RS Aggarwal Solutions for Class 12 Chapter 07 Adjoint and Inverse of a Matrix

Question 1. Find the adjoint of the given matrix and verify in each case that A. (adj A) = (adj A) = m |A|.I.
\( \begin{bmatrix} 2 & 3 \\ 5 & 9 \end{bmatrix} \)
Answer: Here, \( A = \begin{bmatrix} 2 & 3 \\ 5 & 9 \end{bmatrix} \)

To find adj A, we need to find the co-factors:

\( a_{11} \) (co - factor of 2) = \( (-1)^{1+1}(9) = (-1)^2(9) = 9 \)
\( a_{12} \) (co - factor of 3) = \( (-1)^{1+2}(5) = (-1)^3(5) = -5 \)
\( a_{21} \) (co - factor of 5) = \( (-1)^{2+1}(3) = (-1)^3(3) = -3 \)
\( a_{22} \) (co - factor of 9) = \( (-1)^{2+2}(2) = (-1)^4(2) = 2 \)

The co - factor matrix = \( \begin{bmatrix} 9 & -5 \\ -3 & 2 \end{bmatrix} \)

Now, adj A = Transpose of co-factor Matrix

\( \therefore adj A = \begin{bmatrix} 9 & -5 \\ -3 & 2 \end{bmatrix}^T = \begin{bmatrix} 9 & -3 \\ -5 & 2 \end{bmatrix} \)

Calculating A (adj A)

\( A.(adj A) = \begin{bmatrix} 2 & 3 \\ 5 & 9 \end{bmatrix}\begin{bmatrix} 9 & -3 \\ -5 & 2 \end{bmatrix} \)

\( = \begin{bmatrix} 2 \times 9 + 3 \times (-5) & 2 \times (-3) + 3 \times 2 \\ 5 \times 9 + 9 \times (-5) & 5 \times (-3) + 9 \times 2 \end{bmatrix} \)

\( = \begin{bmatrix} 18 - 15 & -6 + 6 \\ 45 - 45 & -15 + 18 \end{bmatrix} \)

\( = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \)

\( = 3\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

\( = 3I \)

Calculating (adj A)A

\( (adj A). A = \begin{bmatrix} 9 & -3 \\ -5 & 2 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ 5 & 9 \end{bmatrix} \)

\( = \begin{bmatrix} 9 \times 2 + (-3) \times 5 & 9 \times 3 + (-3) \times 9 \\ -5 \times 2 + 2 \times 5 & -5 \times 3 + 2 \times 9 \end{bmatrix} \)

\( = \begin{bmatrix} 18 - 15 & 27 - 27 \\ -10 + 10 & -15 + 18 \end{bmatrix} \)

\( = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \)

\( = 3I \)

Calculating |A|.I

\( = (2 \times 9 - 3 \times 5)I \)
\( = (18 - 15)I \)
\( = 3I \)

Thus, A(adj A) = (adj A)A = |A|I = 3I

\( \Rightarrow A(adj A) = (adj A)A = |A|I \)

Hence Proved

Ans. \( \begin{bmatrix} 9 & -3 \\ -5 & 2 \end{bmatrix} \)

Exam Tip: Always verify the property by computing both A(adj A) and (adj A)A separately - examiners check that you show all working, not just the final determinant value.

Question 2. Find the adjoint of the given matrix and verify in each case that A. (adj A) = (adj A) = m |A|.I.
\( \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} \)
Answer: Here, \( A = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} \)

To find adj A, we must first calculate the co-factors:

\( a_{11} \) (co - factor of 3) = \( (-1)^{1+1}(2) = (-1)^2(2) = 2 \)
\( a_{12} \) (co - factor of -5) = \( (-1)^{1+2}(-1) = (-1)^3(-1) = 1 \)
\( a_{21} \) (co - factor of -1) = \( (-1)^{2+1}(-5) = (-1)^3(-5) = 5 \)
\( a_{22} \) (co - factor of 2) = \( (-1)^{2+2}(3) = (-1)^4(3) = 3 \)

Now, adj A = Transpose of co-factor Matrix

\( \therefore adj A = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}^T = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \)

Calculating A (adj A)

\( A.(adj A) = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \)

\( = \begin{bmatrix} 3 \times 2 + (-5) \times 1 & 3 \times 5 + (-5) \times 3 \\ (-1) \times 2 + 2 \times 1 & (-1) \times 5 + 2 \times 3 \end{bmatrix} \)

\( = \begin{bmatrix} 6 - 5 & 15 - 15 \\ -2 + 2 & -5 + 6 \end{bmatrix} \)

\( = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

\( = I \)

Calculating (adj A)A

\( (adj A). A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} \)

\( = \begin{bmatrix} 2 \times 3 + 5 \times (-1) & 2 \times (-5) + 5 \times 2 \\ 1 \times 3 + 3 \times (-1) & 1 \times (-5) + 3 \times 2 \end{bmatrix} \)

\( = \begin{bmatrix} 6 - 5 & -10 + 10 \\ 3 - 3 & -5 + 6 \end{bmatrix} \)

\( = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

\( = I \)

Calculating |A|.I

\( |A|.I = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}I \)

If \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), then determinant of A is given by

\( |A| = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc \)

\( = (3 \times 2 - (-1) \times (-5))I \)
\( = (6 - 5)I \)
\( = I \)

Thus, A(adj A) = (adj A)A = |A|I = I

\( \Rightarrow A(adj A) = (adj A)A = |A|I \)

Hence Proved

Ans. \( \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \)

Exam Tip: When the determinant equals 1, both products simplify to the identity matrix - use this as a quick check that your co-factor matrix is correct.

Question 3. Find the adjoint of the given matrix and verify in each case that A. (adj A) = (adj A) = m |A|.I.
\( \begin{bmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \)
Answer: Here, \( A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \)

To find adj A, we must first determine the co-factors:

\( a_{11} \) (co - factor of \( \cos \alpha \)) = \( (-1)^{1+1}(\cos \alpha) = (-1)^2(\cos \alpha) = \cos \alpha \)
\( a_{12} \) (co - factor of \( \sin \alpha \)) = \( (-1)^{1+2}(\sin \alpha) = (-1)^3(\sin \alpha) = -\sin \alpha \)
\( a_{21} \) (co - factor of \( \sin \alpha \)) = \( (-1)^{2+1}(\sin \alpha) = (-1)^3(\sin \alpha) = -\sin \alpha \)
\( a_{22} \) (co - factor of \( \cos \alpha \)) = \( (-1)^{2+2}(\cos \alpha) = (-1)^4(\cos \alpha) = \cos \alpha \)

The co - factor matrix = \( \begin{bmatrix} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \)

Now, adj A = Transpose of co-factor Matrix

\( \therefore adj A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}^T = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \)

Calculating A (adj A)

\( A.(adj A) = \begin{bmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}\begin{bmatrix} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \)

\( = \begin{bmatrix} \cos \alpha \times \cos \alpha + \sin \alpha \times (-\sin \alpha) & \cos \alpha \times (-\sin \alpha) + \sin \alpha \times \cos \alpha \\ \sin \alpha \times \cos \alpha + \cos \alpha \times (-\sin \alpha) & \sin \alpha \times (-\sin \alpha) + \cos \alpha \times \cos \alpha \end{bmatrix} \)

\( = \begin{bmatrix} \cos^2 \alpha - \sin^2 \alpha & -\cos \alpha \sin \alpha + \cos \alpha \sin \alpha \\ \sin \alpha \cos \alpha - \cos \alpha \sin \alpha & -\sin^2 \alpha + \cos^2 \alpha \end{bmatrix} \)

\( = \begin{bmatrix} \cos^2 \alpha - \sin^2 \alpha & 0 \\ 0 & \cos^2 \alpha - \sin^2 \alpha \end{bmatrix} \)

\( = (\cos^2 \alpha - \sin^2 \alpha) I \)

Calculating (adj A)A

\( (adj A). A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}\begin{bmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \)

\( = \begin{bmatrix} \cos \alpha \times \cos \alpha + (-\sin \alpha) \times \sin \alpha & \cos \alpha \times \sin \alpha + (-\sin \alpha) \times \cos \alpha \\ (-\sin \alpha) \times \cos \alpha + \cos \alpha \times \sin \alpha & (-\sin \alpha) \times \sin \alpha + \cos \alpha \times \cos \alpha \end{bmatrix} \)

\( = \begin{bmatrix} \cos^2 \alpha - \sin^2 \alpha & \cos \alpha \sin \alpha - \cos \alpha \sin \alpha \\ -\sin \alpha \cos \alpha + \cos \alpha \sin \alpha & -\sin^2 \alpha + \cos^2 \alpha \end{bmatrix} \)

\( = \begin{bmatrix} \cos^2 \alpha - \sin^2 \alpha & 0 \\ 0 & \cos^2 \alpha - \sin^2 \alpha \end{bmatrix} \)

\( = (\cos^2 \alpha - \sin^2 \alpha) I \)

Calculating |A|.I

\( |A|.I = \begin{bmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}I \)

If \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), then determinant of A is given by

\( |A| = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc \)

\( = [\cos \alpha \times \cos \alpha - (\sin \alpha) \times (\sin \alpha)]I \)
\( = [\cos^2 \alpha - \sin^2 \alpha] I \)

Thus, A(adj A) = (adj A)A = |A|I = I

\( \Rightarrow A(adj A) = (adj A)A = |A|I \)

Hence Proved

Ans. \( \begin{bmatrix} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \)

Exam Tip: Trigonometric identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \) is key - simplify diagonal entries to get the correct determinant multiple.

