Rd Sharma XII Vol 1 2020 for Class 12 Science Maths Chapter 6 - Adjoint And Inverse Of Matrix

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Page No 6.22:

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 6.22:

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 6.22:

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 6.22:

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 6.23:

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 6.23:

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 6.23:

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 6.23:

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 6.23:

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 6.23:

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 6.23:

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 6.23:

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 6.23:

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 6.23:

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 6.23:

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 6.23:

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 6.23:

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 6.24:

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 6.24:

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 6.24:

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 6.24:

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 6.24:

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 6.24:

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 6.24:

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 6.24:

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 6.24:

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 6.24:

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 6.24:

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 6.24:

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 6.24:

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 6.24:

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 6.24:

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 6.24:

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 6.24:

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 6.25:

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 6.25:

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 6.25:

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 6.25:

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 6.25:

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 6.34:

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 6.34:

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 6.34:

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 6.34:

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 6.34:

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 6.34:

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 6.34:

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 6.34:

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 6.34:

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 6.34:

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 6.34:

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 6.34:

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 6.34:

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 6.34:

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 6.34:

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 6.34:

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 6.34:

Question 17:

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

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

Answer:

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

$[\begin{matrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ - 2 & - 4 & - 5 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A R_{2} \to R_{2} - 2 R_{1} [\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ - 2 & - 4 & - 5 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ - 2 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] R_{3} \to R_{3} + 2 R_{1} [\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ - 2 & 1 & 0 \\ 2 & 0 & 1 \end{matrix}] R_{2} \to R_{2} - R_{3} [\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ - 4 & 1 & - 1 \\ 2 & 0 & 1 \end{matrix}]$

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

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

Page No 6.34:

Question 18:

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

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

Answer:

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

$A = I A \Rightarrow [\begin{matrix} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A Applying R_{1} \leftrightarrow R_{3} \Rightarrow [\begin{matrix} 1 & 1 & - 2 \\ 3 & 2 & - 4 \\ 2 & - 3 & 5 \end{matrix}] = [\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}] A Applying R_{2} \to R_{2} - 3 R_{1} and R_{3} \to R_{3} - 2 R_{1} \Rightarrow [\begin{matrix} 1 & 1 & - 2 \\ 0 & - 1 & 2 \\ 0 & - 5 & 9 \end{matrix}] = [\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & - 3 \\ 1 & 0 & - 2 \end{matrix}] A Applying R_{2} \to (- 1) R_{2} \Rightarrow [\begin{matrix} 1 & 1 & - 2 \\ 0 & 1 & - 2 \\ 0 & - 5 & 9 \end{matrix}] = [\begin{matrix} 0 & 0 & 1 \\ 0 & - 1 & 3 \\ 1 & 0 & - 2 \end{matrix}] A Applying R_{3} \to R_{3} + 5 R_{2} and R_{1} \to R_{1} - R_{2} \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & - 2 \\ 0 & 0 & - 1 \end{matrix}] = [\begin{matrix} 0 & 1 & - 2 \\ 0 & - 1 & 3 \\ 1 & - 5 & 13 \end{matrix}] A Applying R_{3} \to (- 1) R_{3} \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 0 & 1 & - 2 \\ 0 & - 1 & 3 \\ - 1 & 5 & - 13 \end{matrix}] A Applying R_{2} \to R_{2} + 2 R_{3} \Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 0 & 1 & - 2 \\ - 2 & 9 & - 23 \\ - 1 & 5 & - 13 \end{matrix}] A Hence, A^{- 1} = [\begin{matrix} 0 & 1 & - 2 \\ - 2 & 9 & - 23 \\ - 1 & 5 & - 13 \end{matrix}]$

Page No 6.35:

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 6.35:

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 6.35:

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 6.35:

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 6.35:

Question 5:

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

Answer:

Page No 6.35:

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 6.35:

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 6.36:

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 6.36:

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 6.36:

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 6.36:

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 6.36:

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 6.36:

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 6.36:

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 6.36:

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 6.36:

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 6.36:

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 6.36:

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 6.36:

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 6.36:

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 6.36:

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 6.36:

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 6.37:

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 6.37:

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 6.37:

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 6.37:

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 6.37:

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 6.37:

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 6.37:

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 6.37:

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 6.37:

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

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

Page No 6.38:

Question 32:

If A and B are invertible matrices, then which one of the following is not correct?
(a) adj A = $|A|$ A^-1 (b) det (A^-1) = [det(A)]^-1
(c) (AB)^-1 = B^-1A^-1 (d) (A+B)^-1 = B^-1 + A^-1

Answer:

(a) adj A = $|A|$ A⁻¹

As we know,
$A^{- 1} = \frac{adj A}{|A|} \Rightarrow A^{- 1} |A| = adj A$

Thus, adj A = $|A|$ A⁻¹ is correct.

