0 6 - 12 + 9] \Rightarrow 4 A^{- 1} = [3 1 - 1 1 3 1 - 1 1 3] \Rightarrow A^{- 1} = \frac{1}{4} [3 1 - 1 1 3 1 - 1 1 3]$

Page No 7.24:

Question 27:

If $A = \frac{1}{9} [\begin{matrix} - 8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & - 8 & 4 \end{matrix}]$ , prove that $A^{- 1} = A^{3}$ .

Answer:

$A = \frac{1}{9} [- 8 1 4 4 4 7 1 - 8 4] = [\frac{- 8}{9} \frac{1}{9} \frac{4}{9} \frac{4}{9} \frac{4}{9} \frac{7}{9} \frac{1}{9} \frac{- 8}{9} \frac{4}{9}] \Rightarrow A^{T} = [\frac{- 8}{9} \frac{4}{9} \frac{1}{9} \frac{1}{9} \frac{4}{9} \frac{- 8}{9} \frac{4}{9} \frac{7}{9} \frac{4}{9}] = \frac{1}{9} [- 8 4 1 1 4 - 8 4 7 4] . . . (1) |A| = |\frac{- 8}{9} \frac{1}{9} \frac{4}{9} \frac{4}{9} \frac{4}{9} \frac{7}{9} \frac{1}{9} \frac{- 8}{9} \frac{4}{9}| = \frac{1}{9 \times 9 \times 9} |- 8 1 4 4 4 7 1 - 8 4| = \frac{1}{9 \times 9 \times 9} [(- 8 \times 72) - (1 \times 9) + \{4 \times (- 36)\}] = \frac{1}{9 \times 9 \times 9} \times 9 \times \{- 64 - 1 - 16\} = - \frac{9 \times 81}{9 \times 9 \times 9} = - 1 If C_{i j} is a cofactor of a_{i j} such that A = [a_{i j}], then we have C_{11} = \frac{8}{9} C_{12} = \frac{- 1}{9} C_{13} = \frac{- 4}{9} C_{21} = \frac{- 4}{9} C_{22} = \frac{- 4}{9} C_{23} = \frac{- 7}{9} C_{31} = \frac{- 1}{9} C_{32} = \frac{8}{9} C_{33} = \frac{- 4}{9} Now, adj A = {[\frac{8}{9} \frac{- 1}{9} \frac{- 4}{9} \frac{- 4}{9} \frac{- 4}{9} \frac{- 7}{9} \frac{- 1}{9} \frac{8}{9} \frac{- 4}{9}]}^{T} = [\frac{8}{9} \frac{- 4}{9} \frac{- 1}{9} \frac{- 1}{9} \frac{- 4}{9} \frac{8}{9} \frac{- 4}{9} \frac{- 7}{9} \frac{- 4}{9}] ∴ A^{- 1} = \frac{1}{|A|} a d j A = - 1 \times \frac{1}{9} [8 - 4 - 1 - 1 - 4 8 - 4 - 7 - 4] = \frac{1}{9} [- 8 4 1 1 4 - 8 4 7 4] = A^{T} [From (1)] \Rightarrow A^{- 1} = A^{T}$

Page No 7.24:

Question 28:

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

Answer:

$We have, A = [3 - 3 4 2 - 3 4 0 - 1 1] A^{2} = [3 - 3 4 2 - 3 4 0 - 1 1] [3 - 3 4 2 - 3 4 0 - 1 1] = [9 - 6 + 0 - 9 + 9 - 4 12 - 12 + 4 6 - 6 + 0 - 6 + 9 - 4 8 - 12 + 4 0 - 2 + 0 0 + 3 - 1 0 - 4 + 1] = [3 - 4 4 0 - 1 0 - 2 2 - 3] Now, A^{3} = A^{2} \times A^{} = [3 - 4 4 0 - 1 0 - 2 2 - 3] [3 - 3 4 2 - 3 4 0 - 1 1] = [9 - 8 - 9 + 12 - 4 12 - 16 + 4 0 - 2 + 0 0 + 3 + 0 - 4 - 6 + 4 + 0 6 - 6 + 3 - 8 + 8 - 3] = [1 - 1 0 - 2 3 - 4 - 2 3 - 3] Again, A^{3} \times A = [1 - 1 0 - 2 3 - 4 - 2 3 - 3] [3 - 3 4 2 - 3 4 0 - 1 1] = [3 - 2 + 0 - 3 + 3 + 0 4 - 4 + 0 - 6 + 6 6 - 9 + 4 - 8 + 12 - 4 0 0 1] = [1 0 0 0 1 0 0 0 1] = I_{3} [Identity matrix of order 3] \Rightarrow A^{3} \times A = I_{3} \Rightarrow A^{3} = A^{- 1}$

Page No 7.24:

Question 29:

If $A = [\begin{matrix} - 1 & 2 & 0 \\ - 1 & 1 & 1 \\ 0 & 1 & 0 \end{matrix}]$ , show that $A^{2} = A^{- 1} .$

Answer:

$We have, A = [- 1 2 0 - 1 1 1 0 1 0] Now, A^{2} = [- 1 2 0 - 1 1 1 0 1 0] [- 1 2 0 - 1 1 1 0 1 0] = [1 - 2 + 0 - 2 + 2 + 0 0 + 2 + 0 1 - 1 + 0 - 2 + 1 + 1 0 + 1 + 0 0 - 1 + 0 0 + 1 + 0 0 + 1 + 0] = [- 1 0 2 0 0 1 - 1 1 1] and A^{2} \times A = [- 1 0 2 0 0 1 - 1 1 1] [- 1 2 0 - 1 1 1 0 1 0] = [1 + 0 + 0 - 2 + 0 + 2 0 0 + 0 + 0 0 + 0 + 1 0 0 - 2 + 1 + 1 1] = [1 0 0 0 1 0 0 0 1] = I_{3} [Identity matrix of order 3] \Rightarrow A^{2} \times A = I_{3} \Rightarrow A^{2} = A^{- 1}$

Page No 7.24:

Question 30:

Solve the matrix equation $[\begin{matrix} 5 & 4 \\ 1 & 1 \end{matrix}] X = [\begin{matrix} 1 & - 2 \\ 1 & 3 \end{matrix}]$ , where X is a 2 × 2 matrix.

Answer:

$Let : A = [5 4 1 1] B = [1 - 2 1 3] Now, |A| = |5 4 1 1| = 5 - 4 = 1 \neq 0 Hence, A is invertible . If C_{i j} is cofactor of a_{i j} in A, then C_{11} = 1, C_{12} = - 1, C_{21} = - 4 and C_{22} = 5 . \Rightarrow adj A = {[1 - 1 - 4 - 5]}^{T} = [1 - 4 - 1 5] ∴ A^{- 1} = \frac{1}{|A|} adj A = [1 - 4 - 1 5] Now, the given equation becomes A X = B . \Rightarrow A^{- 1} (A X) = A^{- 1} \times B \Rightarrow (A^{- 1} A) X = A^{- 1} \times B \Rightarrow X = A^{- 1} \times B \Rightarrow X = [1 - 4 - 1 5] \times [1 - 2 1 3] \Rightarrow X = [1 - 4 - 2 - 12 - 1 + 5 2 + 15] \Rightarrow X = [- 3 - 14 4 17]$

Page No 7.24:

Question 31:

Find the matrix X satisfying the matrix equation

$X [\begin{matrix} 5 & 3 \\ - 1 & - 2 \end{matrix}] = [\begin{matrix} 14 & 7 \\ 7 & 7 \end{matrix}]$ .

Answer:

$Let : A = [5 3 - 1 - 2] \Rightarrow |A| = |5 3 - 1 - 2| = - 10 + 3 = - 7 \neq 0 Hence, A is invertible . If C_{i j} is a cofactor of a_{i j} in A, then C_{11} = - 2, C_{12} = 1, C_{21} = - 3 and C_{22} = 5 . Now, adj A = {[- 2 1 - 3 5]}^{T} = [- 2 - 3 1 5] \Rightarrow A^{- 1} = \frac{1}{|A|} adj A = \frac{- 1}{7} [- 2 - 3 1 5] Let : B = [14 7 7 7] \Rightarrow |B| = |14 7 7 7| = 98 - 49 = 49 \neq 0 Hence, B is invertible . The given matrix equation becomes X A = B . \Rightarrow (X A) A^{- 1} = B A^{- 1} \Rightarrow X (A A^{- 1}) = [14 7 7 7] \times \frac{- 1}{7} \times [- 2 - 3 1 5] \Rightarrow X = \frac{- 1}{7} [- 28 + 7 - 42 + 35 - 14 + 7 - 21 + 35] \Rightarrow X = \frac{- 1}{7} [- 21 - 7 - 7 14] \Rightarrow X = [3 1 1 - 2]$

Page No 7.24:

Question 32:

Find the matrix X for which

$[\begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix}] X [\begin{matrix} - 1 & 1 \\ - 2 & 1 \end{matrix}] = [\begin{matrix} 2 & - 1 \\ 0 & 4 \end{matrix}]$

Answer:

$Let A = [3 2 7 5], B = [- 1 1 - 2 1] and C = [2 - 1 0 4] Now, |A| = |3 2 7 5| = 15 - 14 = 1 |B| = |- 1 1 - 2 1| = - 1 + 2 = 1 Since, |A| \neq 0 and |B| \neq 0 Hence, A & B are invertible, so A^{- 1} and B^{- 1} exist . Cofactors of matrix A are A_{11} = 5 A_{12} = - 7 A_{21} = - 2 A_{22} = 3 Now, adj A = {[5 - 7 - 2 3]}^{T} = [5 - 2 - 7 3] A^{- 1} = \frac{1}{|A|} adj A = [5 - 2 - 7 3] Cofactors of matrix B are B_{11} = 1 B_{12} = 2 B_{21} = - 1 B_{22} = - 1 Now, adj B = {[1 2 - 1 - 1]}^{T} = [1 - 1 2 - 1] B^{- 1} = \frac{1}{|B|} adj B = [1 - 1 2 - 1] The given equation becomes A X B = C \Rightarrow (A^{- 1} A) X (B B^{- 1}) = A^{- 1} C B^{- 1} \Rightarrow (I) X (I) = A^{- 1} C B^{- 1} \Rightarrow X = [5 - 2 - 7 3] [2 - 1 0 4] [1 - 1 2 - 1] \Rightarrow X = [5 - 2 - 7 3] [2 - 2 - 2 + 1 0 + 8 0 - 4] \Rightarrow X = [5 - 2 - 7 3] [0 - 1 8 - 4] \Rightarrow X = [- 16 3 24 - 5]$

Page No 7.24:

Question 33:

Find the matrix X satisfying the equation

$[\begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix}] X [\begin{matrix} 5 & 3 \\ 3 & 2 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] .$

Answer:

$Let A = [2 1 5 3], B = [5 3 3 2] and I = [1 0 0 1] \Rightarrow |A| = |2 1 5 3| = 6 - 5 = 1 Since, |A| \neq 0 Thus, A is invertible. Also, |B| = |5 3 3 2| = 10 - 9 = 1 Thus, B is invertible. Cofactors of matrices A & B are A_{11} = 3 A_{12} = - 5 A_{21} = - 1 A_{22} = 2 B_{11} = 2 B_{12} = - 3 B_{21} = - 3 B_{22} = 5 Now, adj A = {[3 - 5 - 1 2]}^{T} = [3 - 1 - 5 2] adj B = {[2 - 3 - 3 5]}^{T} = [2 - 3 - 3 5] A^{- 1} = \frac{1}{|A|} adj A = [3 - 1 - 5 2] B^{- 1} = \frac{1}{|B|} adj B = [2 - 3 - 3 5] The given matrix equation becomes A X B = I \Rightarrow A^{- 1} A X B B^{- 1} = I A^{- 1} B^{- 1} \Rightarrow (A^{- 1} A) X (B B^{- 1}) = A^{- 1} B^{- 1} \Rightarrow I X I = A^{- 1} B^{- 1} \Rightarrow X = A^{- 1} B^{- 1} \Rightarrow X = [3 - 1 - 5 2] [2 - 3 - 3 5] = [6 + 3 - 9 - 5 - 10 - 6 15 + 10] = [9 - 14 - 16 25]$