Question 4. Find the adjoint of the given matrix and verify in each case that A. (adj A) = (adj A) = m |A|.I.
\( \begin{bmatrix} 1 & -1 & 2 \\ 3 & 1 & -2 \\ 1 & 0 & 3 \end{bmatrix} \)
Answer: Here, \( A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 1 & -2 \\ 1 & 0 & 3 \end{bmatrix} \)

To find adj A, we need to calculate the co-factors:

\( a_{11} = \begin{vmatrix} 1 & -2 \\ 0 & 3 \end{vmatrix} = 3 - (0) = 3 \)

\( a_{12} = - \begin{vmatrix} 3 & -2 \\ 1 & 3 \end{vmatrix} = -(9 - (-2)) = -(9 + 2) = -11 \)

\( a_{13} = \begin{vmatrix} 3 & 1 \\ 1 & 0 \end{vmatrix} = 0 - 1 = -1 \)

\( a_{21} = - \begin{vmatrix} -1 & 2 \\ 0 & 3 \end{vmatrix} = -(-3 - 0) = 3 \)

\( a_{22} = \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} = 3 - 2 = 1 \)

\( a_{23} = - \begin{vmatrix} 1 & -1 \\ 1 & 0 \end{vmatrix} = -(0 - (-1)) = -1 \)

\( a_{31} = \begin{vmatrix} -1 & 2 \\ 1 & -2 \end{vmatrix} = 2 - 2 = 0 \)

\( a_{32} = - \begin{vmatrix} 1 & 2 \\ 3 & -2 \end{vmatrix} = -(-2 - 6) = 8 \)

\( a_{33} = \begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} = 1 - (-3) = 1 + 3 = 4 \)

\( \therefore adj A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}^T = \begin{bmatrix} 3 & -11 & -1 \\ 3 & 1 & -1 \\ 0 & 8 & 4 \end{bmatrix}^T = \begin{bmatrix} 3 & 3 & 0 \\ -11 & 1 & 8 \\ -1 & -1 & 4 \end{bmatrix} \)

Calculating A (adj A)

\( A.(adj A) = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 1 & -2 \\ 1 & 0 & 3 \end{bmatrix}\begin{bmatrix} 3 & 3 & 0 \\ -11 & 1 & 8 \\ -1 & -1 & 4 \end{bmatrix} \)

\( = \begin{bmatrix} 3 + 11 - 2 & 3 - 1 - 2 & 0 - 8 + 8 \\ 9 - 11 + 2 & 9 + 1 + 2 & 0 + 8 - 8 \\ 3 - 0 - 3 & 3 + 0 - 3 & 0 + 0 + 12 \end{bmatrix} \)

\( = \begin{bmatrix} 12 & 0 & 0 \\ 0 & 12 & 0 \\ 0 & 0 & 12 \end{bmatrix} \)

\( = 12 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = 12I \)

Calculating (adj A)A

\( (adj A). A = \begin{bmatrix} 3 & 3 & 0 \\ -11 & 1 & 8 \\ -1 & -1 & 4 \end{bmatrix}\begin{bmatrix} 1 & -1 & 2 \\ 3 & 1 & -2 \\ 1 & 0 & 3 \end{bmatrix} \)

\( = \begin{bmatrix} 3 + 9 + 0 & -3 + 3 + 0 & 6 - 6 + 0 \\ -11 + 3 + 8 & 11 + 1 + 0 & -22 - 2 + 24 \\ -1 - 3 + 4 & 1 - 1 + 0 & -2 + 2 + 12 \end{bmatrix} \)

\( = \begin{bmatrix} 12 & 0 & 0 \\ 0 & 12 & 0 \\ 0 & 0 & 12 \end{bmatrix} \)

\( = 12 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = 12I \)

Calculating |A|.I

Expanding along C1, we get

\( |A| = a_{11} (-1)^{1+1}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} + a_{21} (-1)^{2+1}\begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix} + a_{31} (-1)^{3+1}\begin{vmatrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{vmatrix} \)

\( |A|.I = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 1 & -2 \\ 1 & 0 & 3 \end{bmatrix}I \)

\( = [1(3 - 0) - (-1){9 - (-2)} + 2(0 - 1)]I \)
\( = [3 + 1(11) + 2(-1)] I \)
\( = (3 + 11 - 2 )I \)
\( = 12I \)

Thus, A(adj A) = (adj A)A = |A|I = 12I

\( \Rightarrow A(adj A) = (adj A)A = |A|I \)

Hence Proved

Ans. \( \begin{bmatrix} 3 & 3 & 0 \\ -11 & 1 & 8 \\ -1 & -1 & 4 \end{bmatrix} \)

Exam Tip: For 3 × 3 matrices, expand the determinant along any row or column - choosing a row/column with zeros reduces computation. Always transpose the co-factor matrix to get adj A.

Question 5. Find the adjoint of the given matrix and verify in each case that A. (adj A) = (adj A) = m |A|.I.
\( \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \)
Answer: Here, \( A = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \)

To find adj A, we must first determine the co-factors:

\( a_{11} = \begin{vmatrix} 6 & -5 \\ -2 & 2 \end{vmatrix} = 12 - (10) = 2 \)

\( a_{12} = - \begin{vmatrix} -15 & -5 \\ 5 & 2 \end{vmatrix} = -(-30 - (-25)) = -(-30 + 25) = 5 \)

\( a_{13} = \begin{vmatrix} -15 & 6 \\ 5 & -2 \end{vmatrix} = 30 - 30 = 0 \)

\( a_{21} = - \begin{vmatrix} -1 & 1 \\ -2 & 2 \end{vmatrix} = -(-2 - (-2)) = 0 \)

\( a_{22} = \begin{vmatrix} 3 & 1 \\ 5 & 2 \end{vmatrix} = 6 - 5 = 1 \)

\( a_{23} = - \begin{vmatrix} 3 & -1 \\ 5 & -2 \end{vmatrix} = -(-6 - (-5)) = -(-6 + 5) = 1 \)

\( a_{31} = \begin{vmatrix} -1 & 1 \\ 6 & -5 \end{vmatrix} = 5 - 6 = -1 \)

\( a_{32} = - \begin{vmatrix} 3 & 1 \\ -15 & -5 \end{vmatrix} = -(-15 - (-15)) = -(-15 + 15) = 0 \)

\( a_{33} = \begin{vmatrix} 3 & -1 \\ -15 & 6 \end{vmatrix} = 18 - 15 = 3 \)

\( \therefore adj A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}^T = \begin{bmatrix} 2 & 5 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 3 \end{bmatrix}^T = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} \)

Calculating A (adj A)

\( A.(adj A) = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} \)

\( = \begin{bmatrix} 6 - 5 + 0 & 0 - 1 + 1 & -3 + 0 + 3 \\ -30 + 30 + 0 & 0 + 6 - 5 & 15 + 0 - 15 \\ 10 - 10 + 0 & 0 - 2 + 2 & -5 + 0 + 6 \end{bmatrix} \)

\( = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = I \)

Calculating (adj A)A

\( (adj A). A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}\begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \)

\( = \begin{bmatrix} 6 + 0 - 5 & -2 + 0 + 2 & 2 + 0 - 2 \\ 15 - 15 + 0 & -5 + 6 + 0 & 5 - 5 + 0 \\ 0 - 15 + 15 & 0 + 6 - 6 & 0 - 5 + 6 \end{bmatrix} \)

\( = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = I \)

Calculating |A|.I

Expanding along C1, we get

\( |A| = a_{11} (-1)^{1+1}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} + a_{21} (-1)^{2+1}\begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix} + a_{31} (-1)^{3+1}\begin{vmatrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{vmatrix} \)

\( |A|.I = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}I \)

\( = [3(12 - 10) - (-15){-2 - (-2)} + 5(5 - 6)]I \)
\( = [3(2) + 15(0) + 5(-1)] I \)
\( = (6 - 5)I \)
\( = I \)

Thus, A(adj A) = (adj A)A = |A|I = I

\( \Rightarrow A(adj A) = (adj A)A = |A|I \)

Hence Proved

Ans. \( \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} \)

Exam Tip: When determinant is 1, the matrix is non-singular - this guarantees that A(adj A) equals I and confirms your co-factor calculations are correct.