(b) det(A⁻¹) = [det(A)]⁻¹

As we know,
$|A^{- 1}| = \frac{1}{|A|} \Rightarrow |A^{- 1}| = {|A|}^{- 1}$

Thus, det(A⁻¹) = [det(A)]⁻¹ is correct.

(c) (AB)⁻¹ = B⁻¹A⁻¹

As we know,
By reversal law of inverse
(AB)⁻¹ = B⁻¹A⁻¹

Thus, (AB)⁻¹ = B⁻¹A⁻¹ is correct.

(d) (A + B)⁻¹ = B⁻¹ + A⁻¹

${(A + B)}^{- 1} = \frac{1}{|A + B|} adj (A + B) \neq \frac{1}{|A|} adj (A) + \frac{1}{|B|} adj (B) \neq A^{- 1} + B^{- 1}$

Thus, (A + B)⁻¹ = B⁻¹ + A⁻¹ is incorrect.

Hence, the correct option is (d).

Page No 6.38:

Question 33:

If A = $[\begin{matrix} 2 & λ & - 3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{matrix}]$ , then A^-1 exists if
(a) λ=2 (b) λ≠2 (c) λ≠-2 (d) none of these

Answer:

Given: A = $[\begin{matrix} 2 & λ & - 3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{matrix}]$

A⁻¹ exists only if |A| ≠ 0.
$|\begin{matrix} 2 & λ & - 3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{matrix}| \neq 0 \Rightarrow 2 (6 - 5) + 1 (5 λ + 6) \neq 0 \Rightarrow 2 (1) + 5 λ + 6 \neq 0 \Rightarrow 2 + 5 λ + 6 \neq 0 \Rightarrow 5 λ + 8 \neq 0 \Rightarrow 5 λ \neq - 8 \Rightarrow λ \neq \frac{- 8}{5}$

Thus, A⁻¹ exists if λ ∈ $R - \{- \frac{8}{5}\}$

Hence, the correct option is (a).

Page No 6.38:

Question 1:

If A is a unit matrix of order n, then A (adj A) = ___________________.

Answer:

As we know that, A(adj A) = |A|I.

But it is given that A is a unit matrix of order n
Therefore, A(adj A) = |I|I = (1)I = I

Hence, if A is a unit matrix of order n, then A (adj A) = I.

Page No 6.38:

Question 2:

If A is a non-singular square matrix such that A³ = I, then A^-1 = _______________.

Answer:

Given: A³ = I

A³ = I
Multiplying both sides by A⁻¹, we get
⇒ A³A⁻¹ = I A⁻¹
⇒ A²(AA⁻¹) = I A⁻¹
⇒ A²(I) = A⁻¹
⇒ A² = A⁻¹

Hence, if A is a non-singular square matrix such that A³ = I, then A⁻¹ = A².

Page No 6.38:

Question 3:

If A and B are square matrices of the same order and AB = 3I, then A^-1= __________________.

Answer:

Given:
A and B are square matrices of the same order
AB = 3I

AB = 3I
Pre-Multiplying both sides by A⁻¹, we get
⇒ A⁻¹(AB) = A⁻¹(3I)
⇒ (A⁻¹A)B = 3(A⁻¹I)
⇒ (I)B = 3A⁻¹
⇒ B = 3A⁻¹
⇒ $\frac{1}{3} B$ = A⁻¹

Hence, if A and B are square matrices of the same order and AB = 3I, then A⁻¹= $\frac{1}{3} B$ .

Page No 6.38:

Question 4:

If the matrix A = $[\begin{matrix} 1 & a & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1 \end{matrix}]$ is not invertible, than a = ___________________.