Page No 7.24:

Question 34:

If $A = [\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}]$ , find $A^{- 1}$ and prove that $A^{2} - 4 A - 5 I = O$

Answer:

$A = [1 2 2 2 1 2 2 2 1] \Rightarrow |A| = |1 2 2 2 1 2 2 2 1| = 1 (1 - 4) - 2 (2 - 4) + 2 (4 - 2) = - 3 + 4 + 4 = 5 Since, |A| \neq 0 Hence, A is invertible . Now, A^{2} = [1 2 2 2 1 2 2 2 1] [1 2 2 2 1 2 2 2 1] = [1 + 4 + 4 2 + 2 + 4 2 + 4 + 2 2 + 2 + 4 4 + 1 + 4 4 + 2 + 2 2 + 4 + 2 4 + 2 + 2 1 + 4 + 4] = [9 8 8 8 9 8 8 8 9] Now, A^{2} - 4 A - 5 I = [9 8 8 8 9 8 8 8 9] - 4 [1 2 2 2 1 2 2 2 1] - 5 [1 0 0 0 1 0 0 0 1] = [9 - 4 - 5 8 - 8 - 0 8 - 8 - 0 8 - 8 - 0 9 - 4 - 5 8 - 8 - 0 8 - 8 - 0 8 - 8 - 0 9 - 4 - 5] = [0 0 0 0 0 0 0 0 0] = O \Rightarrow A^{2} - 4 A - 5 I = O [Proved] Again, A^{2} - 4 A - 5 I = O \Rightarrow A^{- 1} (A^{2} - 4 A - 5 I) = A^{- 1} O [Pre - multiplying with A^{- 1}] \Rightarrow A^{- 1} A^{2} - 4 A^{- 1} A - 5 A^{- 1} = O \Rightarrow A - 4 I = 5 A^{- 1} \Rightarrow 5 A^{- 1} = [1 2 2 2 1 2 2 2 1] - 4 [1 0 0 0 1 0 0 0 1] = [1 - 4 2 - 0 2 - 0 2 - 0 1 - 4 2 - 0 2 - 0 2 - 0 1 - 4] = [- 3 2 2 2 - 3 2 2 2 - 3] \Rightarrow A^{- 1} = \frac{1}{5} [- 3 2 2 2 - 3 2 2 2 - 3]$

Page No 7.25:

Question 35:

Prove that $|A adj A| = {|A|}^{n}$ .

Answer:

$Let A = [a_{i j}] be a square matrix of order n \times n . If C_{i j} is a cofactor of a_{i j} in A, then adj A = {[C_{i j}]}^{T} = [C_{j i}] . Also, it is a matrix of order n \times n . Because A and adj A are matrices of order n \times n, A \times (adj A) exists and is of order n \times n . \Rightarrow {\{A \times (adj A)\}}_{i j} = \sum_{r = 1}^{n} A_{i r} {(adj A)}_{r j} {= \sum}_{r = 1}^{n} a_{i r} C_{r j} = \{\begin{cases} |A| i f i = j \\ 0 otherwise \end{cases} Thus, each diagonal element of A \times (adj A) is |A| . Also, the non - diagonal elements are zero . \Rightarrow A \times (adj A) = [|A| 0 0 \dots \dots \dots \dots 0 0 |A| 0 \dots \dots \dots \dots \dots 0 0 0 |A| 0 \dots \dots \dots 0 ⋮ 0 0 0 \dots \dots \dots \dots \dots \dots \dots \dots \dots |A|] \Rightarrow |A \times (adj A)| = ||A| 0 0 \dots \dots \dots \dots 0 0 |A| 0 \dots \dots \dots \dots \dots 0 0 0 |A| 0 \dots \dots \dots 0 ⋮ 0 0 0 \dots \dots \dots \dots \dots \dots \dots \dots \dots |A|| = {|A|}^{n} |1 0 0 \dots \dots \dots \dots 0 0 1 0 \dots \dots \dots \dots \dots 0 0 0 1 0 \dots \dots \dots 0 ⋮ 0 0 0 \dots \dots \dots \dots \dots \dots \dots \dots \dots 1| = {|A|}^{n} I_{n} = {|A|}^{n} \Rightarrow |A \times (adj A)| = {|A|}^{n} Hence proved .$

Page No 7.25:

Question 36:

$If A^{- 1} = [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}] and B = [\begin{matrix} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{matrix}], find {(A B)}^{- 1} .$

Answer:

We know that (AB)⁻¹ = B⁻¹ A⁻¹.

$B = [\begin{matrix} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{matrix}] B^{- 1} = \frac{1}{|B|} Adj . B Now, |B| = |\begin{matrix} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{matrix}| = 1 (3 + 0) + 1 (2 - 4) = 1 Now, to find Adj . B \begin{matrix} B_{11} = {(- 1)}^{1 + 1} (3) = 3 B_{12} = {(- 1)}^{1 + 2} (- 1) = 1 B_{13} = {(- 1)}^{1 + 3} (2) = 2 & B_{21} = {(- 1)}^{2 + 1} (2 - 4) = 2 B_{22} = {(- 1)}^{2 + 2} (1) = 1 B_{23} = {(- 1)}^{2 + 3} (- 2) = 2 & B_{31} = {(- 1)}^{3 + 1} (6) = 6 B_{32} = {(- 1)}^{3 + 2} (- 2) = 2 B_{33} = {(- 1)}^{3 + 3} (3 + 2) = 5 \end{matrix} Therefore, Adj . B = [\begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{matrix}] Thus, B^{- 1} = [\begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{matrix}] . {(A B)}^{- 1} = B^{- 1} A^{- 1} = [\begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{matrix}] [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}] = [\begin{matrix} 9 - 30 + 30 & - 3 + 12 - 12 & 3 - 10 + 12 \\ 3 - 15 + 10 & - 1 + 6 - 4 & 1 - 5 + 4 \\ 6 - 30 + 25 & - 2 + 12 - 10 & 2 - 10 + 10 \end{matrix}] = [\begin{matrix} 9 & - 3 & 5 \\ - 2 & 1 & 0 \\ 1 & 0 & 2 \end{matrix}] Hence, {(A B)}^{- 1} = [\begin{matrix} 9 & - 3 & 5 \\ - 2 & 1 & 0 \\ 1 & 0 & 2 \end{matrix}]$

Page No 7.25:

Question 37:

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

Answer:

We know that (A^T)⁻¹ = (A⁻¹)^T.

$A = [\begin{matrix} 1 & - 2 & 3 \\ 0 & - 1 & 4 \\ - 2 & 2 & 1 \end{matrix}] A^{- 1} = \frac{1}{|A|} Adj . A Now, |A| = |\begin{matrix} 1 & - 2 & 3 \\ 0 & - 1 & 4 \\ - 2 & 2 & 1 \end{matrix}| = 1 (- 1 - 8) - 2 (- 8 + 3) = - 9 + 10 = 1 Now, to find Adj . A \begin{matrix} A_{11} = {(- 1)}^{1 + 1} (- 9) = - 9 A_{12} = {(- 1)}^{1 + 2} (8) = - 8 A_{13} = {(- 1)}^{1 + 3} (- 2) = - 2 & A_{21} = {(- 1)}^{2 + 1} (- 8) = 8 A_{22} = {(- 1)}^{2 + 2} (7) = 7 A_{23} = {(- 1)}^{2 + 3} (- 2) = 2 & A_{31} = {(- 1)}^{3 + 1} (- 5) = - 5 A_{32} = {(- 1)}^{3 + 2} (4) = - 4 A_{33} = {(- 1)}^{3 + 3} (- 1) = - 1 \end{matrix} Therefore, Adj . A = [\begin{matrix} - 9 & 8 & - 5 \\ - 8 & 7 & - 4 \\ - 2 & 2 & - 1 \end{matrix}] Thus, A^{- 1} = [\begin{matrix} - 9 & 8 & - 5 \\ - 8 & 7 & - 4 \\ - 2 & 2 & - 1 \end{matrix}] . {(A^{T})}^{- 1} = {(A^{- 1})}^{T} = {[\begin{matrix} - 9 & 8 & - 5 \\ - 8 & 7 & - 4 \\ - 2 & 2 & - 1 \end{matrix}]}^{T} = [\begin{matrix} - 9 & - 8 & - 2 \\ 8 & 7 & 2 \\ - 5 & - 4 & - 1 \end{matrix}] Hence, {(A^{T})}^{- 1} = [\begin{matrix} - 9 & - 8 & - 2 \\ 8 & 7 & 2 \\ - 5 & - 4 & - 1 \end{matrix}] .$

Page No 7.25:

Question 38:

Find the adjoint of the matrix $A = [\begin{matrix} - 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1 \end{matrix}]$ and hence show that $A (adj A) = |A| I_{3}$ .

Answer:

$A = [\begin{matrix} - 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1 \end{matrix}] Now, to find Adj . A \begin{matrix} A_{11} = {(- 1)}^{1 + 1} (- 3) = - 3 A_{12} = {(- 1)}^{1 + 2} (6) = - 6 A_{13} = {(- 1)}^{1 + 3} (- 6) = - 6 & A_{21} = {(- 1)}^{2 + 1} (- 6) = 6 A_{22} = {(- 1)}^{2 + 2} (3) = 3 A_{23} = {(- 1)}^{2 + 3} (6) = - 6 & A_{31} = {(- 1)}^{3 + 1} (6) = 6 A_{32} = {(- 1)}^{3 + 2} (6) = - 6 A_{33} = {(- 1)}^{3 + 3} (3) = 3 \end{matrix} Therefore, Adj . A = [\begin{matrix} - 3 & 6 & 6 \\ - 6 & 3 & - 6 \\ - 6 & - 6 & 3 \end{matrix}] Now, |A| = |\begin{matrix} - 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1 \end{matrix}| = - 1 (1 - 4) - 2 (- 2 - 4) + 2 (4 + 2) = 3 + 12 + 12 = 27 To show : A (a d j A) = |A| I_{3} LHS = [\begin{matrix} - 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1 \end{matrix}] [\begin{matrix} - 3 & 6 & 6 \\ - 6 & 3 & - 6 \\ - 6 & - 6 & 3 \end{matrix}] = [\begin{matrix} 3 + 12 + 12 & - 6 - 6 + 12 & - 6 + 12 - 6 \\ - 6 - 6 + 12 & 12 + 3 + 12 & 12 - 6 - 6 \\ - 6 + 12 - 6 & 12 - 6 - 6 & 12 + 12 + 3 \end{matrix}] = [\begin{matrix} 27 & 0 & 0 \\ 0 & 27 & 0 \\ 0 & 0 & 27 \end{matrix}] = 27 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = |A| I_{3} = RHS Hence, A (adj A) = |A| I_{3} .$

Page No 7.25:

Question 39:

$If A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}], find A^{- 1} and show that A^{- 1} = \frac{1}{2} (A^{2} - 3 I) .$

Answer:

$A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] A^{- 1} = \frac{1}{|A|} Adj . A Now, |A| = |\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}| = - 1 (- 1) + 1 (1) = 2 Now, to find Adj . A \begin{matrix} A_{11} = {(- 1)}^{1 + 1} (- 1) = - 1 A_{12} = {(- 1)}^{1 + 2} (- 1) = 1 A_{13} = {(- 1)}^{1 + 3} (1) = 1 & A_{21} = {(- 1)}^{2 + 1} (- 1) = 1 A_{22} = {(- 1)}^{2 + 2} (- 1) = - 1 A_{23} = {(- 1)}^{2 + 3} (- 1) = 1 & A_{31} = {(- 1)}^{3 + 1} (1) = 1 A_{32} = {(- 1)}^{3 + 2} (- 1) = 1 A_{33} = {(- 1)}^{3 + 3} (- 1) = - 1 \end{matrix} Therefore, Adj . A = [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] Thus, A^{- 1} = \frac{1}{2} [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] . Now, A^{2} = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] = [\begin{matrix} 0 + 1 + 1 & 0 + 0 + 1 & 0 + 1 + 0 \\ 0 + 0 + 1 & 1 + 0 + 1 & 1 + 0 + 0 \\ 0 + 1 + 0 & 1 + 0 + 0 & 1 + 1 + 0 \end{matrix}] = [\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] Now, to show A^{- 1} = \frac{1}{2} (A^{2} - 3 I) RHS = \frac{1}{2} (A^{2} - 3 I) = \frac{1}{2} ([\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] - 3 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]) = \frac{1}{2} [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] = A^{- 1} = LHS Hence, A^{- 1} = \frac{1}{2} (A^{2} - 3 I) .$

Page No 7.34:

Question 1:

Find the inverse of each of the following matrices by using elementary row transformations:

$[\begin{matrix} 7 & 1 \\ 4 & - 3 \end{matrix}]$

Answer:

$A = [7 1 4 - 3] We know A = I A \Rightarrow [7 1 4 - 3] = [1 0 0 1] A \Rightarrow [1 \frac{1}{7} 4 - 3] = [\frac{1}{7} 0 0 1] A (Applying R_{1} \to \frac{1}{7} R_{1}) \Rightarrow [1 \frac{1}{7} 0 - \frac{25}{7}] = [\frac{1}{7} 0 - \frac{4}{7} 1] A (Applying R_{2} \to R_{2} - 4 R_{1}) \Rightarrow [1 \frac{1}{7} 0 1] = [\frac{1}{7} 0 \frac{4}{25} - \frac{7}{25}] A (Applying R_{2} \to - \frac{7}{25} R_{2}) \Rightarrow [1 0 0 1] = [\frac{3}{25} \frac{1}{25} \frac{4}{25} - \frac{7}{25}] A (Applying R_{1} \to R_{1} - \frac{1}{7} R_{2}) ∴ A^{- 1} = \frac{1}{25} [3 1 4 - 7]$

Page No 7.34:

Question 2:

Find the inverse of each of the following matrices by using elementary row transformations:

$[\begin{matrix} 5 & 2 \\ 2 & 1 \end{matrix}]$

Answer:

$A = [5 2 2 1] We know A = I A \Rightarrow [5 2 2 1] = [1 0 0 1] A \Rightarrow [5 - 4 2 - 2 2 1] = [1 - 0 0 - 2 0 1] A [Applying R_{1} \to R_{1} - 2 R_{2}] \Rightarrow [1 0 2 1] = [1 - 2 0 1] A \Rightarrow [1 0 2 - 2 1] = [1 - 2 0 - 2 1 + 4] A [Applying R_{2} \to R_{2} - 2 R_{1}] \Rightarrow [1 0 0 1] = [1 - 2 - 2 5] A \Rightarrow A^{- 1} = [1 - 2 - 2 5]$

Page No 7.34:

Question 3:

Find the inverse of each of the following matrices by using elementary row transformations:

$[\begin{matrix} 1 & 6 \\ - 3 & 5 \end{matrix}]$

Answer:

$A = [1 6 - 3 5] We know A = I A \Rightarrow [1 6 - 3 5] = [1 0 0 1] A \Rightarrow [1 6 - 3 + 3 5 + 18] = [1 0 0 + 3 1 + 0] A [Applying R_{2} \to R_{2} + 3 R_{1}] \Rightarrow [1 6 0 23] = [1 0 3 1] A \Rightarrow [1 6 - 6 0 23] = [1 - \frac{18}{23} 0 - \frac{6}{23} 3 1] A [Applying R_{1} \to R_{1} - \frac{6}{23} R_{2}] \Rightarrow [1 0 0 1] = [\frac{5}{23} \frac{- 6}{23} \frac{3}{23} \frac{1}{23}] A [Applying R_{2} \to \frac{1}{23} R_{2}] \Rightarrow A^{- 1} = [\frac{5}{23} \frac{- 6}{23} \frac{3}{23} \frac{1}{23}] = \frac{1}{23} [5 - 6 3 1] \Rightarrow A^{- 1} = \frac{1}{23} [5 - 6 3 1]$

Page No 7.34:

Question 4:

Find the inverse of each of the following matrices by using elementary row transformations:

$[\begin{matrix} 2 & 5 \\ 1 & 3 \end{matrix}]$

Answer:

$A = [2 5 1 3] We know A = I A \Rightarrow [2 5 1 3] = [1 0 0 1] A \Rightarrow [2 - 1 5 - 3 1 3] = [1 - 0 0 - 1 0 1] A [Applying R_{1} \to R_{1} - R_{2}] \Rightarrow [1 2 1 3] = [1 - 1 0 1] A \Rightarrow [1 2 1 - 1 3 - 2] = [1 - 1 0 - 1 1 + 1] A [Applying R_{2} \to R_{2} - R_{1}] \Rightarrow [1 2 0 1] = [1 - 1 - 1 2] A \Rightarrow [1 0 0 1] = [1 + 2 - 1 - 4 - 1 2] A [Applying R_{1} \to R_{1} - 2 R_{2}] \Rightarrow [1 0 0 1] = [3 - 5 - 1 2] A \Rightarrow A^{- 1} = [3 - 5 - 1 2]$

Page No 7.34:

Question 5:

Find the inverse of each of the following matrices by using elementary row transformations:

$[\begin{matrix} 3 & 10 \\ 2 & 7 \end{matrix}]$

Answer:

$A = [3 10 2 7] We know A = I A \Rightarrow [3 10 2 7] = [1 0 0 1] A \Rightarrow [3 - 2 10 - 7 2 7] = [1 - 0 0 - 1 0 1] A [Applying R_{1} \to R_{1} - R_{2}] \Rightarrow [1 3 2 7] = [1 - 1 0 1] A \Rightarrow [1 3 2 - 2 7 - 6] = [1 - 1 0 - 2 1 + 2] [Applying R_{2} \to R_{2} - 2 R_{1}] \Rightarrow [1 3 0 1] = [1 - 1 - 2 3] A \Rightarrow [1 3 - 3 0 1] = [1 + 6 - 1 - 9 - 2 3] A [Applying R_{1} \to R_{1} - 3 R_{2}] \Rightarrow [1 0 0 1] = [7 - 10 - 2 3] A \Rightarrow A^{- 1} = [7 - 10 - 2 3]$

Page No 7.34:

Question 6:

Find the inverse of each of the following matrices by using elementary row transformations:

$[\begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix}]$

Answer:

$A = [\begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix}] We know A = I A \Rightarrow [\begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A \Rightarrow [\begin{matrix} 0 & 1 & 2 \\ - 2 & 1 & 2 \\ 3 & 1 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \end{matrix}] A [Applying R_{2} \to R_{2} - R_{3}] \Rightarrow [\begin{matrix} - 3 & 0 & 1 \\ - 2 & 1 & 2 \\ 3 & 1 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & - 1 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \end{matrix}] A [Applying R_{1} \to R_{1} - R_{3}] \Rightarrow [\begin{matrix} - 3 & 0 & 1 \\ - 2 & 1 & 2 \\ 0 & 1 & 2 \end{matrix}] = [\begin{matrix} 1 & 0 & - 1 \\ 0 & 1 & - 1 \\ 1 & 0 & 0 \end{matrix}] A [Applying R_{3} \to R_{3} + R_{1}] \Rightarrow [\begin{matrix} - 3 & 0 & 1 \\ 0 & 3 & 4 \\ 0 & 1 & 2 \end{matrix}] = [\begin{matrix} 1 & 0 & - 1 \\ - 2 & 3 & - 1 \\ 1 & 0 & 0 \end{matrix}] A [Applying R_{2} \to 3 R_{2} - 2 R_{1}] \Rightarrow [\begin{matrix} - 3 & 0 & 1 \\ 0 & 3 & 4 \\ 0 & - 2 & - 2 \end{matrix}] = [\begin{matrix} 1 & 0 & - 1 \\ - 2 & 3 & - 1 \\ 3 & - 3 & 1 \end{matrix}] A [Applying R_{3} \to R_{3} - R_{2}] \Rightarrow [\begin{matrix} - 3 & 0 & 1 \\ 0 & 3 & 4 \\ 0 & 1 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & - 1 \\ - 2 & 3 & - 1 \\ \frac{- 3}{2} & \frac{3}{2} & \frac{- 1}{2} \end{matrix}] A [Applying R_{3} \to - \frac{1}{2} R_{3}] \Rightarrow [\begin{matrix} - 3 & 0 & 1 \\ 0 & - 1 & 0 \\ 0 & 1 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & - 1 \\ 4 & - 3 & 1 \\ \frac{- 3}{2} & \frac{3}{2} & \frac{- 1}{2} \end{matrix}] A [Applying R_{2} \to R_{2} - 4 R_{3}] \Rightarrow [\begin{matrix} - 3 & 0 & 1 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & - 1 \\ 4 & - 3 & 1 \\ \frac{5}{2} & \frac{- 3}{2} & \frac{1}{2} \end{matrix}] A [Applying R_{3} \to R_{3} + R_{2}] \Rightarrow [\begin{matrix} - 3 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} - \frac{3}{2} & \frac{3}{2} & - \frac{3}{2} \\ 4 & - 3 & 1 \\ \frac{5}{2} & \frac{- 3}{2} & \frac{1}{2} \end{matrix}] A [Applying R_{1} \to R_{1} - R_{3}] \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{2} & \frac{- 1}{2} & \frac{1}{2} \\ 4 & - 3 & 1 \\ \frac{5}{2} & \frac{- 3}{2} & \frac{1}{2} \end{matrix}] A [Applying R_{1} \to \frac{- 1}{3} R_{1}] \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{2} & \frac{- 1}{2} & \frac{1}{2} \\ - 4 & 3 & - 1 \\ \frac{5}{2} & \frac{- 3}{2} & \frac{1}{2} \end{matrix}] A [Applying R_{2} \to - R_{2}]$

Page No 7.34:

Question 7:

Find the inverse of each of the following matrices by using elementary row transformations:

$[\begin{matrix} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix}]$

Answer:

$A = [\begin{matrix} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix}] We know A = I A \Rightarrow [\begin{matrix} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A \Rightarrow [\begin{matrix} 1 & 0 & \frac{- 1}{2} \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A [Applying R_{1} \to \frac{1}{2} R_{1}] \Rightarrow [\begin{matrix} 1 & 0 & \frac{- 1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & 3 \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 & 0 \\ \frac{- 5}{2} & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A [Applying R_{2} \to R_{2} - 5 R_{1}] \Rightarrow [\begin{matrix} 1 & 0 & \frac{- 1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & \frac{1}{2} \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 & 0 \\ \frac{- 5}{2} & 1 & 0 \\ \frac{5}{2} & - 1 & 1 \end{matrix}] A [Applying R_{3} \to R_{3} - R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & \frac{- 1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 & 0 \\ - \frac{5}{2} & 1 & 0 \\ 5 & - 2 & 2 \end{matrix}] A [Applying R_{3} \to 2 R_{3}] \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}] A [Applying R_{1} \to R_{1} + \frac{1}{2} R_{3} and R_{2} \to R_{2} - \frac{5}{2} R_{3}] \Rightarrow A^{- 1} = [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}]$

Page No 7.34:

Question 8:

Find the inverse of each of the following matrices by using elementary row transformations:

$[\begin{matrix} 2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{matrix}]$

Answer:

$A = [\begin{matrix} 2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{matrix}] We know A = I A \Rightarrow [\begin{matrix} 2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A \Rightarrow [\begin{matrix} 1 & \frac{3}{2} & \frac{1}{2} \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A [Applying R_{1} \to \frac{1}{2} R_{1}] \Rightarrow [\begin{matrix} 1 & \frac{3}{2} & \frac{1}{2} \\ 0 & 1 & 0 \\ 0 & \frac{5}{2} & \frac{1}{2} \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 & 0 \\ - 1 & 1 & 0 \\ \frac{- 3}{2} & 0 & 1 \end{matrix}] A [Applying R_{2} \to R_{2} - 2 R_{1} and R_{3} \to R_{3} - 3 R_{1}] \Rightarrow [\begin{matrix} 1 & 0 & \frac{1}{2} \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{2} \end{matrix}] = [\begin{matrix} 2 & \frac{- 3}{2} & 0 \\ - 1 & 1 & 0 \\ 1 & \frac{- 5}{2} & 1 \end{matrix}] A [Applying R_{1} \to R_{1} - \frac{3}{2} R_{2} and R_{3} \to R_{3} - \frac{5}{2} R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & \frac{1}{2} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 2 & \frac{- 3}{2} & 0 \\ - 1 & 1 & 0 \\ 2 & - 5 & 2 \end{matrix}] A [Applying R_{3} \to 2 R_{3}] \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 1 & - 1 \\ - 1 & 1 & 0 \\ 2 & - 5 & 2 \end{matrix}] A [Applying R_{1} \to R_{1} - \frac{1}{2} R_{3}] ∴ A^{- 1} = [\begin{matrix} 1 & 1 & - 1 \\ - 1 & 1 & 0 \\ 2 & - 5 & 2 \end{matrix}]$

Page No 7.34:

Question 9:

Find the inverse of each of the following matrices by using elementary row transformations:

$[\begin{matrix} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{matrix}]$

Answer:

$A = [\begin{matrix} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{matrix}] We know A = I A \Rightarrow [\begin{matrix} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A \Rightarrow [\begin{matrix} 1 & - 1 & \frac{4}{3} \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A [Applying R_{1} \to \frac{1}{3} R_{1}] \Rightarrow [\begin{matrix} 1 & - 1 & \frac{4}{3} \\ 0 & - 1 & \frac{4}{3} \\ 0 & - 1 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{3} & 0 & 0 \\ \frac{- 2}{3} & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A [Applying R_{2} \to R_{2} - 2 R_{1}] \Rightarrow [\begin{matrix} 1 & - 1 & \frac{4}{3} \\ 0 & 1 & \frac{- 4}{3} \\ 0 & - 1 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{3} & 0 & 0 \\ \frac{2}{3} & - 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A [Applying R_{2} \to - R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & \frac{- 4}{3} \\ 0 & 0 & \frac{- 1}{3} \end{matrix}] = [\begin{matrix} 1 & - 1 & 0 \\ \frac{2}{3} & - 1 & 0 \\ \frac{2}{3} & - 1 & 1 \end{matrix}] A [Applying R_{1} \to R_{1} + R_{2} and R_{3} \to R_{3} + R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & \frac{- 4}{3} \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & - 1 & 0 \\ \frac{2}{3} & - 1 & 0 \\ - 2 & 3 & - 3 \end{matrix}] A [Applying R_{3} \to - 3 R_{3}] \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & - 1 & 0 \\ - 2 & 3 & - 4 \\ - 2 & 3 & - 3 \end{matrix}] A [Applying R_{2} \to R_{2} + \frac{4}{3} R_{3}] ∴ A^{- 1} = [\begin{matrix} 1 & - 1 & 0 \\ - 2 & 3 & - 4 \\ - 2 & 3 & - 3 \end{matrix}]$

Page No 7.34:

Question 10:

Find the inverse of each of the following matrices by using elementary row transformations:

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

Answer:

$A = [\begin{matrix} 1 & 2 & 0 \\ 2 & 3 & - 1 \\ 1 & - 1 & 3 \end{matrix}] We know A = I A \Rightarrow [\begin{matrix} 1 & 2 & 0 \\ 2 & 3 & - 1 \\ 1 & - 1 & 3 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A \Rightarrow [\begin{matrix} 1 & 2 & 0 \\ 0 & - 1 & - 1 \\ 0 & - 3 & 3 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ - 2 & 1 & 0 \\ - 1 & 0 & 1 \end{matrix}] A [Applying R_{2} \to R_{2} - 2 R_{1} and R_{3} \to R_{3} - R_{1}] \Rightarrow [\begin{matrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & - 3 & 3 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 2 & - 1 & 0 \\ - 1 & 0 & 1 \end{matrix}] A [Applying R_{2} \to - R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & - 2 \\ 0 & 1 & 1 \\ 0 & 0 & 6 \end{matrix}] = [\begin{matrix} - 3 & 2 & 0 \\ 2 & - 1 & 0 \\ 5 & - 3 & 1 \end{matrix}] A [Applying R_{1} \to R_{1} - 2 R_{2} and R_{3} \to R_{3} + 3 R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & - 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} - 3 & 2 & 0 \\ 2 & - 1 & 0 \\ \frac{5}{6} & - \frac{1}{2} & \frac{1}{6} \end{matrix}] A [Applying R_{3} \to \frac{1}{6} R_{3}] \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} - \frac{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & - \frac{1}{2} & \frac{- 1}{6} \\ \frac{5}{6} & - \frac{1}{2} & \frac{1}{6} \end{matrix}] A [Applying R_{1} \to R_{1} + 2 R_{3} and R_{2} \to R_{2} - R_{3}] ∴ A^{- 1} = [\begin{matrix} - \frac{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & - \frac{1}{2} & \frac{- 1}{6} \\ \frac{5}{6} & - \frac{1}{2} & \frac{1}{6} \end{matrix}]$

Page No 7.34:

Question 11:

Find the inverse of each of the following matrices by using elementary row transformations:

$[\begin{matrix} 2 & - 1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1 \end{matrix}]$

Answer:

$A = [\begin{matrix} 2 & - 1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1 \end{matrix}] We know A = I A \Rightarrow [\begin{matrix} 2 & - 1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A \Rightarrow [\begin{matrix} 1 & - \frac{1}{2} & \frac{3}{2} \\ 1 & 2 & 4 \\ 3 & 1 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A [Applying R_{1} \to \frac{1}{2} R_{1}] \Rightarrow [\begin{matrix} 1 & - \frac{1}{2} & \frac{3}{2} \\ 0 & \frac{5}{2} & \frac{5}{2} \\ 0 & \frac{5}{2} & \frac{- 7}{2} \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 & 0 \\ - \frac{1}{2} & 1 & 0 \\ - \frac{3}{2} & 0 & 1 \end{matrix}] A [Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - 3 R_{1}] \Rightarrow [\begin{matrix} 1 & - \frac{1}{2} & \frac{3}{2} \\ 0 & 1 & 1 \\ 0 & \frac{5}{2} & \frac{- 7}{2} \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 & 0 \\ - \frac{1}{5} & \frac{2}{5} & 0 \\ - \frac{3}{2} & 0 & 1 \end{matrix}] A [Applying R_{2} \to \frac{2}{5} R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & - 6 \end{matrix}] = [\begin{matrix} \frac{2}{5} & \frac{1}{5} & 0 \\ - \frac{1}{5} & \frac{2}{5} & 0 \\ - 1 & - 1 & 1 \end{matrix}] A [Applying R_{1} \to R_{1} + \frac{1}{2} R_{2} and R_{3} \to R_{3} - \frac{5}{2} R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \frac{2}{5} & \frac{1}{5} & 0 \\ - \frac{1}{5} & \frac{2}{5} & 0 \\ \frac{1}{6} & \frac{1}{6} & - \frac{1}{6} \end{matrix}] A [Applying R_{3} \to - \frac{1}{6} R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{15} & \frac{- 2}{15} & \frac{1}{3} \\ - \frac{11}{30} & \frac{7}{30} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} & - \frac{1}{6} \end{matrix}] A [Applying R_{2} \to R_{2} - R_{3} and R_{1} \to R_{1} - 2 R_{3}] \Rightarrow A^{- 1} = - \frac{1}{30} [\begin{matrix} - 2 & 4 & - 10 \\ 11 & - 7 & - 5 \\ - 5 & - 5 & 5 \end{matrix}]$

Page No 7.34:

Question 12:

Find the inverse of each of the following matrices by using elementary row transformations:

$[\begin{matrix} 1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1 \end{matrix}]$

Answer:

$A = [\begin{matrix} 1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1 \end{matrix}] We know A = I A \Rightarrow [\begin{matrix} 1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A \Rightarrow [\begin{matrix} 1 & 1 & 2 \\ 0 & - 2 & - 5 \\ 0 & 1 & - 3 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ - 3 & 1 & 0 \\ - 2 & 0 & 1 \end{matrix}] A [Applying R_{2} \to R_{2} - 3 R_{1} and R_{3} \to R_{3} - 2 R_{1}] \Rightarrow [\begin{matrix} 1 & 1 & 2 \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & - 3 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ \frac{3}{2} & - \frac{1}{2} & 0 \\ - 2 & 0 & 1 \end{matrix}] A [Applying R_{2} \to \frac{- 1}{2} R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & - \frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & - \frac{11}{2} \end{matrix}] = [\begin{matrix} - \frac{1}{2} & \frac{1}{2} & 0 \\ \frac{3}{2} & - \frac{1}{2} & 0 \\ - \frac{7}{2} & \frac{1}{2} & 1 \end{matrix}] A [Applying R_{1} \to R_{1} - R_{2} and R_{3} \to R_{3} - R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & - \frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} - \frac{1}{2} & \frac{1}{2} & 0 \\ \frac{3}{2} & - \frac{1}{2} & 0 \\ \frac{7}{11} & \frac{- 1}{11} & \frac{- 2}{11} \end{matrix}] A [Applying R_{3} \to - \frac{2}{11} R_{3}] \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \frac{- 2}{11} & \frac{5}{11} & \frac{- 1}{11} \\ \frac{- 1}{11} & \frac{- 3}{11} & \frac{5}{11} \\ \frac{7}{11} & \frac{- 1}{11} & \frac{- 2}{11} \end{matrix}] A [Applying R_{2} \to R_{2} - \frac{5}{2} R_{3} and R_{1} \to R_{1} + \frac{1}{2} R_{3}] \Rightarrow A^{- 1} = \frac{1}{11} [\begin{matrix} - 2 & 5 & - 1 \\ - 1 & - 3 & 5 \\ 7 & - 1 & - 2 \end{matrix}]$

Page No 7.34:

Question 13:

Find the inverse of each of the following matrices by using elementary row transformations:

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

Answer:

$A = [\begin{matrix} 2 & - 1 & 4 \\ 4 & 0 & 2 \\ 3 & - 2 & 7 \end{matrix}] We know A = I A \Rightarrow [\begin{matrix} 2 & - 1 & 4 \\ 4 & 0 & 2 \\ 3 & - 2 & 7 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A \Rightarrow [\begin{matrix} 1 & - \frac{1}{2} & 2 \\ 4 & 0 & 2 \\ 3 & - 2 & 7 \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A [Applying R_{1} \to \frac{1}{2} R_{1}] \Rightarrow [\begin{matrix} 1 & - \frac{1}{2} & 2 \\ 0 & 2 & - 6 \\ 0 & - \frac{1}{2} & 1 \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 & 0 \\ - 2 & 1 & 0 \\ \frac{- 3}{2} & 0 & 1 \end{matrix}] A [Applying R_{2} \to R_{2} - 4 R_{1} and R_{3} \to R_{3} - 3 R_{1}] \Rightarrow [\begin{matrix} 1 & - \frac{1}{2} & 2 \\ 0 & 1 & - 3 \\ 0 & - \frac{1}{2} & 1 \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 & 0 \\ - 1 & \frac{1}{2} & 0 \\ \frac{- 3}{2} & 0 & 1 \end{matrix}] A [Applying R_{2} \to \frac{1}{2} R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & \frac{1}{2} \\ 0 & 1 & - 3 \\ 0 & 0 & - \frac{1}{2} \end{matrix}] = [\begin{matrix} 0 & \frac{1}{4} & 0 \\ - 1 & \frac{1}{2} & 0 \\ - 2 & \frac{1}{4} & 1 \end{matrix}] A [Applying R_{1} \to R_{1} + \frac{1}{2} R_{2} and R_{3} \to R_{3} + \frac{1}{2} R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & \frac{1}{2} \\ 0 & 1 & - 3 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 0 & \frac{1}{4} & 0 \\ - 1 & \frac{1}{2} & 0 \\ 4 & - \frac{1}{2} & - 2 \end{matrix}] A [Applying R_{3} \to - 2 R_{3}] \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} - 2 & \frac{1}{2} & 1 \\ 11 & - 1 & - 6 \\ 4 & - \frac{1}{2} & - 2 \end{matrix}] A [Applying R_{1} \to R_{1} - \frac{1}{2} R_{3} and R_{2} \to R_{2} + 3 R_{3}] \Rightarrow A^{- 1} = [\begin{matrix} - 2 & \frac{1}{2} & 1 \\ 11 & - 1 & - 6 \\ 4 & - \frac{1}{2} & - 2 \end{matrix}]$

Page No 7.34:

Question 14:

Find the inverse of each of the following matrices by using elementary row transformations:

$[\begin{matrix} 3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{matrix}]$

Answer:

$A = [\begin{matrix} 3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{matrix}] We know A = I A \Rightarrow [\begin{matrix} 3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A \Rightarrow [\begin{matrix} 1 & 0 & - \frac{1}{3} \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A [Applying R_{1} \to \frac{1}{3} R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & - \frac{1}{3} \\ 0 & 3 & \frac{2}{3} \\ 0 & 4 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{3} & 0 & 0 \\ - \frac{2}{3} & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A [Applying R_{2} \to R_{2} - 2 R_{1}] \Rightarrow [\begin{matrix} 1 & 0 & - \frac{1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 4 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{3} & 0 & 0 \\ - \frac{2}{9} & \frac{1}{3} & 0 \\ 0 & 0 & 1 \end{matrix}] A [Applying R_{2} \to \frac{1}{3} R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & - \frac{1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & \frac{1}{9} \end{matrix}] = [\begin{matrix} \frac{1}{3} & 0 & 0 \\ - \frac{2}{9} & \frac{1}{3} & 0 \\ \frac{8}{9} & \frac{- 4}{3} & 1 \end{matrix}] A [Applying R_{3} \to R_{3} - 4 R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & - \frac{1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{3} & 0 & 0 \\ - \frac{2}{9} & \frac{1}{3} & 0 \\ 8 & - 12 & 9 \end{matrix}] A [Applying R_{3} \to 9 R_{3}] \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 3 & - 4 & 3 \\ - 2 & 3 & - 2 \\ 8 & - 12 & 9 \end{matrix}] A [Applying R_{2} \to R_{2} - \frac{2}{9} R_{3} and R_{1} \to R_{1} + \frac{1}{3} R_{3}] ∴ A^{- 1} = [\begin{matrix} 3 & - 4 & 3 \\ - 2 & 3 & - 2 \\ 8 & - 12 & 9 \end{matrix}]$

Page No 7.34:

Question 15:

Find the inverse of each of the following matrices by using elementary row transformations:

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

Answer:

$A = [\begin{matrix} 1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0 \end{matrix}] We know A = I A \Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A \Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ 0 & 9 & - 7 \\ 0 & - 5 & 4 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ - 2 & 0 & 1 \end{matrix}] A [Applying R_{2} \to R_{2} + 3 R_{1} and R_{3} \to R_{3} - 2 R_{1}] \Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ 0 & 1 & - \frac{7}{9} \\ 0 & - 5 & 4 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ - 2 & 0 & 1 \end{matrix}] A [Applying R_{2} \to \frac{1}{9} R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & \frac{1}{3} \\ 0 & 1 & - \frac{7}{9} \\ 0 & 0 & \frac{1}{9} \end{matrix}] = [\begin{matrix} 0 & - \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ - \frac{1}{3} & \frac{5}{9} & 1 \end{matrix}] A [Applying R_{1} \to R_{1} - 3 R_{2} and R_{3} \to R_{3} + 5 R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & \frac{1}{3} \\ 0 & 1 & - \frac{7}{9} \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 0 & - \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ - 3 & 5 & 9 \end{matrix}] A [Applying R_{3} \to 9 R_{3}] \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & - 2 & - 3 \\ - 2 & 4 & 7 \\ - 3 & 5 & 9 \end{matrix}] A [Applying R_{2} \to R_{2} + \frac{7}{9} R_{3} and R_{1} \to R_{1} - \frac{1}{3} R_{3}] \Rightarrow A^{- 1} = [\begin{matrix} 1 & - 2 & - 3 \\ - 2 & 4 & 7 \\ - 3 & 5 & 9 \end{matrix}]$

Page No 7.34:

Question 16:

Find the inverse of each of the following matrices by using elementary row transformations:

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

Answer:

$Let A = [\begin{matrix} - 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix}] . To find inverse, first write A = I A . i . e ., [\begin{matrix} - 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A \Rightarrow [\begin{matrix} 1 & - 1 & - 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix}] = [\begin{matrix} - 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A [Applying R_{1} \to (- 1) R_{1}] \Rightarrow [\begin{matrix} 1 & - 1 & - 2 \\ 0 & 3 & 5 \\ 0 & 4 & 7 \end{matrix}] = [\begin{matrix} - 1 & 0 & 0 \\ 1 & 1 & 0 \\ 3 & 0 & 1 \end{matrix}] A [Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - 3 R_{1}] \Rightarrow [\begin{matrix} 1 & - 1 & - 2 \\ 0 & 1 & \frac{5}{3} \\ 0 & 4 & 7 \end{matrix}] = [\begin{matrix} - 1 & 0 & 0 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ 3 & 0 & 1 \end{matrix}] A [Applying R_{2} \to \frac{1}{3} R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & \frac{- 1}{3} \\ 0 & 1 & \frac{5}{3} \\ 0 & 0 & \frac{1}{3} \end{matrix}] = [\begin{matrix} \frac{- 2}{3} & \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ \frac{5}{3} & - \frac{4}{3} & 1 \end{matrix}] A [Applying R_{3} \to R_{3} - 4 R_{2} and R_{1} \to R_{1} + R_{2}] \Rightarrow [\begin{matrix} 1 & 0 & \frac{- 1}{3} \\ 0 & 1 & \frac{5}{3} \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \frac{- 2}{3} & \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ 5 & - 4 & 3 \end{matrix}] A [Applying R_{3} \to 3 R_{3}] \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & - 1 & 1 \\ - 8 & 7 & - 5 \\ 5 & - 4 & 3 \end{matrix}] A [Applying R_{2} \to R_{2} - \frac{5}{3} R_{3} and R_{1} \to R_{1} + \frac{1}{3} R_{3}] Hence, A^{- 1} = [\begin{matrix} 1 & - 1 & 1 \\ - 8 & 7 & - 5 \\ 5 & - 4 & 3 \end{matrix}] .$

Page No 7.35:

Question 1:

Write the adjoint of the matrix $A = [\begin{matrix} - 3 & 4 \\ 7 & - 2 \end{matrix}] .$

Answer:

$Let C_{i j} be a cofactor of a_{i j} in A . Now, C_{11} = - 2 C_{12} = - 7 C_{21} = - 4 C_{22} = - 3 ∴ a d j A = {[\begin{matrix} - 2 & - 7 \\ - 4 & - 3 \end{matrix}]}^{T} = [\begin{matrix} - 2 & - 4 \\ - 7 & - 3 \end{matrix}]$

Page No 7.35:

Question 2:

If A is a square matrix such that A (adj A) 5I, where I denotes the identity matrix of the same order. Then, find the value of |A|.

Answer:

We know
$A (a d j A) = |A| I$
Here,
$A (a d j A) = 5 I ∴ |A| = 5$

Page No 7.35:

Question 3:

If A is a square matrix of order 3 such that |A| = 5, write the value of |adj A|.

Answer:

For any square matrix of order n,
$|a d j A| = {|A|}^{n - 1}$
$\Rightarrow |a d j A| = {|A|}^{2} = 5^{2} = 25$

Page No 7.35:

Question 4:

If A is a square matrix of order 3 such that |adj A| = 64, find |A|.

Answer:

For any square matrix of order n,
$|a d j A| = {|A|}^{n - 1} \Rightarrow 64 = {|A|}^{2} [∵ |a d j A| = 64] \Rightarrow |A| = \pm 8$

Page No 7.35:

Question 5:

If A is a non-singular square matrix such that |A| = 10, find |A⁻¹|.

Answer:

For any non-singular matrix A,
$|A^{- 1}| = \frac{1}{|A|} ∴ |A^{- 1}| = \frac{1}{10} [∵ |A| = 10]$

Page No 7.35:

Question 6:

If A, B, C are three non-null square matrices of the same order, write the condition on A such that AB = AC ⇒ B = C.

Answer:

Consider $A B = A C$ .
On multiplying both sides by $A^{- 1}$ , we get

$A A^{- 1} B = A A^{- 1} C$
$\Rightarrow I B = I C [Because A A^{- 1} = I where I is the identity matrix] \Rightarrow B = C$

Therefore, the required condition is A must be invertible or $|A| \neq 0$ .

Page No 7.35:

Question 7:

If A is a non-singular square matrix such that $A^{- 1} = [\begin{matrix} 5 & 3 \\ - 2 & - 1 \end{matrix}]$ , then find ${(A^{T})}^{- 1} .$

Answer:

For any invertible matrix A,
$(A^{T})^{- 1} = (A^{- 1})^{T}$

$We have A^{- 1} = [\begin{matrix} 5 & 3 \\ - 2 & - 1 \end{matrix}] \Rightarrow (A^{T})^{- 1} = [\begin{matrix} 5 & - 2 \\ 3 & - 1 \end{matrix}]$

Page No 7.35:

Question 8:

If adj $A = [\begin{matrix} 2 & 3 \\ 4 & - 1 \end{matrix}] and adj B = [\begin{matrix} 1 & - 2 \\ - 3 & 1 \end{matrix}]$ , find adj AB.

Answer:

Given:

$a d j A = [\begin{matrix} 2 & 3 \\ 4 & - 1 \end{matrix}] a d j B = [\begin{matrix} 1 & - 2 \\ - 3 & 1 \end{matrix}]$

For any two non-singular matrices,

$a d j (A B) = (a d j B) \times (a d j A) \Rightarrow a d j (A B) = [\begin{matrix} - 6 & 5 \\ - 2 & - 10 \end{matrix}]$

Page No 7.35:

Question 9:

If A is symmetric matrix, write whether A^T is symmetric or skew-symmetric.

Answer:

For any symmetric matrix, $A^{T} = A$ .

Hence, $A^{T}$ is also symmetric.

Page No 7.35:

Question 10:

If A is a square matrix of order 3 such that |A| = 2, then write the value of adj (adj A).

Answer:

For any square matrix A, we have

$a d j (a d j A) = {|A|}^{n - 2} A \Rightarrow a d j (a d j A) = 2 A [∵ |A| = 2 and n = 3]$

Page No 7.35:

Question 11:

If A is a square matrix of order 3 such that |A| = 3, then write the value of adj (adj A).