Question 6. Find the adjoint of the given matrix and verify in each case that A. (adj A) = (adj A) = m |A|.I.
\( \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} \)
Answer: Here, \( A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} \)

To find adj A, we must calculate the co-factors:

\( a_{11} = \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} = 2 - 3 = -1 \)

\( a_{12} = - \begin{vmatrix} 1 & 3 \\ 3 & 1 \end{vmatrix} = -(1 - 9) = 8 \)

\( a_{13} = \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = 1 - 6 = -5 \)

\( a_{21} = - \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} = -(1 - 2) = 1 \)

\( a_{22} = \begin{vmatrix} 0 & 2 \\ 3 & 1 \end{vmatrix} = 0 - 6 = -6 \)

\( a_{23} = - \begin{vmatrix} 0 & 1 \\ 3 & 1 \end{vmatrix} = -(0 - 3) = 3 \)

\( a_{31} = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = 3 - 4 = -1 \)

\( a_{32} = - \begin{vmatrix} 0 & 2 \\ 1 & 3 \end{vmatrix} = -(0 - 2) = 2 \)

\( a_{33} = \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} = 0 - 1 = -1 \)

\( \therefore adj A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}^T = \begin{bmatrix} -1 & 8 & -5 \\ 1 & -6 & 3 \\ -1 & 2 & -1 \end{bmatrix}^T = \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix} \)

Calculating A (adj A)

\( A.(adj A) = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}\begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix} \)

\( = \begin{bmatrix} 0 + 8 - 10 & 0 - 6 + 6 & 0 + 2 - 2 \\ -1 + 16 - 15 & 1 - 12 + 9 & -1 + 4 - 3 \\ -3 + 8 - 5 & 3 - 6 + 3 & -3 + 2 - 1 \end{bmatrix} \)

\( = \begin{bmatrix} -2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2 \end{bmatrix} \)

\( = -2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = -2I \)

Calculating (adj A)A

\( (adj A). A = \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix}\begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} \)

\( = \begin{bmatrix} 0 + 1 - 3 & -1 + 2 - 1 & -2 + 3 - 1 \\ 0 - 6 + 6 & 8 - 12 + 2 & 16 - 18 + 2 \\ 0 + 3 - 3 & -5 + 6 - 1 & -10 + 9 - 1 \end{bmatrix} \)

\( = \begin{bmatrix} -2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2 \end{bmatrix} \)

\( = -2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = -2I \)

Calculating |A|.I

Expanding along C1, we get

\( |A| = a_{11} (-1)^{1+1}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} + a_{21} (-1)^{2+1}\begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix} + a_{31} (-1)^{3+1}\begin{vmatrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{vmatrix} \)

\( |A|.I = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}I \)

\( = [0(2 - 3) - (1){1 - 2} + 3(3 - 4)]I \)
\( = [0 - 1(-1) + 3(-1)] I \)
\( = (1 - 3)I \)
\( = -2I \)

Thus, A(adj A) = (adj A)A = |A|I = -2I

\( \Rightarrow A(adj A) = (adj A)A = |A|I \)

Hence Proved

Ans. \( \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix} \)

Exam Tip: When determinant is negative, A(adj A) gives -|A| times the identity - keep track of sign throughout to verify the property correctly.

Question 7. Find the adjoint of the given matrix and verify in each case that A. (adj A) = (adj A) = m |A|.I.
\( \begin{bmatrix} 9 & 7 & 3 \\ 5 & -1 & 4 \\ 6 & 8 & 2 \end{bmatrix} \)
Answer: Here, \( A = \begin{bmatrix} 9 & 7 & 3 \\ 5 & -1 & 4 \\ 6 & 8 & 2 \end{bmatrix} \)

To find adj A, we must determine the co-factors:

\( a_{11} = \begin{vmatrix} -1 & 4 \\ 8 & 2 \end{vmatrix} = -2 - 32 = -34 \)

\( a_{12} = - \begin{vmatrix} 5 & 4 \\ 6 & 2 \end{vmatrix} = -(10 - 24) = -(-14) = 14 \)

\( a_{13} = \begin{vmatrix} 5 & -1 \\ 6 & 8 \end{vmatrix} = 40 - (-6) = 40 + 6 = 46 \)

\( a_{21} = - \begin{vmatrix} 7 & 3 \\ 8 & 2 \end{vmatrix} = -(14 - 24) = 10 \)

\( a_{22} = \begin{vmatrix} 9 & 3 \\ 6 & 2 \end{vmatrix} = 18 - 18 = 0 \)

\( a_{23} = - \begin{vmatrix} 9 & 7 \\ 6 & 8 \end{vmatrix} = -(72 - 42) = -30 \)

\( a_{31} = \begin{vmatrix} 7 & 3 \\ -1 & 4 \end{vmatrix} = 28 - (-3) = 31 \)

\( a_{32} = - \begin{vmatrix} 9 & 3 \\ 5 & 4 \end{vmatrix} = -(36 - 15) = -21 \)

\( a_{33} = \begin{vmatrix} 9 & 7 \\ 5 & -1 \end{vmatrix} = -9 - 35 = -44 \)

\( \therefore adj A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}^T = \begin{bmatrix} -34 & 14 & 46 \\ 10 & 0 & -30 \\ 31 & -21 & -44 \end{bmatrix}^T = \begin{bmatrix} -34 & 10 & 31 \\ 14 & 0 & -21 \\ 46 & -30 & -44 \end{bmatrix} \)

Calculating A (adj A)

\( A.(adj A) = \begin{bmatrix} 9 & 7 & 3 \\ 5 & -1 & 4 \\ 6 & 8 & 2 \end{bmatrix}\begin{bmatrix} -34 & 10 & 31 \\ 14 & 0 & -21 \\ 46 & -30 & -44 \end{bmatrix} \)

\( = \begin{bmatrix} -306 + 98 + 138 & 90 + 0 - 90 & 279 - 147 - 132 \\ -170 - 14 + 184 & 50 + 0 - 120 & 155 + 21 - 176 \\ -204 + 112 + 92 & 60 + 0 - 60 & 186 - 168 - 88 \end{bmatrix} \)

\( = \begin{bmatrix} -70 & 0 & 0 \\ 0 & -70 & 0 \\ 0 & 0 & -70 \end{bmatrix} \)

\( = -70 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = -70 I \)

Calculating (adj A)A

\( (adj A). A = \begin{bmatrix} -34 & 10 & 31 \\ 14 & 0 & -21 \\ 46 & -30 & -44 \end{bmatrix}\begin{bmatrix} 9 & 7 & 3 \\ 5 & -1 & 4 \\ 6 & 8 & 2 \end{bmatrix} \)

\( = \begin{bmatrix} -306 + 50 + 186 & -238 - 10 + 248 & -102 + 40 + 62 \\ 126 + 0 - 126 & 98 + 0 - 168 & 42 + 0 - 42 \\ 414 - 150 - 264 & 322 + 30 - 352 & 138 - 120 - 88 \end{bmatrix} \)

\( = \begin{bmatrix} -70 & 0 & 0 \\ 0 & -70 & 0 \\ 0 & 0 & -70 \end{bmatrix} \)

\( = -70 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = -70 I \)

Calculating |A|.I

Expanding along C1, we get

\( |A| = a_{11} (-1)^{1+1}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} + a_{21} (-1)^{2+1}\begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix} + a_{31} (-1)^{3+1}\begin{vmatrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{vmatrix} \)

\( |A|.I = \begin{bmatrix} 9 & 7 & 3 \\ 5 & -1 & 4 \\ 6 & 8 & 2 \end{bmatrix}I \)

\( = [9(-2 - 32) - (5){14 - 24} + 6(28 - (-3))]I \)
\( = [9(-34) - 5(-10) + 6(31)] I \)
\( = (-306 + 50 + 186)I \)
\( = -70 I \)

Thus, A(adj A) = (adj A)A = |A|I = -70 I

\( \Rightarrow A(adj A) = (adj A)A = |A|I \)

Hence Proved

Ans. \( \begin{bmatrix} -34 & 10 & 31 \\ 14 & 0 & -21 \\ 46 & -30 & -44 \end{bmatrix} \)

Exam Tip: Always compute the determinant carefully - it becomes the multiplier for the identity matrix in the final step. Verify all nine entries of A(adj A) to avoid sign errors.