Answer:

Given: A = $[\begin{matrix} 1 & a & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1 \end{matrix}]$

A is not invertible if |A| = 0.
$|\begin{matrix} 1 & α & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1 \end{matrix}| = 0 \Rightarrow 1 (2 - 5) - 1 (α - 2) + 2 (5 α - 4) = 0 \Rightarrow 1 (- 3) - 1 α + 2 + 10 α - 8 = 0 \Rightarrow - 3 - α + 2 + 10 α - 8 = 0 \Rightarrow 9 α - 9 = 0 \Rightarrow 9 α = 9 \Rightarrow α = 1$

Hence, if the matrix A = $[\begin{matrix} 1 & a & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1 \end{matrix}]$ is not invertible, than a = 1.

Page No 6.38:

Question 5:

If A is a singular matrix, then A (adj A) = ____________________.

Answer:

As we know that, A(adj A) = |A|I.

But it is given that A is a singular matrix
Thus, |A| = 0.
Therefore, A(adj A) = 0I = O, where O is the zero matrix.

Hence, if A is a singular matrix, then A (adj A) = O.

Page No 6.38:

Question 6:

Let A be a square matrix of order 3 such that $|A|$ = 11 and B be the matrix of confactors of elements of A. Then, ${|B|}^{2}$ = ________________.

Answer:

Given:
A be a square matrix of order 3
$|A|$ = 11
B be the matrix of cofactors of elements of A

Since, B be the matrix of cofactors of elements of A
$\Rightarrow B = {(adj A)}^{T} \Rightarrow |B| = |{(adj A)}^{T}| \Rightarrow |B| = |adj A| (∵ |A^{T}| = |A|) \Rightarrow {|B|}^{2} = {|adj A|}^{2}$

As we know,
$|adj A| = {|A|}^{n - 1}, where n is the order of A \Rightarrow |B| = {|A|}^{n - 1} \Rightarrow |B| = {|A|}^{3 - 1} (∵ Order of A is 3) \Rightarrow |B| = {|A|}^{2} \Rightarrow |B| = {(11)}^{2} (∵ |A| = 11) \Rightarrow |B| = 121 \Rightarrow {|B|}^{2} = {(121)}^{2} \Rightarrow {|B|}^{2} = 14641$

Hence, ${|B|}^{2}$ = 14641.

Page No 6.38:

Question 7:

If A is a square matrix of order 2 such that A (adj A) = $[\begin{matrix} 10 & 0 \\ 0 & 10 \end{matrix}], then |A|$ =______________.

Answer:

As we know that, A(adj A) = |A|I.

But it is given that A (adj A) = $[\begin{matrix} 10 & 0 \\ 0 & 10 \end{matrix}]$
$\Rightarrow A (adj A) = 10 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow A (adj A) = 10 I \Rightarrow |A| I = 10 I \Rightarrow |A| = 10$

Hence, |A| = 10.

Page No 6.38:

Question 8:

If A is an invertible matrix of order 3 and $|A| = 3, then |a d j A|$ = ___________________.

Answer:

Given:
A is an invertible matrix of order 3
$|A|$ = 3

As we know,
$|adj A| = {|A|}^{n - 1}, where n is the order of A \Rightarrow |adj A| = {|A|}^{3 - 1} (∵ Order of A is 3) \Rightarrow |adj A| = {|A|}^{2} \Rightarrow |adj A| = {(3)}^{2} (∵ |A| = 3) \Rightarrow |adj A| = 9$

Hence, |adj A| = 9.

Page No 6.38:

Question 9:

If A is an invertible matrix of order 3 and $|A| = 5, then adj (a d j A)$ = __________________.

Answer:

Given:
A is an invertible matrix of order 3
$|A|$ = 5

As we know,
$adj (adj A) = {|A|}^{n - 2} A, where n is the order of A \Rightarrow adj (adj A) = {|A|}^{3 - 2} A (∵ Order of A is 3) \Rightarrow adj (adj A) = {|A|}^{1} A \Rightarrow adj (adj A) = 5 A (∵ |A| = 5) \Rightarrow adj (adj A) = 5 A$

Hence, adj (adj A) = 5A.

Page No 6.38:

Question 10:

If A is an invertible matrix of order 3 and $|A| = 4, then |adj (a d j A)|$ =__________________.