Answer:

For any square matrix A, we have

$|a d j (a d j A)| = {|A|}^{{(n - 1)}^{2}} \Rightarrow |a d j (a d j A)| = {(3)}^{4} = 81$

Page No 7.35:

Question 12:

If A is a square matrix of order 3 such that adj (2A) = k adj (A), then write the value of k.

Answer:

$For any matrtix A o f order n, adj (λ A) = λ^{n - 1} (adj A), where λ is a constant . Thus, for matrix A of order 3, we have adj (2 A) = 2^{3 - 1} (adj A) \Rightarrow adj (2 A) = 2^{2} (adj A) \Rightarrow adj (2 A) = 4 adj (A) \Rightarrow k adj (A) = 4 adj (A) [∵ adj (2 A) = k adj (A)] \Rightarrow k = 4$

Page No 7.35:

Question 13:

If A is a square matrix, then write the matrix adj (A^T) − (adj A)^T.

Answer:

$In a non - singular matrix, adj A^{T} = {(adj A)}^{T} . \Rightarrow (adj A^{T}) - {(adj A)}^{T} = Null matrix$

Page No 7.35:

Question 14:

Let A be a 3 × 3 square matrix, such that A (adj A) = 2 I, where I is the identity matrix. Write the value of |adj A|.

Answer:

$We know that for a matix A of order n, A . (a d j A) = |A| I_{n}, where I is the identity matrix . Given : A . (a d j A) = 2 I \Rightarrow |A| I = 2 I \Rightarrow |A| = 2 Now, |a d j A| = {|A|}^{n - 1} \Rightarrow |a d j A| = 2^{2} = 4$

Page No 7.35:

Question 15:

If A is a non-singular symmetric matrix, write whether A⁻¹ is symmetric or skew-symmetric.

Answer:

$Let A be an invertible symmetric matrix . Then, |A| \neq 0 and A^{T} = A Now, {(A^{- 1})}^{T} = {(A^{T})}^{- 1} \Rightarrow {(A^{- 1})}^{T} = A^{- 1} [∵ A^{T} = A] Thus, A^{- 1} is symmetric matrix .$

Page No 7.35:

Question 16:

If $A = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] and A (adj A =) [\begin{matrix} k & 0 \\ 0 & k \end{matrix}]$ , then find the value of k.

Answer:

$A = [\cos θ \sin θ - \sin θ \cos θ] ∴ |A| = |\cos θ \sin θ - \sin θ \cos θ| = \cos^{2} θ + \sin^{2} θ = 1 \neq 0 Thus, A^{- 1} exists . Now, A^{- 1} = \frac{adj A}{|A|} = [\cos θ - \sin θ \sin θ \cos θ] \Rightarrow A^{- 1} = adj A \Rightarrow A A^{- 1} = A adj A \Rightarrow A A^{- 1} = [k 0 0 k] \Rightarrow [1 0 0 1] = [k 0 0 k] [∵ A A^{- 1} = I] \Rightarrow k = 1$

Page No 7.35:

Question 17:

If A is an invertible matrix such that |A⁻¹| = 2, find the value of |A|.

Answer:

$We know, {|A|}^{- 1} = \frac{1}{|A|} \Rightarrow 2 = \frac{1}{|A|} [∵ {|A|}^{- 1} = 2] \Rightarrow |A| = \frac{1}{2}$

Page No 7.35:

Question 18:

If A is a square matrix such that $A (adj A) = [\begin{matrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{matrix}]$ , then write the value of |adj A|.

Answer:

$Given : A (adj A) = [5 0 0 0 5 0 0 0 5] \Rightarrow |A| I_{n} = 5 [1 0 0 0 1 0 0 0 1] \Rightarrow |A| = 5 Now, |adj A| = {|A|}^{n - 1} = 5^{3 - 1} = 25$

Page No 7.35:

Question 19:

If $A = [\begin{matrix} 2 & 3 \\ 5 & - 2 \end{matrix}]$ be such that $A^{- 1} = k A,$ then find the value of k.

Answer:

$A = [2 3 5 - 2] ∴ |A| = |2 3 5 - 2| = - 14 - 15 = - 19 The value is non - zero, so A^{- 1} exists . By definition, we have A^{- 1} A = I [I is the identity matrix] \Rightarrow k A . A = I [Substituting A^{- 1} = k A] \Rightarrow k [2 3 5 - 2] [2 3 5 - 2] = [1 0 0 1] \Rightarrow k [4 + 15 6 - 6 10 - 10 15 + 4] = [1 0 0 1] \Rightarrow k [19 0 0 19] = [1 0 0 1] \Rightarrow k = \frac{1}{19}$

Page No 7.35:

Question 20:

Let A be a square matrix such that $A^{2} - A + I = O$ , then write $A^{- 1}$ interms of A.

Answer:

$Given : A^{2} - A + I = O A^{- 1} (A^{2} - A + I) = A^{- 1} O (Pre - multiplying both sides because A^{- 1} exists) (A^{- 1} A^{2}) - (A^{- 1} A) + A^{- 1} I = O (A^{- 1} O = O) \Rightarrow A - I + A^{- 1} = O (A^{- 1} I = A^{- 1}) \Rightarrow A^{- 1} = I - A$

Page No 7.36:

Question 21:

If C_ij is the cofactor of the element a_ij of the matrix $A = [\begin{matrix} 2 & - 3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & - 7 \end{matrix}]$ , then write the value of a₃₂C₃₂.

Answer:

In the given matrix $A = [\begin{matrix} 2 & - 3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & - 7 \end{matrix}]$ ,
C₃₂ = (−1)^{3 + 2} (8 − 30) = 22

Therefore, a₃₂C₃₂ = 5 × 22 = 110.

Hence, the value of a₃₂C₃₂ is 110.

Page No 7.36:

Question 22:

Find the inverse of the matrix $[\begin{matrix} 3 & - 2 \\ - 7 & 5 \end{matrix}] .$

Answer:

$|A| = |\begin{matrix} 3 & - 2 \\ - 7 & 5 \end{matrix}| = 1 \neq 0 A is a non - singular matrix . Therefore, it is invertible . Let C_{i j} be a cofactor of a_{i j} i n A . The cofactors of element A are given by C_{11} = 5 C_{12} = 7 C_{21} = 2 C_{22} = 3 ∴ A^{- 1} = \frac{1}{|A|} {[\begin{matrix} 5 & 7 \\ 2 & 3 \end{matrix}]}^{T} = [\begin{matrix} 5 & 2 \\ 7 & 3 \end{matrix}]$

Page No 7.36:

Question 23:

Find the inverse of the matrix $[\begin{matrix} c o θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]$ .

Answer:

$A = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] ∴ |A| = \cos^{2} θ + \sin^{2} θ = 1 \neq 0 A is a singular matrix . Therefore, it is invertible . Let C_{i j} be a cofactor of a_{i j} in A . Th e cofactors of element A are given by C_{11} = \cos θ C_{12} = \sin θ C_{21} = - \sin θ C_{22} = \cos θ Now, adj A = {[\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]}^{T} = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}] ∴ A^{- 1} = \frac{1}{|A|} adj A = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]$

Page No 7.36:

Question 24:

If $A = [\begin{matrix} 1 & - 3 \\ 2 & 0 \end{matrix}]$ , write adj A.

Answer:

$|A| = |\begin{matrix} 1 & - 3 \\ 2 & 0 \end{matrix}| = 6 \neq 0 A is a non - singular matrix . Therefore, it is invertible . Let C_{i j} be a cofactor of a_{i j} in A . The cofactors of element A are given by C_{11} = 0 C_{12} = - 2 C_{21} = 3 C_{22} = 1 ∴ adj A = {[\begin{matrix} 0 & - 2 \\ 3 & 1 \end{matrix}]}^{T} = [\begin{matrix} 0 & 3 \\ - 2 & 1 \end{matrix}]$

Page No 7.36:

Question 25:

If $A = [\begin{matrix} a & b \\ c & d \end{matrix}], B = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ , find adj (AB).

Answer:

$A \times B = [\begin{matrix} a & b \\ c & d \end{matrix}] A \times B is a non - singular matrix . Therefore, it is invertible . Let C_{i j} be a cofactor of a_{i j} in A . Th e cofactors of element A are given by C_{11} = d C_{12} = - c C_{21} = - b C_{22} = a ∴ adj A = {[\begin{matrix} d & - c \\ - b & a \end{matrix}]}^{T} = [\begin{matrix} d & - b \\ - c & a \end{matrix}]$

Page No 7.36:

Question 26:

If $A = [\begin{matrix} 3 & 1 \\ 2 & - 3 \end{matrix}]$ , then find |adj A|.

Answer:

$|A| = |\begin{matrix} 3 & 1 \\ 2 & - 3 \end{matrix}| = - 11 ∴ |adj A| = {|A|}^{n - 1} = {(- 11)}^{2 - 1} = - 11$

Page No 7.36:

Question 27:

If $A = [\begin{matrix} 2 & 3 \\ 5 & - 2 \end{matrix}]$ , write $A^{- 1}$ in terms of A.

Answer:

$|A| = |\begin{matrix} 2 & 3 \\ 5 & - 2 \end{matrix}| = - 19 \neq 0 A is a non - singular matri x . Therefore, it is invertible . Let C_{i j} be a cofactor of a_{i j} in A . The cofactors of element A are given by C_{11} = - 2 C_{12} = - 5 C_{21} = - 3 C_{22} = 2 adj A = {[\begin{matrix} - 2 & - 5 \\ - 3 & 2 \end{matrix}]}^{T} = [\begin{matrix} - 2 & - 3 \\ - 5 & 2 \end{matrix}] ∴ A^{- 1} = \frac{1}{|A|} adj A = [\begin{matrix} 2 / 19 & 3 / 19 \\ 5 / 19 & - 2 / 19 \end{matrix}]$

$\Rightarrow A^{- 1} = \frac{1}{19} A$

Page No 7.36:

Question 28:

Write $A^{- 1} for A = [\begin{matrix} 2 & 5 \\ 1 & 3 \end{matrix}]$

Answer:

$|A| = |\begin{matrix} 2 & 5 \\ 1 & 3 \end{matrix}| = 1 \neq 0 Let C_{i j} be the cofactor of a_{i j} i n A . The cofactors of element A are given by C_{11} = 3 C_{12} = - 1 C_{21} = - 5 C_{22} = 2 a d j A = {[\begin{matrix} 3 & - 1 \\ - 5 & 2 \end{matrix}]}^{T} = [\begin{matrix} 3 & - 5 \\ - 1 & 2 \end{matrix}] |A| = 6 - 5 = 1 ∴ A^{- 1} = \frac{1}{|A|} a d j A = [\begin{matrix} 3 & - 5 \\ - 1 & 2 \end{matrix}]$

Page No 7.36:

Question 29:

Use elementary column operation C₂ → C₂ + 2C₁ in the following matrix equation :

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

Answer:

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

Applying C₂ → C₂ + 2C₁, we get

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

Page No 7.36:

Question 30:

In the following matrix equation use elementary operation R₂ → R₂ + R₁and the equation thus obtained:

$[\begin{matrix} 2 & 3 \\ 1 & 4 \end{matrix}] [\begin{matrix} 1 & 0 \\ 2 & - 1 \end{matrix}] = [\begin{matrix} 8 & - 3 \\ 9 & - 4 \end{matrix}]$

Answer:

$[\begin{matrix} 2 & 3 \\ 1 & 4 \end{matrix}] [\begin{matrix} 1 & 0 \\ 2 & - 1 \end{matrix}] = [\begin{matrix} 8 & - 3 \\ 9 & - 4 \end{matrix}]$

By applying elementary operation R₂ → R₂ + R₁, we get

$[\begin{matrix} 2 & 3 \\ 3 & 7 \end{matrix}] [\begin{matrix} 1 & 0 \\ 2 & - 1 \end{matrix}] = [\begin{matrix} 8 & - 3 \\ 17 & - 7 \end{matrix}]$ (Every row operation is equivalent to left-multiplication by an elementary matrix.)