Question 8. Find the adjoint of the given matrix and verify in each case that A. (adj A) = (adj A) = m |A|.I.
\( \begin{bmatrix} 4 & 5 & 3 \\ 1 & 0 & 6 \\ 2 & 7 & 9 \end{bmatrix} \)
Answer: Here, \( A = \begin{bmatrix} 4 & 5 & 3 \\ 1 & 0 & 6 \\ 2 & 7 & 9 \end{bmatrix} \)

To find adj A, we must calculate the co-factors:

\( a_{11} = \begin{vmatrix} 0 & 6 \\ 7 & 9 \end{vmatrix} = 0 - 42 = -42 \)

\( a_{12} = - \begin{vmatrix} 1 & 6 \\ 2 & 9 \end{vmatrix} = -(9 - 12) = 3 \)

\( a_{13} = \begin{vmatrix} 1 & 0 \\ 2 & 7 \end{vmatrix} = 7 - 0 = 7 \)

\( a_{21} = - \begin{vmatrix} 5 & 3 \\ 7 & 9 \end{vmatrix} = -(45 - 21) = -24 \)

\( a_{22} = \begin{vmatrix} 4 & 3 \\ 2 & 9 \end{vmatrix} = 36 - 6 = 30 \)

\( a_{23} = - \begin{vmatrix} 4 & 5 \\ 2 & 7 \end{vmatrix} = -(28 - 10) = -18 \)

\( a_{31} = \begin{vmatrix} 5 & 3 \\ 0 & 6 \end{vmatrix} = 30 - 0 = 30 \)

\( a_{32} = - \begin{vmatrix} 4 & 3 \\ 1 & 6 \end{vmatrix} = -(24 - 3) = -(21) = -21 \)

\( a_{33} = \begin{vmatrix} 4 & 5 \\ 1 & 0 \end{vmatrix} = 0 - 5 = -5 \)

\( \therefore adj A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}^T = \begin{bmatrix} -42 & 3 & 7 \\ -24 & 30 & -18 \\ 30 & -21 & -5 \end{bmatrix}^T = \begin{bmatrix} -42 & -24 & 30 \\ 3 & 30 & -21 \\ 7 & -18 & -5 \end{bmatrix} \)

Calculating A (adj A)

\( A.(adj A) = \begin{bmatrix} 4 & 5 & 3 \\ 1 & 0 & 6 \\ 2 & 7 & 9 \end{bmatrix}\begin{bmatrix} -42 & -24 & 30 \\ 3 & 30 & -21 \\ 7 & -18 & -5 \end{bmatrix} \)

\( = \begin{bmatrix} -168 + 15 + 21 & -96 + 150 - 54 & 120 - 105 - 15 \\ -42 + 0 + 42 & -24 + 0 - 108 & 30 + 0 - 30 \\ -84 + 21 + 63 & -48 + 210 - 162 & 60 - 147 - 45 \end{bmatrix} \)

\( = \begin{bmatrix} -132 & 0 & 0 \\ 0 & -132 & 0 \\ 0 & 0 & -132 \end{bmatrix} \)

\( = -132 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = -132I \)

Calculating (adj A)A

\( (adj A). A = \begin{bmatrix} -42 & -24 & 30 \\ 3 & 30 & -21 \\ 7 & -18 & -5 \end{bmatrix}\begin{bmatrix} 4 & 5 & 3 \\ 1 & 0 & 6 \\ 2 & 7 & 9 \end{bmatrix} \)

\( = \begin{bmatrix} -168 - 24 + 60 & -210 + 0 + 210 & -126 - 144 + 270 \\ 12 + 30 - 42 & 15 + 0 - 147 & 9 + 180 - 189 \\ 28 - 18 - 10 & 35 + 0 - 35 & 21 - 108 - 45 \end{bmatrix} \)

\( = \begin{bmatrix} -132 & 0 & 0 \\ 0 & -132 & 0 \\ 0 & 0 & -132 \end{bmatrix} \)

\( = -132 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = -132I \)

Calculating |A|.I

Expanding along C1, we get

\( |A| = a_{11} (-1)^{1+1}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} + a_{21} (-1)^{2+1}\begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix} + a_{31} (-1)^{3+1}\begin{vmatrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{vmatrix} \)

\( |A|.I = \begin{bmatrix} 4 & 5 & 3 \\ 1 & 0 & 6 \\ 2 & 7 & 9 \end{bmatrix}I \)

\( = [4(0 - 42) - (1){45 - 21} + 2(30 - 0 )]I \)
\( = [4(-42) - 1(24) + 2(30)] I \)
\( = (-168 - 24 + 60)I \)
\( = -132I \)

Thus, A(adj A) = (adj A)A = |A|I = -132I

\( \Rightarrow A(adj A) = (adj A)A = |A|I \)

Hence Proved

Ans. \( \begin{bmatrix} -42 & -24 & 30 \\ 3 & 30 & -21 \\ 7 & -18 & -5 \end{bmatrix} \)

Exam Tip: When the determinant is a large negative number, double-check all sign calculations in the co-factor matrix - a single error propagates through both matrix products.

Question 9. Find the adjoint of the given matrix and verify in each case that A. (adj A) = (adj A) = m |A|.I.
\( \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
Answer: Here, \( A = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

To find adj A, we must determine the co-factors:

\( a_{11} = \begin{vmatrix} \cos \alpha & 0 \\ 0 & 1 \end{vmatrix} = \cos \alpha \)

\( a_{12} = - \begin{vmatrix} \sin \alpha & 0 \\ 0 & 1 \end{vmatrix} = -\sin \alpha \)

\( a_{13} = \begin{vmatrix} \sin \alpha & \cos \alpha \\ 0 & 0 \end{vmatrix} = 0 \)

\( a_{21} = - \begin{vmatrix} -\sin \alpha & 0 \\ 0 & 1 \end{vmatrix} = \sin \alpha \)

\( a_{22} = \begin{vmatrix} \cos \alpha & 0 \\ 0 & 1 \end{vmatrix} = \cos \alpha \)

\( a_{23} = - \begin{vmatrix} \cos \alpha & -\sin \alpha \\ 0 & 0 \end{vmatrix} = 0 \)

\( a_{31} = \begin{vmatrix} -\sin \alpha & 0 \\ \cos \alpha & 0 \end{vmatrix} = 0 \)

\( a_{32} = - \begin{vmatrix} \cos \alpha & 0 \\ \sin \alpha & 0 \end{vmatrix} = 0 \)

\( a_{33} = \begin{vmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{vmatrix} = [\cos^2 \alpha - \{-\sin^2 \alpha\}] = [\cos^2 \alpha + \sin^2 \alpha] = 1 \)

\( [\because \cos^2 \alpha + \sin^2 \alpha = 1] \)

\( \therefore adj A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}^T = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}^T = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

Calculating A (adj A)

\( A.(adj A) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & \cos \alpha \sin \alpha - \sin \alpha \cos \alpha & 0 \\ \sin \alpha \cos \alpha - \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & 0 & 0 \\ 0 & \cos^2 \alpha + \sin^2 \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( [\because \cos^2 \alpha + \sin^2 \alpha = 1] \)

\( = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = I \)

Calculating (adj A)A

\( (adj A). A = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & -\sin \alpha \cos \alpha + \cos \alpha \sin \alpha & 0 \\ -\sin \alpha \cos \alpha + \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & 0 & 0 \\ 0 & \cos^2 \alpha + \sin^2 \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( [\because \cos^2 \alpha + \sin^2 \alpha = 1] \)

\( = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( = I \)

Calculating |A|.I

Expanding along C1, we get

\( |A| = a_{11} (-1)^{1+1}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} + a_{21} (-1)^{2+1}\begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix} + a_{31} (-1)^{3+1}\begin{vmatrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{vmatrix} \)

\( |A|.I = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}I \)

Thus, A(adj A) = (adj A)A = |A|I = I

\( \Rightarrow A(adj A) = (adj A)A = |A|I \)

Hence Proved

Ans. \( \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

Exam Tip: Block matrices with zeros simplify greatly - use this structure to compute minors quickly, and always apply the trigonometric identity when working with cosine and sine terms.

Question 10. If A = \(\begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}\), show that adj A = A.
Answer: We start with the given matrix \(A = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}\).

To demonstrate that adj A equals A, we first determine the co-factors. For the element at position (1,1): \(a_{11} = \begin{vmatrix} 0 & 1 \\ 4 & 3 \end{vmatrix} = 0 - 4 = -4\). For position (1,2): \(a_{12} = -\begin{vmatrix} 1 & 1 \\ 4 & 3 \end{vmatrix} = -(3 - 4) = -(-1) = 1\). For position (1,3): \(a_{13} = \begin{vmatrix} 1 & 0 \\ 4 & 4 \end{vmatrix} = 4 - 0 = 4\).

Continuing with row 2: \(a_{21} = -\begin{vmatrix} -3 & -3 \\ 4 & 3 \end{vmatrix} = -(-9 + 12) = -3\). Position (2,2): \(a_{22} = \begin{vmatrix} -4 & -3 \\ 4 & 3 \end{vmatrix} = -12 + 12 = 0\). Position (2,3): \(a_{23} = -\begin{vmatrix} -4 & -3 \\ 4 & 4 \end{vmatrix} = -(-16 + 12) = 4\).

For row 3: \(a_{31} = \begin{vmatrix} -3 & -3 \\ 0 & 1 \end{vmatrix} = -3 + 0 = -3\). Position (3,2): \(a_{32} = -\begin{vmatrix} -4 & -3 \\ 1 & 1 \end{vmatrix} = -(-4 + 3) = 1\). Position (3,3): \(a_{33} = \begin{vmatrix} -4 & -3 \\ 1 & 0 \end{vmatrix} = 0 + 3 = 3\).

The co-factor matrix is \(\begin{bmatrix} -4 & 1 & 4 \\ -3 & 0 & 4 \\ -3 & 1 & 3 \end{bmatrix}\). Taking the transpose gives us adj A = \(\begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix} = A\).

Therefore, adj A = A has been shown.

Exam Tip: When verifying adj A = A, always compute each co-factor methodically using the correct signs, then check that the transposed co-factor matrix matches the original exactly.

Question 11. If A = \(\begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}\), show that adj A = 3A'.
Answer: We have the matrix \(A = \begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}\).