Answer:

Given:
A is an invertible matrix of order 3
$|A|$ = 4

As we know,
$|adj (adj A)| = {|A|}^{{(n - 1)}^{2}}, where n is the order of A \Rightarrow |adj (adj A)| = {|A|}^{{(3 - 1)}^{2}} (∵ Order of A is 3) \Rightarrow |adj (adj A)| = {|A|}^{{(2)}^{2}} \Rightarrow |adj (adj A)| = {|A|}^{4} \Rightarrow |adj (adj A)| = 4^{4} (∵ |A| = 4) \Rightarrow |adj (adj A)| = 256$

Hence, |adj (adj A)| = 256.

Page No 6.38:

Question 11:

If A = diag (1, 2, 3), then $|a d j (a d j A)|$ =________________.

Answer:

Given:
A = diag (1, 2, 3)
⇒ $|A|$ = 1 × 2 × 3 = 6

As we know,
$|adj (adj A)| = {|A|}^{{(n - 1)}^{2}}, where n is the order of A \Rightarrow |adj (adj A)| = {|A|}^{{(3 - 1)}^{2}} (∵ Order of A is 3) \Rightarrow |adj (adj A)| = {|A|}^{{(2)}^{2}} \Rightarrow |adj (adj A)| = {|A|}^{4} \Rightarrow |adj (adj A)| = 6^{4} (∵ |A| = 6) \Rightarrow |adj (adj A)| = 1296$

Hence, |adj (adj A)| = 1296.

Page No 6.38:

Question 12:

If A is a square matrix of order 3 such that $|A| = \frac{5}{2}, then |A^{- 1}|$ = ________________.

Answer:

Given:
A is a square matrix of order 3
|A| = $\frac{5}{2}$

As we know,
$|A^{- 1}| = {|A|}^{- 1} \Rightarrow |A^{- 1}| = \frac{1}{|A|} \Rightarrow |A^{- 1}| = \frac{1}{\frac{5}{2}} \Rightarrow |A^{- 1}| = \frac{2}{5}$

Hence, $|A^{- 1}| = \frac{2}{5}$ .

Page No 6.38:

Question 13:

If A is a square matrix such that A (adj A) = 10I, then $|A|$ = ____________________.

Answer:

Given:
A is a square matrix
A(adj A) = 10I

As we know,
$A (adj A) = |A| I \Rightarrow 10 I = |A| I \Rightarrow |A| = 10$

Hence, $|A| = 10$ .

Page No 6.38:

Question 14:

Let A be a square matrix of order 3 and B = $|A| A^{- 1} . If |A| = - 5, then |B|$ = _________________.

Answer:

Given:
A is a square matrix of order 3
B = |A|A⁻¹
|A| = −5

Now,
$B = |A| A^{- 1} \Rightarrow |B| = ||A| A^{- 1}| \Rightarrow |B| = {|A|}^{3} |A^{- 1}| (∵ order of A is 3) \Rightarrow |B| = {|A|}^{3} \frac{1}{|A|} (∵ |A^{- 1}| = \frac{1}{|A|}) \Rightarrow |B| = {|A|}^{2} \Rightarrow |B| = {(- 5)}^{2} (∵ |A| = - 5) \Rightarrow |B| = 25$

Hence, $|B| = 25$ .

Page No 6.38:

Question 15:

If k is a scalar and I is a unit matrix of order 3, then adj (kI) = ________________.

Answer:

Given:
I is a unit matrix of order 3

As we know,
$A (adj A) = |A| I \Rightarrow A^{- 1} A (adj A) = A^{- 1} |A| I \Rightarrow (A^{- 1} A) (adj A) = A^{- 1} |A| I \Rightarrow I (adj A) = |A| (A^{- 1} I) \Rightarrow adj A = |A| A^{- 1} \Rightarrow adj (k I) = |k I| {(k I)}^{- 1} \Rightarrow adj (k I) = k^{3} |I| {(k I)}^{- 1} (∵ order of I is 3) \Rightarrow adj (k I) = k^{3} \times 1 \times {(k I)}^{- 1} \Rightarrow adj (k I) = k^{3} k^{- 1} {(I)}^{- 1} \Rightarrow adj (k I) = k^{2} {(I)}^{- 1} \Rightarrow adj (k I) = k^{2} I (∵ I^{- 1} = I) \Rightarrow adj (k I) = k^{2} I$

Hence, adj (kI) = k²I.