Page No 7.37:

Question 1:

If A is an invertible matrix, then which of the following is not true
(a) ${(A^{2})}^{- 1} = {(A^{- 1})}^{2}$
(b) $|A^{- 1}| = {|A|}^{- 1}$
(c) ${(A^{T})}^{- 1} = {(A^{- 1})}^{T}$
(d) $|A| \neq 0$

Answer:

(a) ${(A^{2})}^{- 1} = {(A^{- 1})}^{2}$

We have, $|A^{- 1}| = {|A|}^{- 1}$ , ${(A^{T})}^{- 1} = {(A^{- 1})}^{T}$ and $|A| \neq 0$ all are the properties of the inverse of a matrix A

Page No 7.37:

Question 2:

If A is an invertible matrix of order 3, then which of the following is not true
(a) $|adj A| = {|A|}^{2}$
(b) ${(A^{- 1})}^{- 1} = A$
(c) If $B A = C A, than B \neq C$ , where B and C are square matrices of order 3
(d) ${(A B)}^{- 1} = B^{- 1} A^{- 1}, where B \neq {[b_{i j}]}_{3 \times 3} and |B| \neq 0$

Answer:

(c) If $B A = C A$ , then $B \neq C$ where B and C are square matrices of order 3.

If A is an invertible matrix, then $A^{- 1}$ exists.

Now,
$B A = C A$
On multiplying both sides by $A^{- 1}$ , we get
$B A A^{- 1} = C A A^{- 1}$

$\Rightarrow B I = C I [∵ A A^{- 1} = I where I is the identity matrix] \Rightarrow B = C$

Therefore, the statement given in (c) is not true.

Page No 7.37:

Question 3:

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

(a) is a skew-symmetric matrix
(b) A⁻¹ + B⁻¹
(c) does not exist
(d) none of these

Answer:

(d) none of these

$We have (A + B) = [\begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix}] ∴ |A + B| = - 1 \neq 0 Thus, {(A + B)}^{- 1} exists . Now, {(A + B)}^{T} = [\begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix}] Here, {(A + B)}^{T} \neq - (A + B) Hence, it is not a skew symmetric matrix . We also know that A^{- 1} + B^{- 1} is not the same as {(A + B)}^{- 1} .$

Page No 7.37:

Question 4:

If $S = [\begin{matrix} a & b \\ c & d \end{matrix}]$ , then adj A is

(a) $[\begin{matrix} - d & - b \\ - c & a \end{matrix}]$

(b) $[\begin{matrix} d & - b \\ - c & a \end{matrix}]$

(c) $[\begin{matrix} d & b \\ c & a \end{matrix}]$

(d) $[\begin{matrix} d & c \\ b & a \end{matrix}]$

Answer:

(b) $[\begin{matrix} d & - b \\ - c & a \end{matrix}]$

Adjoint of a square matrix of order 2 is obtained by interchanging the diagonal elements and changing the signs of off-diagonal elements.

Here,

$A = [a b c d] \Rightarrow a d j A = [d - b - c a]$

Page No 7.37:

Question 5:

If A is a singular matrix, then adj A is
(a) non-singular
(b) singular
(c) symmetric
(d) not defined

Answer:

(b) singular

$A is singular, so |A| = 0 . By definition, we have A adj (A) = O \Rightarrow |A adj (A)| = |O| \Rightarrow |A| |adj (A)| = 0 \Rightarrow |adj (A)| = 0 Hence, adj (A) is singular .$

Page No 7.37:

Question 6:

If A, B are two n × n non-singular matrices, then
(a) AB is non-singular
(b) AB is singular
(c) ${(A B)}^{- 1} A^{- 1} B^{- 1}$
(d) (AB)⁻¹ does not exist

Answer:

(a) AB is non-singular

$A and B are non - singular matrices of order n \times n . ∴ |A| \neq 0 and |B| \neq 0 . . . (1) A and B are of the same order, so A B is defined and is of the same order . Thus, |AB| = |A| |B| \Rightarrow |AB| \neq 0 [Using (1)] Thus, A B is non - singular .$

Page No 7.37:

Question 7:

If $A = [\begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix}]$ , then the value of |adj A| is

(a) a²⁷
(b) a⁹
(c) a⁶
(d) a²

Answer:

(c) a⁶

$A = [a 0 0 0 a 0 0 0 a] ∴ |A| = |a 0 0 0 a 0 0 0 a| = a^{3} \neq 0 and n = 3 Thus, we have |a d j A| = {|A|}^{n - 1} = {(a^{3})}^{2} = a^{6}$

Page No 7.37:

Question 8:

If $A = [\begin{matrix} 1 & 2 & - 1 \\ - 1 & 1 & 2 \\ 2 & - 1 & 1 \end{matrix}]$ , then ded (adj (adj A)) is

(a) 14⁴
(b) 14³
(c) 14²
(d) 14

Answer:

(a) 14⁴

$Given : A = [1 2 - 1 - 1 1 2 2 - 1 1] ∴ |A| = |1 2 - 1 - 1 1 2 2 - 1 1| = 1 (1 + 2) - 2 (- 1 - 4) - 1 (1 - 2) = 3 + 10 + 1 = 14 We have |adj (adj A)| = {|A|}^{{(n - 1)}^{2}} \Rightarrow |adj (adj A)| = {(14)}^{2^{2}} = 14^{4}$

Page No 7.37:

Question 9:

If B is a non-singular matrix and A is a square matrix, then det (B⁻¹ AB) is equal to
(a) Det (A⁻¹)
(b) Det (B⁻¹)
(c) Det (A)
(d) Det (B)

Answer:

(c) Det (A)

$B is non - singular . This implies that |B| \neq 0, that B is invertible and that B^{- 1} exists . Here, B is invertible . ∴ |B^{- 1}| = {|B|}^{- 1} = \frac{1}{|B|} \Rightarrow |B^{- 1} A B| = |B^{- 1}| |A B| \Rightarrow |B^{- 1} A B| = {|B|}^{- 1} |A| |B| \Rightarrow |B^{- 1} A B| = \frac{1}{|B|} |A| |B| \Rightarrow |B^{- 1} A B| = |A|$

Page No 7.37:

Question 10:

For any 2 × 2 matrix, if $A (a d j A) = [\begin{matrix} 10 & 0 \\ 0 & 10 \end{matrix}]$ , then |A| is equal to
(a) 20
(c) 100
(d) 10
(d) 0

Answer:

$(c) 10 A (a d j A) = [10 0 0 10] By definition, we have A (a d j A) = |A| I = (a d j A) A (Where I is the identity matrix) \Rightarrow |A| I = A (a d j A) \Rightarrow |A| I = 10 [1 0 0 1] \Rightarrow |A| = 10$

Page No 7.37:

Question 11:

If A⁵ = O such that $A^{n} \neq I for 1 \leq n \leq 4, then {(I - A)}^{- 1}$ equals
(a) A⁴
(b) A³
(c) I + A
(d) none of these

Answer:

$(d) none of the these I - A^{5} = (I - A) (I + A + A^{2} + A^{3} + A^{4}) Now, A^{5} = 0 \Rightarrow I = (I - A) (I + A + A^{2} + A^{3} + A^{4}) \Rightarrow \frac{I}{(I - A)} = (I + A + A^{2} + A^{3} + A^{4}) \Rightarrow {(I - A)}^{- 1} = I + A + A^{2} + A^{3} + A^{4}$

Page No 7.37:

Question 12:

If A satisfies the equation $x^{3} - 5 x^{2} + 4 x + λ = 0$ then A⁻¹ exists if
(a) $λ \neq 1$
(b) $λ \neq 2$
(c) $λ \neq - 1$
(d) $λ \neq 0$

Answer:

$(d) λ \neq 0 A satisfies x^{3} - 5 x^{2} + 4 x + λ = 0 . \Rightarrow A^{3} - 5 A^{2} + 4 A = - λ Assuming A^{- 1} exists, we get A^{- 1} (A^{3} - 5 A^{2} + 4 A) = - λ A^{- 1} \Rightarrow A^{2} - 5 A + 4 = - A^{- 1} λ \Rightarrow A^{- 1} = \frac{- (A^{2} - 5 A + 4)}{λ} Thus, A^{- 1} exists if λ \neq 0 .$

Page No 7.37:

Question 13:

If for the matrix A, A³ = I, then A⁻¹ =
(a) A²
(b) A³
(c) A
(d) none of these

Answer:

$(a) A^{2} Given : A^{3} = I \Rightarrow A^{3} A^{- 1} = I A^{- 1} [Multiplying both sides by A^{- 1}] \Rightarrow A^{2} = A^{- 1}$

Page No 7.38:

Question 14:

If A and B are square matrices such that B = − A⁻¹ BA, then (A + B)² =
(a) O
(b) A² + B²
(c) A² + 2AB + B²
(d) A + B

Answer:

(b) $A^{2} + B^{2}$

$B = - A^{- 1} B A \Rightarrow A B = - A A^{- 1} B A \Rightarrow A B = - B A . . . (1) (∵ A A^{- 1} = I) Now, {(A + B)}^{2} = (A + B) (A + B) \Rightarrow {(A + B)}^{2} = A^{2} + A B + B A + B^{2} \Rightarrow {(A + B)}^{2} = A^{2} - B A + B A + B^{2} [Using (1)] \Rightarrow {(A + B)}^{2} = A^{2} + B^{2}$

Page No 7.38:

Question 15:

If $A = [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}], then A^{5} =$

(a) 5A
(b) 10A
(c) 16A
(d) 32A

Answer:

(c) 16A

$A = [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}] \Rightarrow A = 2 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow A = 2 I \Rightarrow A^{5} = {(2 I)}^{5} \Rightarrow A^{5} = 16 \times 2 I \Rightarrow A^{5} = 16 [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}] \Rightarrow A^{5} = 16 A$

Page No 7.38:

Question 16:

For non-singular square matrix A, B and C of the same order $(A B^{- 1} C) =$
(a) $A^{- 1} B C^{- 1}$
(b) $C^{- 1} B^{- 1} A^{- 1}$
(c) $C B A^{- 1}$
(d) $C^{- 1} B A^{- 1}$

Answer:

Disclaimer: In Quesion, We are to find the inverse of $(A B^{- 1} C)$ . The inverse is missing in the question.

(d) $C^{- 1} B A^{- 1}$

We have,
${(A B^{- 1} C)}^{- 1} = C^{- 1} {(B^{- 1})}^{- 1} A^{- 1} = C^{- 1} B A^{- 1}$

Page No 7.38:

Question 17:

The matrix $[\begin{matrix} 5 & 10 & 3 \\ - 2 & - 4 & 6 \\ - 1 & - 2 & b \end{matrix}]$ is a singular matrix, if the value of b is
(a) − 3
(b) 3
(c) 0
(d) non-existent

Answer:

(d) non-existent
For any singular matrix, the value of the determinant is 0.
Here,

$A = [\begin{matrix} 5 & 10 & 3 \\ - 2 & - 4 & 6 \\ - 1 & - 2 & b \end{matrix}] |A| = 5 (- 4 b + 12) - 10 (- 2 b + 6) + 3 (4 - 4) = 0 \Rightarrow - 20 b + 60 + 20 b - 12 = 0$

Hence, b is non-existent.