To demonstrate adj A = 3A', we first find the transpose. The transpose is \(A' = \begin{bmatrix} -1 & 2 & 2 \\ -2 & 1 & -2 \\ -2 & -2 & 1 \end{bmatrix}\).

Next, we compute the co-factors of A. For (1,1): \(a_{11} = \begin{vmatrix} 1 & -2 \\ -2 & 1 \end{vmatrix} = 1 - 4 = -3\). For (1,2): \(a_{12} = -\begin{vmatrix} 2 & -2 \\ 2 & 1 \end{vmatrix} = -(2 + 4) = -6\). For (1,3): \(a_{13} = \begin{vmatrix} 2 & 1 \\ 2 & -2 \end{vmatrix} = -4 - 2 = -6\).

For row 2: \(a_{21} = -\begin{vmatrix} -2 & -2 \\ -2 & 1 \end{vmatrix} = -(-2 + 4) = 6\). Position (2,2): \(a_{22} = \begin{vmatrix} -1 & -2 \\ 2 & 1 \end{vmatrix} = -1 + 4 = 3\). Position (2,3): \(a_{23} = -\begin{vmatrix} -1 & -2 \\ 2 & -2 \end{vmatrix} = -(2 + 4) = -6\).

For row 3: \(a_{31} = \begin{vmatrix} -2 & -2 \\ 1 & -2 \end{vmatrix} = 4 + 2 = 6\). Position (3,2): \(a_{32} = -\begin{vmatrix} -1 & -2 \\ 2 & 1 \end{vmatrix} = -(2 + 4) = -6\). Position (3,3): \(a_{33} = \begin{vmatrix} -1 & -2 \\ 2 & 1 \end{vmatrix} = -1 + 4 = 3\).

The co-factor matrix is \(\begin{bmatrix} -3 & -6 & -6 \\ 6 & 3 & -6 \\ 6 & -6 & 3 \end{bmatrix}\). The adjugate is adj A = \(\begin{bmatrix} -3 & 6 & 6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{bmatrix}\).

We can factor this as adj A = \(3\begin{bmatrix} -1 & 2 & 2 \\ -2 & 1 & -2 \\ -2 & -2 & 1 \end{bmatrix} = 3A'\).

Thus adj A = 3A' has been proven.

Exam Tip: When proving relationships like adj A = kA', compute both sides separately and verify they match exactly; this confirms the relationship clearly.

Question 12. If A = \(\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}\), find A\(^{-1}\).
Answer: We have A = \(\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}\). To find A\(^{-1}\), we use the formula \(A^{-1} = \frac{\text{adj A}}{|A|}\).

First, we determine the co-factors. The co-factor of 3 is \((-1)^{1+1}(2) = 2\). The co-factor of -5 is \((-1)^{1+2}(-1) = 1\). The co-factor of -1 is \((-1)^{2+1}(-5) = 5\). The co-factor of 2 is \((-1)^{2+2}(3) = 3\).

The co-factor matrix is \(\begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}\). Taking the transpose, we get adj A = \(\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}\).

Next, we find the determinant: \(|A| = 3 \times 2 - (-1) \times (-5) = 6 - 5 = 1\).

Therefore, \(A^{-1} = \frac{\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}}{1} = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}\).

Exam Tip: Always verify that |A| ≠ 0 before attempting to find the inverse; if the determinant is zero, the matrix is singular and has no inverse.

Question 13. If A = \(\begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}\), find A\(^{-1}\).
Answer: We have A = \(\begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}\). To find A\(^{-1}\), we use \(A^{-1} = \frac{\text{adj A}}{|A|}\).

We determine the co-factors. The co-factor of 4 is \((-1)^{1+1}(3) = 3\). The co-factor of 1 is \((-1)^{1+2}(2) = -2\). The co-factor of 2 is \((-1)^{2+1}(1) = -1\). The co-factor of 3 is \((-1)^{2+2}(4) = 4\).

The co-factor matrix is \(\begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}\). Taking the transpose gives adj A = \(\begin{bmatrix} 3 & -1 \\ -2 & 4 \end{bmatrix}\).

Now we find the determinant: \(|A| = 4 \times 3 - 1 \times 2 = 12 - 2 = 10\).

Therefore, \(A^{-1} = \frac{1}{10}\begin{bmatrix} 3 & -1 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} \frac{3}{10} & -\frac{1}{10} \\ -\frac{1}{5} & \frac{2}{5} \end{bmatrix}\).

Exam Tip: Express the final inverse matrix with all entries in simplest fractional form; avoid leaving fractions in the denominator of matrix entries.

Question 14. If A = \(\begin{bmatrix} 2 & -3 \\ 4 & 6 \end{bmatrix}\), find A\(^{-1}\).
Answer: We have A = \(\begin{bmatrix} 2 & -3 \\ 4 & 6 \end{bmatrix}\). To find A\(^{-1}\), we apply \(A^{-1} = \frac{\text{adj A}}{|A|}\).

First, we find the co-factors. The co-factor of 2 is \((-1)^{1+1}(6) = 6\). The co-factor of -3 is \((-1)^{1+2}(4) = -4\). The co-factor of 4 is \((-1)^{2+1}(-3) = 3\). The co-factor of 6 is \((-1)^{2+2}(2) = 2\).

The co-factor matrix is \(\begin{bmatrix} 6 & -4 \\ 3 & 2 \end{bmatrix}\). Taking the transpose gives adj A = \(\begin{bmatrix} 6 & 3 \\ -4 & 2 \end{bmatrix}\).

Next, we compute the determinant: \(|A| = 2 \times 6 - (-3) \times 4 = 12 + 12 = 24\).

Therefore, \(A^{-1} = \frac{1}{24}\begin{bmatrix} 6 & 3 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{4} & \frac{1}{8} \\ -\frac{1}{6} & \frac{1}{12} \end{bmatrix}\).

Exam Tip: Always reduce each fraction entry to lowest terms; check that no entry can be simplified further before writing the final answer.

Question 15. If A = \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\), when (ad - bc) ≠ 0, find A\(^{-1}\).
Answer: We have the general 2×2 matrix A = \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\). To find A\(^{-1}\), we use \(A^{-1} = \frac{\text{adj A}}{|A|}\).

We determine the co-factors. The co-factor of a is \((-1)^{1+1}(d) = d\). The co-factor of b is \((-1)^{1+2}(c) = -c\). The co-factor of c is \((-1)^{2+1}(b) = -b\). The co-factor of d is \((-1)^{2+2}(a) = a\).

The co-factor matrix is \(\begin{bmatrix} d & -c \\ -b & a \end{bmatrix}\). Taking the transpose gives adj A = \(\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\).

The determinant is: \(|A| = a \times d - c \times b = ad - bc\).

Therefore, the general formula for the inverse of a 2×2 matrix is \(A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\), provided that (ad - bc) ≠ 0.

Exam Tip: This general formula is fundamental - memorizing it can save time when finding inverses of 2×2 matrices in quick calculations.

Question 16. If A = \(\begin{bmatrix} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{bmatrix}\), find A\(^{-1}\).
Answer: We have A = \(\begin{bmatrix} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{bmatrix}\). To find A\(^{-1}\), we use \(A^{-1} = \frac{\text{adj A}}{|A|}\).

First, we find the determinant by expanding along C1:

\(|A| = 1 \begin{vmatrix} -1 & -1 \\ 3 & -1 \end{vmatrix} - 1 \begin{vmatrix} 2 & 5 \\ 3 & -1 \end{vmatrix} + 2 \begin{vmatrix} 2 & 5 \\ -1 & -1 \end{vmatrix}\)

\(= 1(1 + 3) - 1(-2 - 15) + 2(-2 + 5)\)

\(= 4 + 17 + 6 = 27\)

Next, we find all co-factors. \(a_{11} = \begin{vmatrix} -1 & -1 \\ 3 & -1 \end{vmatrix} = 1 + 3 = 4\). \(a_{12} = -\begin{vmatrix} 1 & -1 \\ 2 & -1 \end{vmatrix} = -(-1 + 2) = -1\). \(a_{13} = \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = 3 + 2 = 5\).

\(a_{21} = -\begin{vmatrix} 2 & 5 \\ 3 & -1 \end{vmatrix} = -(-2 - 15) = 17\). \(a_{22} = \begin{vmatrix} 1 & 5 \\ 2 & -1 \end{vmatrix} = -1 - 10 = -11\). \(a_{23} = -\begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = -(3 - 4) = 1\).

\(a_{31} = \begin{vmatrix} 2 & 5 \\ -1 & -1 \end{vmatrix} = -2 + 5 = 3\). \(a_{32} = -\begin{vmatrix} 1 & 5 \\ 1 & -1 \end{vmatrix} = -(-1 - 5) = 6\). \(a_{33} = \begin{vmatrix} 1 & 2 \\ 1 & -1 \end{vmatrix} = -1 - 2 = -3\).

The adjugate is adj A = \(\begin{bmatrix} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{bmatrix}\).

Therefore, \(A^{-1} = \frac{1}{27}\begin{bmatrix} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{bmatrix}\).