Page No 6.38:

Question 16:

If A = $[\begin{matrix} \cos x & \sin x \\ - \sin x & \cos x \end{matrix}]$ and A (adj A) = $[\begin{matrix} k & 0 \\ 0 & k \end{matrix}]$ , then k = _________________.

Answer:

Given:
A = $[\begin{matrix} \cos x & \sin x \\ - \sin x & \cos x \end{matrix}]$
A(adj A) = $[\begin{matrix} k & 0 \\ 0 & k \end{matrix}]$

Now,
$A = [\begin{matrix} \cos x & \sin x \\ - \sin x & \cos x \end{matrix}] \Rightarrow |A| = |\begin{matrix} \cos x & \sin x \\ - \sin x & \cos x \end{matrix}| \Rightarrow |A| = \cos^{2} x + \sin^{2} x \Rightarrow |A| = 1$

As we know,
$A (adj A) = |A| I \Rightarrow [\begin{matrix} k & 0 \\ 0 & k \end{matrix}] = |A| I \Rightarrow k [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = |A| I \Rightarrow k I = |A| I \Rightarrow k = |A| \Rightarrow k = 1 (∵ |A| = 1)$

Hence, $k = 1$ .

Page No 6.38:

Question 17:

If A is a non-singular matrix of order 3, then adj (adj A) is equal to ________________.

Answer:

Given:
A is a non-singular matrix of order 3

As we know,
$adj (adj A) = {|A|}^{n - 2} A, where n is the order of A \Rightarrow adj (adj A) = {|A|}^{3 - 2} A (∵ Order of A is 3) \Rightarrow adj (adj A) = {|A|}^{1} A \Rightarrow adj (adj A) = |A| A$

Hence, adj (adj A) is equal to |A|A.

Page No 6.39:

Question 18:

If A = [a_ij]2×2, where a_ij = $\{\begin{cases} i + j, if i \neq j \\ i^{2} - 2 j, if i = j \end{cases}$ , then A^-1 = ____________________.

Answer:

Given:
A = [a_ij]_2×2, where a_ij = $\{\begin{cases} i + j, if i \neq j \\ i^{2} - 2 j, if i = j \end{cases}$

$a_{i j} = \{\begin{cases} i + j, if i \neq j \\ i^{2} - 2 j, if i = j \end{cases} a_{11} = {(1)}^{2} - 2 (1) = - 1 a_{12} = 1 + 2 = 3 a_{21} = 2 + 1 = 3 a_{22} = {(2)}^{2} - 2 (2) = 0 Thus, A = [\begin{matrix} - 1 & 3 \\ 3 & 0 \end{matrix}] Therefore, A^{- 1} = \frac{1}{0 - 9} [\begin{matrix} 0 & - 3 \\ - 3 & - 1 \end{matrix}] = - \frac{1}{9} [\begin{matrix} 0 & - 3 \\ - 3 & - 1 \end{matrix}] = \frac{1}{9} [\begin{matrix} 0 & 3 \\ 3 & 1 \end{matrix}]$

Hence, $A^{- 1} = \frac{1}{9} [\begin{matrix} 0 & 3 \\ 3 & 1 \end{matrix}] .$

Page No 6.39:

Question 19:

If A = $[\begin{matrix} 0 & 3 \\ 2 & 0 \end{matrix}]$ and A^-1 = λ (adj A), then λ = _____________________.

Answer:

Given:
A = $[\begin{matrix} 0 & 3 \\ 2 & 0 \end{matrix}]$
A⁻¹ = λ (adj A)

$Now, A = [\begin{matrix} 0 & 3 \\ 2 & 0 \end{matrix}] \Rightarrow |A| = |\begin{matrix} 0 & 3 \\ 2 & 0 \end{matrix}| = - 6 As we know that, A^{- 1} = \frac{adj A}{|A|} \Rightarrow λ (adj A) = \frac{1}{|A|} (adj A) \Rightarrow λ = \frac{1}{|A|} \Rightarrow λ = \frac{1}{- 6} \Rightarrow λ = - \frac{1}{6}$

Hence, $λ = - \frac{1}{6}$ .

Page No 6.39:

Question 20:

If A is a 3×3 non-singular matrix such that AA^T = A^TA and B = A^-1A^T, then BB^T = ______________.