Page No 7.38:

Question 18:

If d is the determinant of a square matrix A of order n, then the determinant of its adjoint is
(a) dⁿ
(b) dⁿ⁻¹
(c) dⁿ⁺¹
(d) d

Answer:

(b) dⁿ⁻¹

We know,
$|adj A| = {|A|}^{n - 1}$
$\Rightarrow |adj A| = d^{n - 1}$

Page No 7.38:

Question 19:

If A is a matrix of order 3 and |A| = 8, then |adj A| =
(a) 1
(b) 2
(c) 2³
(d) 2⁶

Answer:

(d) $2^{6}$

$|a d j A| = {|A|}^{n - 1} = 8^{2} = 2^{6}$

Page No 7.38:

Question 20:

If $A^{2} - A + I = 0$ , then the inverse of A is
(a) A⁻²
(b) A + I
(c) I − A
(d) A − I

Answer:

(c) I − A

$Given : A^{2} - A + I = O A^{- 1} (A^{2} - A + I) = A^{- 1} O [multiplying both sides by A^{- 1}] \Rightarrow (A^{- 1} A^{2}) - (A^{- 1} A) + A^{- 1} I = O [∵ A^{- 1} O = O] \Rightarrow A - I + A^{- 1} = O [∵ A^{- 1} I = A^{- 1}] \Rightarrow A^{- 1} = I - A$

Page No 7.38:

Question 21:

If A and B are invertible matrices, which of the following statement is not correct.
(a) $adj A = |A| A^{- 1}$
(b) $\det (A^{- 1}) = {(\det A)}^{- 1}$
(c) ${(A + B)}^{- 1} = A^{- 1} + B^{- 1}$
(d) ${(A B)}^{- 1} = B^{- 1} A^{- 1}$

Answer:

(c) ${(A + B)}^{- 1} = A^{- 1} + B^{- 1}$

We have, $adj A = |A| A^{- 1}$ , $\det (A^{- 1}) = {(\det A)}^{- 1}$ and ${(A B)}^{- 1} = B^{- 1} A^{- 1}$ all are the properites of inverse of a matrix.

Page No 7.38:

Question 22:

If A is a square matrix such that A² = I, then A⁻¹ is equal to
(a) A + I
(b) A
(c) 0
(d) 2A

Answer:

(b) A

$Given : A^{2} = I$
On multiplying both sides by $A^{- 1}$ , we get

$A^{- 1} A^{2} = A^{- 1} I \Rightarrow A = A^{- 1} I \Rightarrow A = A^{- 1}$

Page No 7.38:

Question 23:

Let $A = [\begin{matrix} 1 & 2 \\ 3 & - 5 \end{matrix}] and B = [\begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix}]$ and X be a matrix such that A = BX, then X is equal to

(a) $\frac{1}{2} [\begin{matrix} 2 & 4 \\ 3 & - 5 \end{matrix}]$

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

(c) $[\begin{matrix} 2 & 4 \\ 3 & - 5 \end{matrix}]$

(d) none of these.

Answer:

(a) $\frac{1}{2} [\begin{matrix} 2 & 4 \\ 3 & - 5 \end{matrix}]$

$A = B X \Rightarrow B^{- 1} A = B^{- 1} B X \Rightarrow B^{- 1} A = I X \Rightarrow X = B^{- 1} A . . . (1) Now, B = [\begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix}] adj B = [\begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix}] |B| = 2 ∴ B^{- 1} = \frac{1}{|B|} adj B = \frac{1}{2} [\begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix}] On putting the value of B^{- 1} in eq . (1), we get X = \frac{1}{2} [\begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & 2 \\ 3 & - 5 \end{matrix}] \Rightarrow X = \frac{1}{2} [\begin{matrix} 2 & 4 \\ 3 & - 5 \end{matrix}]$

Page No 7.38:

Question 24:

If $A = [\begin{matrix} 2 & 3 \\ 5 & - 2 \end{matrix}]$ be such that $A^{- 1} = k A$ , then k equals
(a) 19
(b) 1/19
(c) − 19
(d) − 1/19

Answer:

(b) 1/19

$adj A = [\begin{matrix} - 2 & - 3 \\ - 5 & 2 \end{matrix}] |A| = - 19 ∴ A^{- 1} = \frac{1}{|A|} adj A \Rightarrow A^{- 1} = - \frac{1}{19} [\begin{matrix} - 2 & - 3 \\ - 5 & 2 \end{matrix}] Now, A^{- 1} = k A \Rightarrow - \frac{1}{19} [\begin{matrix} - 2 & - 3 \\ - 5 & 2 \end{matrix}] = k A \Rightarrow \frac{1}{19} [\begin{matrix} 2 & 3 \\ 5 & - 2 \end{matrix}] = k A \Rightarrow \frac{1}{19} A = k A \Rightarrow k = \frac{1}{19}$

Page No 7.38:

Question 25:

If $A = \frac{1}{3} [\begin{matrix} 1 & 1 & 2 \\ 2 & 1 & - 2 \\ x & 2 & y \end{matrix}]$ is orthogonal, then x + y =
(a) 3
(b) 0
(c) − 3
(d) 1

Answer:

$We have, A = \frac{1}{3} [\begin{matrix} 1 & 1 & 2 \\ 2 & 1 & - 2 \\ x & 2 & y \end{matrix}] \Rightarrow A^{T} = \frac{1}{3} [\begin{matrix} 1 & 2 & x \\ 1 & 1 & 2 \\ 2 & - 2 & y \end{matrix}] Now, A^{T} A = I \Rightarrow [\begin{matrix} x^{2} + 5 & 2 x + 3 & x y - 2 \\ 3 + 2 x & 6 & 2 y \\ x y - 6 & 2 y & y^{2} + 8 \end{matrix}] = [\begin{matrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{matrix}] The corresponding elements of two equal matrices are not equal . Thus, the matrix A is not orhtogonal .$

Page No 7.38:

Question 26:

If $A = [\begin{matrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2 \end{matrix}], then a I + b A + 2 A^{2}$ equals
(a) A
(b) − A
(c) ab A
(d) none of these

Answer:

(d) none of these

$A = [\begin{matrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 + a & b & 3 \\ a & b & 2 \\ 3 a & 2 b & a + b + 4 \end{matrix}] Now, a I + b A + 2 A^{2} = [\begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix}] + [\begin{matrix} b & 0 & b \\ 0 & 0 & b \\ a b & b^{2} & 2 b \end{matrix}] + [\begin{matrix} 2 + 2 a & 2 b & 6 \\ 2 a & 2 b & 4 \\ 6 a & 6 b & 2 a + 2 b + 8 \end{matrix}] = [\begin{matrix} 3 a + b + 2 & 2 b & b + 6 \\ 2 a & a + 2 b & b + 4 \\ a b + 6 a & b^{2} + 6 b & 3 a + 4 b + 8 \end{matrix}]$

Page No 7.38:

Question 27:

If $[\begin{matrix} 1 & - \tan θ \\ \tan θ & 1 \end{matrix}] [\begin{matrix} 1 & \tan θ \\ - \tan θ & 1 \end{matrix}] - 1 = [\begin{matrix} a & - b \\ b & a \end{matrix}]$ , then

(a) $a = 1, b = 1$
(b) $a = \cos 2 θ, b = \sin 2 θ$
(c) $a = \sin 2 θ, b = \cos 2 θ$
(d) none of these

Answer:

(b) $a = \cos 2 θ, b = \sin 2 θ$

${[\begin{matrix} 1 & \tan θ \\ - \tan θ & 1 \end{matrix}]}^{- 1} = \frac{1}{\sec^{2} θ} [\begin{matrix} 1 & - \tan θ \\ \tan θ & 1 \end{matrix}] Given : [\begin{matrix} 1 & - \tan θ \\ \tan θ & 1 \end{matrix}] {[\begin{matrix} 1 & \tan θ \\ - \tan θ & 1 \end{matrix}]}^{- 1} = [\begin{matrix} a & - b \\ b & a \end{matrix}] \Rightarrow [\begin{matrix} 1 & - \tan θ \\ \tan θ & 1 \end{matrix}] \frac{1}{\sec^{2} θ} [\begin{matrix} 1 & - \tan θ \\ \tan θ & 1 \end{matrix}] = [\begin{matrix} a & - b \\ b & a \end{matrix}] \Rightarrow \frac{1}{\sec^{2} θ} [\begin{matrix} 1 & - \tan θ \\ \tan θ & 1 \end{matrix}] [\begin{matrix} 1 & - \tan θ \\ \tan θ & 1 \end{matrix}] = [\begin{matrix} a & - b \\ b & a \end{matrix}] \Rightarrow [\begin{matrix} \frac{1 - \tan^{2} θ}{\sec^{2} θ} & \frac{- 2 \tan θ}{\sec^{2} θ} \\ \frac{2 \tan θ}{\sec^{2} θ} & \frac{1 - \tan^{2} θ}{\sec^{2} θ} \end{matrix}] = [\begin{matrix} a & - b \\ b & a \end{matrix}] On comparing, we get a = \frac{1 - \tan^{2} θ}{\sec^{2} θ} and b = \frac{2 \tan θ}{\sec^{2} θ} \Rightarrow a = \cos^{2} θ - \sin^{2} θ and b = 2 \sin θ \cos θ \Rightarrow a = \cos 2 θ and b = \sin 2 θ$

Page No 7.39:

Question 28:

If a matrix A is such that 3 $A^{3} + 2 A^{2} + 5 A + I = 0, then A^{- 1}$ equal to
(a) $- (3 A^{2} + 2 A + 5)$
(b) $3 A^{2} + 2 A + 5$
(c) $3 A^{2} - 2 A - 5$
(d) none of these

Answer:

(d) none of these

$3 A^{3} + 2 A^{2} + 5 A + I = 0 \Rightarrow 3 A^{3} + 2 A^{2} + 5 A = - I \Rightarrow A^{- 1} (3 A^{3} + 2 A^{2} + 5 A) = - I A^{- 1} \Rightarrow 3 A^{2} + 2 A + 5 I = - A^{- 1} \Rightarrow A^{- 1} = - 3 A^{2} - 2 A - 5 I$

Page No 7.39:

Question 29:

If A is an invertible matrix, then det (A⁻¹) is equal to
(a) $\det (A)$
(b) $\frac{1}{d e t (A)}$
(c) 1
(d) none of these

Answer:

(b) $\frac{1}{d e t (A)}$

We know that for any invertible matrix A, $|A^{- 1}|$ = $\frac{1}{|A|}$ .

Page No 7.39:

Question 30:

If $A = [\begin{matrix} 2 & - 1 \\ 3 & - 2 \end{matrix}], then A^{n} =$

(a) $A = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ , if n is an even natural number

(b) $A = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ , if n is an odd natural number

(c) $A = [\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}], if n \in N$

(d) none of these

Answer:

Disclaimer: In all option, the power of A (i.e. n is missing)

(a) $A^{n} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ , if n is an even natural number

$A = [\begin{matrix} 2 & - 1 \\ 3 & - 2 \end{matrix}] A^{2} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow A \times A = I \Rightarrow A^{- 1} = A$

Generally,

$A^{n} = (A A^{- 1})^{n / 2} when n is even . A^{n} = A (A A^{- 1})^{n / 2} = A when n is odd . Thus, A^{n} = I when n is even .$

Page No 7.39:

Question 31:

If x, y, z are non-zero real numbers, then the inverse of the matrix $A = [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}]$ , is
(a) $[\begin{matrix} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{matrix}]$

(b) $x y z [\begin{matrix} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{matrix}]$

(c) $\frac{1}{x y z} [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}]$

(d) $\frac{1}{x y z} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

Answer:

(a) $[\begin{matrix} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{matrix}]$

$A = I A \Rightarrow [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{matrix}] A [Applying R_{1} = \frac{1}{x} R_{1}, R_{2} = \frac{1}{y} R_{2} and R_{3} = \frac{1}{z} R_{3}] \Rightarrow A^{- 1} = [\begin{matrix} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{matrix}]$

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