Exam Tip: For 3×3 matrices, expand the determinant along the column or row with the most zeros to simplify calculations; then systematically compute each co-factor using the correct alternating signs.

Question 17. If A = \(\begin{bmatrix} 2 & -1 & 1 \\ 3 & 0 & -1 \\ 2 & 6 & 0 \end{bmatrix}\), find A\(^{-1}\).
Answer: We have A = \(\begin{bmatrix} 2 & -1 & 1 \\ 3 & 0 & -1 \\ 2 & 6 & 0 \end{bmatrix}\). To find A\(^{-1}\), we use \(A^{-1} = \frac{\text{adj A}}{|A|}\).

First, we compute the determinant by expanding along C1:

\(|A| = 2 \begin{vmatrix} 0 & -1 \\ 6 & 0 \end{vmatrix} - 3 \begin{vmatrix} -1 & 1 \\ 6 & 0 \end{vmatrix} + 2 \begin{vmatrix} -1 & 1 \\ 0 & -1 \end{vmatrix}\)

\(= 2(0 + 6) - 3(0 - 6) + 2(1 - 0)\)

\(= 12 + 18 + 2 = 32\)

Next, we compute all co-factors. \(a_{11} = \begin{vmatrix} 0 & -1 \\ 6 & 0 \end{vmatrix} = 0 + 6 = 6\). \(a_{12} = -\begin{vmatrix} 3 & -1 \\ 2 & 0 \end{vmatrix} = -(0 + 2) = -2\). \(a_{13} = \begin{vmatrix} 3 & 0 \\ 2 & 6 \end{vmatrix} = 18 - 0 = 18\).

\(a_{21} = -\begin{vmatrix} -1 & 1 \\ 6 & 0 \end{vmatrix} = -(0 - 6) = 6\). \(a_{22} = \begin{vmatrix} 2 & 1 \\ 2 & 0 \end{vmatrix} = 0 - 2 = -2\). \(a_{23} = -\begin{vmatrix} 2 & -1 \\ 2 & 6 \end{vmatrix} = -(12 + 2) = -14\).

\(a_{31} = \begin{vmatrix} -1 & 1 \\ 0 & -1 \end{vmatrix} = 1 - 0 = 1\). \(a_{32} = -\begin{vmatrix} 2 & 1 \\ 3 & -1 \end{vmatrix} = -(-2 - 3) = 5\). \(a_{33} = \begin{vmatrix} 2 & -1 \\ 3 & 0 \end{vmatrix} = 0 + 3 = 3\).

The adjugate is adj A = \(\begin{bmatrix} 6 & 6 & 1 \\ -2 & -2 & 5 \\ 18 & -14 & 3 \end{bmatrix}\).

Therefore, \(A^{-1} = \frac{1}{32}\begin{bmatrix} 6 & 6 & 1 \\ -2 & -2 & 5 \\ 18 & -14 & 3 \end{bmatrix}\).

Exam Tip: Always choose the expansion row or column wisely - picking the one with the most zeros reduces the computational workload significantly.

Question 18. If A = \(\begin{bmatrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{bmatrix}\), find A\(^{-1}\).
Answer: We have A = \(\begin{bmatrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{bmatrix}\). To find A\(^{-1}\), we apply \(A^{-1} = \frac{\text{adj A}}{|A|}\).

First, we compute the determinant by expanding along C1:

\(|A| = 2 \begin{vmatrix} 2 & 3 \\ -2 & 2 \end{vmatrix} - 2 \begin{vmatrix} -3 & 3 \\ -2 & 2 \end{vmatrix} + 3 \begin{vmatrix} -3 & 3 \\ 2 & 3 \end{vmatrix}\)

\(= 2(4 + 6) - 2(-6 + 6) + 3(-9 - 6)\)

\(= 20 - 0 - 45 = -25\)

Next, we find all co-factors. \(a_{11} = \begin{vmatrix} 2 & 3 \\ -2 & 2 \end{vmatrix} = 4 + 6 = 10\). \(a_{12} = -\begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} = -(4 - 9) = 5\). \(a_{13} = \begin{vmatrix} 2 & 2 \\ 3 & -2 \end{vmatrix} = -4 - 6 = -10\).

\(a_{21} = -\begin{vmatrix} -3 & 3 \\ -2 & 2 \end{vmatrix} = -(-6 + 6) = 0\). \(a_{22} = \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} = 4 - 9 = -5\). \(a_{23} = -\begin{vmatrix} 2 & -3 \\ 3 & -2 \end{vmatrix} = -(-4 + 9) = -5\).

\(a_{31} = \begin{vmatrix} -3 & 3 \\ 2 & 3 \end{vmatrix} = -9 - 6 = -15\). \(a_{32} = -\begin{vmatrix} 2 & 3 \\ 2 & 3 \end{vmatrix} = -(6 - 6) = 0\). \(a_{33} = \begin{vmatrix} 2 & -3 \\ 2 & 2 \end{vmatrix} = 4 + 6 = 10\).

The adjugate is adj A = \(\begin{bmatrix} 10 & 0 & -15 \\ 5 & -5 & 0 \\ -10 & -5 & 10 \end{bmatrix}\).

Therefore, \(A^{-1} = \frac{1}{-25}\begin{bmatrix} 10 & 0 & -15 \\ 5 & -5 & 0 \\ -10 & -5 & 10 \end{bmatrix} = -\frac{1}{25}\begin{bmatrix} 10 & 0 & -15 \\ 5 & -5 & 0 \\ -10 & -5 & 10 \end{bmatrix}\).

Exam Tip: When the determinant is negative, be careful with the sign - include it in the scalar divisor and verify a few entries in your final answer to catch any sign errors.

Question 19. If A = \(\begin{bmatrix} 0 & 0 & -1 \\ 3 & 4 & 5 \\ -2 & -4 & -7 \end{bmatrix}\), find A\(^{-1}\).
Answer: We have A = \(\begin{bmatrix} 0 & 0 & -1 \\ 3 & 4 & 5 \\ -2 & -4 & -7 \end{bmatrix}\). To find A\(^{-1}\), we use \(A^{-1} = \frac{\text{adj A}}{|A|}\).

First, we find the determinant by expanding along C1 (which has two zeros):

\(|A| = 0 - 3 \begin{vmatrix} 0 & -1 \\ -4 & -7 \end{vmatrix} + (-2) \begin{vmatrix} 0 & -1 \\ 4 & 5 \end{vmatrix}\)

\(= 0 - 3(0 - 4) - 2(0 + 4)\)

\(= 12 - 8 = 4\)

Next, we compute all co-factors. \(a_{11} = \begin{vmatrix} 4 & 5 \\ -4 & -7 \end{vmatrix} = -28 + 20 = -8\). \(a_{12} = -\begin{vmatrix} 3 & 5 \\ -2 & -7 \end{vmatrix} = -(-21 + 10) = 11\). \(a_{13} = \begin{vmatrix} 3 & 4 \\ -2 & -4 \end{vmatrix} = -12 + 8 = -4\).

\(a_{21} = -\begin{vmatrix} 0 & -1 \\ -4 & -7 \end{vmatrix} = -(0 - 4) = 4\). \(a_{22} = \begin{vmatrix} 0 & -1 \\ -2 & -7 \end{vmatrix} = 0 - 2 = -2\). \(a_{23} = -\begin{vmatrix} 0 & 0 \\ -2 & -4 \end{vmatrix} = 0\).

\(a_{31} = \begin{vmatrix} 0 & -1 \\ 4 & 5 \end{vmatrix} = 0 + 4 = 4\). \(a_{32} = -\begin{vmatrix} 0 & -1 \\ 3 & 5 \end{vmatrix} = -(0 + 3) = -3\). \(a_{33} = \begin{vmatrix} 0 & 0 \\ 3 & 4 \end{vmatrix} = 0\).

The adjugate is adj A = \(\begin{bmatrix} -8 & 4 & 4 \\ 11 & -2 & -3 \\ -4 & 0 & 0 \end{bmatrix}\).

Therefore, \(A^{-1} = \frac{1}{4}\begin{bmatrix} -8 & 4 & 4 \\ 11 & -2 & -3 \\ -4 & 0 & 0 \end{bmatrix}\).

Exam Tip: Look for rows or columns with zeros to simplify determinant expansion; this strategy cuts down unnecessary calculation effort.

Question 20. If A = \(\begin{bmatrix} 0 & 0 & -1 \\ 3 & 4 & 5 \\ -2 & -4 & -7 \end{bmatrix}\), find A\(^{-1}\).
Answer: We have A = \(\begin{bmatrix} 0 & 0 & -1 \\ 3 & 4 & 5 \\ -2 & -4 & -7 \end{bmatrix}\). To find A\(^{-1}\), we use \(A^{-1} = \frac{\text{adj A}}{|A|}\).