Answer:

Given:
A is a 3×3 non-singular matrix
AA^T = A^TA
B = A⁻¹A^T

$Now, B B^{T} = (A^{- 1} A^{T}) {(A^{- 1} A^{T})}^{T} = (A^{- 1} A^{T}) {(A^{T})}^{T} {(A^{- 1})}^{T} = (A^{- 1} A^{T}) (A) {(A^{T})}^{- 1} = A^{- 1} (A^{T} A) {(A^{T})}^{- 1} = A^{- 1} (A A^{T}) {(A^{T})}^{- 1} (∵ A^{T} A = A A^{T}) = (A^{- 1} A) (A^{T} {(A^{T})}^{- 1}) = (I) (I) = I$

Hence, BB^T = I.

Page No 6.39:

Question 21:

If A and B are two square matrices of the same order such that B = -A^-1BA, then (A+B)² = ______________.

Answer:

Given:
$B = - A^{- 1} B A \Rightarrow A B = - A A^{- 1} B A \Rightarrow A B = - I B A \Rightarrow A B = - B A$

$Now, {(A + B)}^{2} = A^{2} + A B + B A + B^{2} = A^{2} - B A + B A + B^{2} (∵ A B = - B A) = A^{2} + B^{2}$

Hence, (A + B)² = A² + B².

Page No 6.39:

Question 22:

If A is a non-singular matrix of order 3×3, then (A³)^-1 = _____________.

Answer:

ans

Page No 6.39:

Question 23:

If A be a square matrix such that $|a d j A| = {|A|}^{2}$ , then the order of A is __________________.

Answer:

Given:
A is a square matrix
$|adj A| = {|A|}^{2}$

As we know,
$|adj A| = {|A|}^{n - 1}, where n is the order of A \Rightarrow {|A|}^{2} = {|A|}^{n - 1} \Rightarrow 2 = n - 1 \Rightarrow n = 3$

Hence, the order of A is 3.

Page No 6.39:

Question 24:

If A = $[\begin{matrix} x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z \end{matrix}]$ ,xyz = 80, 3x + 2y + 10z = 20 and A adj A = kI, then k = _________________.

Answer:

Given:
A = $[\begin{matrix} x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z \end{matrix}]$
xyz = 80
3x + 2y + 10z = 20
A (adj A) = kI

Now,
$A = [\begin{matrix} x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z \end{matrix}] \Rightarrow |A| = |\begin{matrix} x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z \end{matrix}| \Rightarrow |A| = x (y z - 3) - 2 (5 z - 2) + 1 (15 - 2 y) \Rightarrow |A| = x y z - 3 x - 10 z + 4 + 15 - 2 y \Rightarrow |A| = x y z - (3 x + 2 y + 10 z) + 19 \Rightarrow |A| = 80 - (20) + 19 (∵ x y z = 80 and 3 x + 2 y + 10 z = 20) \Rightarrow |A| = 60 + 19 \Rightarrow |A| = 79$

As we know,
$A (adj A) = |A| I \Rightarrow k I = 79 I \Rightarrow k = 79$

Hence, k = 79.

Page No 6.39:

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 6.39:

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 6.39:

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 6.39:

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 6.39:

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 6.39:

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 6.39:

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 6.39:

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 6.39:

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 6.40:

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 6.40:

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 6.40:

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 6.40:

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 6.40:

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 6.40:

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 6.40:

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 6.40:

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 6.40:

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 6.40:

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:

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

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Question 20:

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

Answer:

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

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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.

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Question 22:

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

Answer:

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

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Question 23:

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

Answer:

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

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Question 24:

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

Answer:

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

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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:

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

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Question 26:

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

Answer:

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

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Question 27:

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

Answer:

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

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

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Question 28:

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

Answer:

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

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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

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

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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.)

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Question 31:

If A is a square matrix with $|A|$ = 4 then find the value of $|A . (a d j A)| .$

Answer:

Given:
A is a square matrix
$|A|$ = 4

Now,
$A (adj A) = |A| I \Rightarrow |A (adj A)| = ||A| I| \Rightarrow |A (adj A)| = {|A|}^{n} |I|, where n is the order of I \Rightarrow |A (adj A)| = {|A|}^{n} \times 1 \Rightarrow |A (adj A)| = 4^{n} (∵ |A| = 4)$

Hence, $|A (adj A)| = 4^{n} .$

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