First, we determine the determinant by expanding along C1:

\(|A| = 0 \begin{vmatrix} 4 & 5 \\ -4 & -7 \end{vmatrix} - 3 \begin{vmatrix} 0 & -1 \\ -4 & -7 \end{vmatrix} + (-2) \begin{vmatrix} 0 & -1 \\ 4 & 5 \end{vmatrix}\)

\(= 0 - 3(0 - 4) - 2(0 + 4)\)

\(= 12 - 8 = 4\)

Now we find all co-factors. \(a_{11} = \begin{vmatrix} 4 & 5 \\ -4 & -7 \end{vmatrix} = -28 + 20 = -8\). \(a_{12} = -\begin{vmatrix} 3 & 5 \\ -2 & -7 \end{vmatrix} = -(-21 + 10) = 11\). \(a_{13} = \begin{vmatrix} 3 & 4 \\ -2 & -4 \end{vmatrix} = -12 + 8 = -4\).

\(a_{21} = -\begin{vmatrix} 0 & -1 \\ -4 & -7 \end{vmatrix} = -(0 - 4) = 4\). \(a_{22} = \begin{vmatrix} 0 & -1 \\ -2 & -7 \end{vmatrix} = 0 - 2 = -2\). \(a_{23} = -\begin{vmatrix} 0 & 0 \\ -2 & -4 \end{vmatrix} = 0\).

\(a_{31} = \begin{vmatrix} 0 & -1 \\ 4 & 5 \end{vmatrix} = 0 + 4 = 4\). \(a_{32} = -\begin{vmatrix} 0 & -1 \\ 3 & 5 \end{vmatrix} = -(0 + 3) = -3\). \(a_{33} = \begin{vmatrix} 0 & 0 \\ 3 & 4 \end{vmatrix} = 0\).

The adjugate is adj A = \(\begin{bmatrix} -8 & 4 & 4 \\ 11 & -2 & -3 \\ -4 & 0 & 0 \end{bmatrix}\).

Therefore, \(A^{-1} = \frac{1}{4}\begin{bmatrix} -8 & 4 & 4 \\ 11 & -2 & -3 \\ -4 & 0 & 0 \end{bmatrix}\).

Exam Tip: When a determinant yields a small integer like 4, divide each matrix entry by that value carefully to avoid arithmetic mistakes in the final result.

Question 21. If A = \(\begin{bmatrix} 2 & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}\), find A\(^{-1}\).
Answer: We have A = \(\begin{bmatrix} 2 & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}\). To find A\(^{-1}\), we use \(A^{-1} = \frac{\text{adj A}}{|A|}\).

First, we find the determinant by expanding along C1:

\(|A| = 2 \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} - (-3) \begin{vmatrix} -1 & 4 \\ 1 & 2 \end{vmatrix} + (-1) \begin{vmatrix} -1 & 4 \\ 0 & 1 \end{vmatrix}\)

\(= 2(0 - 1) + 3(-2 - 4) - 1(-1 - 0)\)

\(= -2 - 18 + 1 = -19\)

Next, we compute all co-factors. \(a_{11} = \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} = 0 - 1 = -1\). \(a_{12} = -\begin{vmatrix} -3 & 1 \\ -1 & 2 \end{vmatrix} = -(-6 + 1) = 5\). \(a_{13} = \begin{vmatrix} -3 & 0 \\ -1 & 1 \end{vmatrix} = -3 - 0 = -3\).

\(a_{21} = -\begin{vmatrix} -1 & 4 \\ 1 & 2 \end{vmatrix} = -(-2 - 4) = 6\). \(a_{22} = \begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix} = 4 + 4 = 8\). \(a_{23} = -\begin{vmatrix} 2 & -1 \\ -1 & 1 \end{vmatrix} = -(2 - 1) = -1\).

\(a_{31} = \begin{vmatrix} -1 & 4 \\ 0 & 1 \end{vmatrix} = -1 - 0 = -1\). \(a_{32} = -\begin{vmatrix} 2 & 4 \\ -3 & 1 \end{vmatrix} = -(2 + 12) = -14\). \(a_{33} = \begin{vmatrix} 2 & -1 \\ -3 & 0 \end{vmatrix} = 0 - 3 = -3\).

The adjugate is adj A = \(\begin{bmatrix} -1 & 6 & -1 \\ 5 & 8 & -14 \\ -3 & -1 & -3 \end{bmatrix}\).

Therefore, \(A^{-1} = \frac{1}{-19}\begin{bmatrix} -1 & 6 & -1 \\ 5 & 8 & -14 \\ -3 & -1 & -3 \end{bmatrix} = \frac{1}{19}\begin{bmatrix} 1 & -6 & 1 \\ -5 & -8 & 14 \\ 3 & 1 & 3 \end{bmatrix}\).

Exam Tip: When the determinant is negative, factor the negative sign out of the scalar and adjust the matrix entries accordingly to keep the final answer clear and organized.

Question 22. If A = \(\begin{bmatrix} 8 & -4 & 1 \\ 10 & 0 & 6 \\ 8 & 1 & 6 \end{bmatrix}\), find A\(^{-1}\).
Answer: We have A = \(\begin{bmatrix} 8 & -4 & 1 \\ 10 & 0 & 6 \\ 8 & 1 & 6 \end{bmatrix}\). To find A\(^{-1}\), we apply \(A^{-1} = \frac{\text{adj A}}{|A|}\).

First, we compute the determinant by expanding along C1:

\(|A| = 8 \begin{vmatrix} 0 & 6 \\ 1 & 6 \end{vmatrix} - 10 \begin{vmatrix} -4 & 1 \\ 1 & 6 \end{vmatrix} + 8 \begin{vmatrix} -4 & 1 \\ 0 & 6 \end{vmatrix}\)

\(= 8(0 - 6) - 10(-24 - 1) + 8(-24 - 0)\)

\(= -48 + 250 - 192 = 10\)

Next, we find all co-factors. \(a_{11} = \begin{vmatrix} 0 & 6 \\ 1 & 6 \end{vmatrix} = 0 - 6 = -6\). \(a_{12} = -\begin{vmatrix} 10 & 6 \\ 8 & 6 \end{vmatrix} = -(60 - 48) = -12\). \(a_{13} = \begin{vmatrix} 10 & 0 \\ 8 & 1 \end{vmatrix} = 10 - 0 = 10\).

\(a_{21} = -\begin{vmatrix} -4 & 1 \\ 1 & 6 \end{vmatrix} = -(-24 - 1) = 25\). \(a_{22} = \begin{vmatrix} 8 & 1 \\ 8 & 6 \end{vmatrix} = 48 - 8 = 40\). \(a_{23} = -\begin{vmatrix} 8 & -4 \\ 8 & 1 \end{vmatrix} = -(8 + 32) = -40\).

\(a_{31} = \begin{vmatrix} -4 & 1 \\ 0 & 6 \end{vmatrix} = -24 - 0 = -24\). \(a_{32} = -\begin{vmatrix} 8 & 1 \\ 10 & 6 \end{vmatrix} = -(48 - 10) = -38\). \(a_{33} = \begin{vmatrix} 8 & -4 \\ 10 & 0 \end{vmatrix} = 0 + 40 = 40\).

The adjugate is adj A = \(\begin{bmatrix} -6 & 25 & -24 \\ -12 & 40 & -38 \\ 10 & -40 & 40 \end{bmatrix}\).

Therefore, \(A^{-1} = \frac{1}{10}\begin{bmatrix} -6 & 25 & -24 \\ -12 & 40 & -38 \\ 10 & -40 & 40 \end{bmatrix}\).

Exam Tip: After finding the adjugate matrix, check if any entries share a common factor with the determinant - simplifying first can make the final matrix cleaner.

Question 23. If A = \(\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}\), show that A\(^{-1}\) = \(\frac{1}{19}\)A.
Answer: We have A = \(\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}\). To prove that A\(^{-1}\) = \(\frac{1}{19}\)A, we find A\(^{-1}\) and verify it equals the claimed form.

First, we find the co-factors. The co-factor of 2 is \((-1)^{1+1}(-2) = -2\). The co-factor of 3 is \((-1)^{1+2}(5) = -5\). The co-factor of 5 is \((-1)^{2+1}(3) = -3\). The co-factor of -2 is \((-1)^{2+2}(2) = 2\).

The co-factor matrix is \(\begin{bmatrix} -2 & -5 \\ -3 & 2 \end{bmatrix}\). Taking the transpose gives adj A = \(\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}\).

Next, we find the determinant: \(|A| = 2 \times (-2) - 3 \times 5 = -4 - 15 = -19\).

Therefore, \(A^{-1} = \frac{\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}}{-19} = -\frac{1}{19}\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix} = \frac{1}{19}\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix} = \frac{1}{19}A\).

Thus, A\(^{-1}\) = \(\frac{1}{19}\)A has been shown.

Exam Tip: When verifying inverse relationships, compute the determinant carefully and then check that scaling the adjugate by the reciprocal of the determinant yields the claimed result.

Question 23. If A = \( \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \), show that A^-1 = A².
Answer: We have \( A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \).
To demonstrate: A^-1 = A²

First, we need to find A^-1 and A^-1 = \( \frac{\text{adj } A}{|A|} \)

Calculating |A|

Expanding |A| along C₁, we get
\( |A| = 1 \begin{vmatrix} -1 & 0 \\ 0 & 0 \end{vmatrix} - 2 \begin{vmatrix} -1 & 1 \\ 0 & 0 \end{vmatrix} + 1 \begin{vmatrix} -1 & 1 \\ -1 & 0 \end{vmatrix} \)
\( = 1(0) - 2(0) + 1(0 - (-1)) \)
\( = 1(1) \)
\( = 1 \)

Now, we need to find adj A and to do that we must find cofactors:

\( a_{11} = \begin{vmatrix} -1 & 0 \\ 0 & 0 \end{vmatrix} = 0 \)

\( a_{12} = -\begin{vmatrix} 2 & 0 \\ 1 & 0 \end{vmatrix} = 0 \)

\( a_{13} = \begin{vmatrix} 2 & -1 \\ 1 & 0 \end{vmatrix} = 0 - (-1) = 1 \)

\( a_{21} = -\begin{vmatrix} -1 & 1 \\ 0 & 0 \end{vmatrix} = -0 = 0 \)

\( a_{22} = \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = 0 - 1 = -1 \)

\( a_{23} = -\begin{vmatrix} 1 & -1 \\ 1 & 0 \end{vmatrix} = -(0 - (-1)) = -1 \)

\( a_{31} = \begin{vmatrix} -1 & 1 \\ -1 & 0 \end{vmatrix} = 0 - (-1) = 1 \)

\( a_{32} = -\begin{vmatrix} 1 & 1 \\ 2 & 0 \end{vmatrix} = -(0 - 2) = 2 \)

\( a_{33} = \begin{vmatrix} 1 & -1 \\ 2 & -1 \end{vmatrix} = -1 - (-2) = -1 + 2 = 1 \)

\( \therefore \text{adj } A = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & -1 \\ 1 & 2 & 1 \end{bmatrix}^T = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix} \)

\( \therefore A^{-1} = \frac{\text{adj } A}{|A|} = \frac{\begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix}}{1} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix} \) ... (i)

Calculating A²

\( A^2 = A \cdot A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 1 - 2 + 1 & -1 + 1 + 0 & 1 + 0 + 0 \\ 2 - 2 + 0 & -2 + 1 + 0 & 2 + 0 + 0 \\ 1 + 0 + 0 & -1 + 0 + 0 & 1 + 0 + 0 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix} \)
\( = A^{-1} \) [from eq. (i)]

Hence, A² = A^-1
In simple words: When you multiply this matrix by itself, you get the same result as when you compute its inverse. This shows the elegant connection between the powers and inverse of this particular matrix.

Exam Tip: Always verify that A² equals A^-1 by computing both separately and comparing results - this ensures accuracy and confirms the relationship.

Question 24. If A = \( \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \), prove that A^-1 = A³.
Answer: We have \( A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \)

To demonstrate: A^-1 = A³

First, we need to find A^-1 and A^-1 = \( \frac{\text{adj } A}{|A|} \)

Calculating |A|

Expanding |A| along C₁, we get
\( |A| = 3 \begin{vmatrix} -3 & 4 \\ -1 & 1 \end{vmatrix} - 2 \begin{vmatrix} -3 & 4 \\ -1 & 1 \end{vmatrix} + 0 \begin{vmatrix} -3 & 4 \\ -3 & 4 \end{vmatrix} \)
\( = 3(-3 - (-4)) - 2(-3 - (-4)) + 0 \)
\( = 3(-3 + 4) - 2(-3 + 4) \)
\( = 3(1) - 2(1) \)
\( = 3 - 2 \)
\( = 1 \)

Now, we need to find adj A and to do that we must find cofactors:

\( a_{11} = \begin{vmatrix} -3 & 4 \\ -1 & 1 \end{vmatrix} = -3 - (-4) = -3 + 4 = 1 \)

\( a_{12} = -\begin{vmatrix} 2 & 4 \\ 0 & 1 \end{vmatrix} = -(2 - 0) = -2 \)

\( a_{13} = \begin{vmatrix} 2 & -3 \\ 0 & -1 \end{vmatrix} = -2 - 0 = -2 \)

\( a_{21} = -\begin{vmatrix} -3 & 4 \\ -1 & 1 \end{vmatrix} = -(-3 - (-4)) = -(-3 + 4) = -1 \)

\( a_{22} = \begin{vmatrix} 3 & 4 \\ 0 & 1 \end{vmatrix} = 3 - 0 = 3 \)

\( a_{23} = -\begin{vmatrix} 3 & -3 \\ 0 & -1 \end{vmatrix} = -(-3 - 0) = 3 \)

\( a_{31} = \begin{vmatrix} -3 & 4 \\ -3 & 4 \end{vmatrix} = -12 - (-12) = 0 \)

\( a_{32} = -\begin{vmatrix} 3 & 4 \\ 2 & 4 \end{vmatrix} = -(12 - 8) = -4 \)

\( a_{33} = \begin{vmatrix} 3 & -3 \\ 2 & -3 \end{vmatrix} = (-9 - (-6)) = -9 + 6 = -3 \)

\( \therefore \text{adj } A = \begin{bmatrix} 1 & -2 & -2 \\ -1 & 3 & 3 \\ 0 & -4 & -3 \end{bmatrix}^T = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix} \)

\( \therefore A^{-1} = \frac{\text{adj } A}{|A|} = \frac{\begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}}{1} = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix} \) ... (i)

Calculating A³

\( A^2 = A \cdot A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 9 - 6 + 0 & -9 + 9 - 4 & 12 - 12 + 4 \\ 6 - 6 + 0 & -6 + 9 - 4 & 8 - 12 + 4 \\ 0 - 2 + 0 & 0 + 3 - 1 & 0 - 4 + 1 \end{bmatrix} \)
\( = \begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix} \)

\( A^3 = A^2 \cdot A = \begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 9 - 8 + 0 & -9 + 12 - 4 & 12 - 16 + 4 \\ 0 - 2 + 0 & 0 + 3 + 0 & 0 - 4 + 0 \\ -6 + 4 + 0 & 6 - 6 + 3 & -8 + 8 - 3 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix} \)
\( = A^{-1} \) [from eq. (i)]

Hence, A³ = A^-1
In simple words: Multiplying the matrix by itself three times produces the same result as finding its inverse. This elegant property reveals a deep relationship between the cube and the inverse of this specific matrix.

Exam Tip: Verify this by independently computing A³ step by step, then comparing with the computed A^-1 - matching results confirms the proof.

Question 25. If A = \( \frac{1}{9}\begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix} \), show that A^-1 = A'.
Answer: We have \( A = \frac{1}{9}\begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix} = \begin{bmatrix} -\frac{8}{9} & \frac{1}{9} & \frac{4}{9} \\ \frac{4}{9} & \frac{4}{9} & \frac{7}{9} \\ \frac{1}{9} & -\frac{8}{9} & \frac{4}{9} \end{bmatrix} \)

To demonstrate: A^-1 = A'

First, we find the transpose of A, denoted as A'.

For any matrix, the transpose is obtained by swapping rows and columns:
\( \text{Transpose of } \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = \begin{bmatrix} a & d & g \\ b & e & h \\ c & f & i \end{bmatrix} \)

Here, \( A = \begin{bmatrix} -\frac{8}{9} & \frac{1}{9} & \frac{4}{9} \\ \frac{4}{9} & \frac{4}{9} & \frac{7}{9} \\ \frac{1}{9} & -\frac{8}{9} & \frac{4}{9} \end{bmatrix} \)

\( \text{So, } A' = \begin{bmatrix} -\frac{8}{9} & \frac{4}{9} & \frac{1}{9} \\ \frac{1}{9} & \frac{4}{9} & -\frac{8}{9} \\ \frac{4}{9} & \frac{7}{9} & \frac{4}{9} \end{bmatrix} \) ... (i)

Now, we need to find A^-1 and \( A^{-1} = \frac{\text{adj } A}{|A|} \)

Calculating |A|

Expanding |A| along C₁, we get
\( |A| = \left(-\frac{8}{9}\right) \begin{vmatrix} \frac{4}{9} & \frac{7}{9} \\ -\frac{8}{9} & \frac{4}{9} \end{vmatrix} - \left(\frac{4}{9}\right) \begin{vmatrix} \frac{1}{9} & \frac{4}{9} \\ -\frac{8}{9} & \frac{4}{9} \end{vmatrix} + \left(\frac{1}{9}\right) \begin{vmatrix} \frac{1}{9} & \frac{4}{9} \\ \frac{4}{9} & \frac{7}{9} \end{vmatrix} \)

\( = \frac{8}{9}\left[\frac{4}{9} \times \frac{9}{9} - \left(\frac{7}{9} \times \left(-\frac{8}{9}\right)\right)\right] - \frac{4}{9}\left[\frac{1}{9} \times \frac{9}{9} - \left(\frac{4}{9} \times \left(-\frac{8}{9}\right)\right)\right] + \frac{1}{9}\left[\frac{1}{9} \times \frac{7}{9} - \frac{4}{9} \times \frac{4}{9}\right] \)

\( = \frac{8}{9}\left[\frac{16 + 56}{81}\right] - \frac{4}{9}\left[\frac{4 + 32}{81}\right] + \frac{1}{9}\left[\frac{7 - 16}{81}\right] \)

\( = \frac{8}{9} \times \frac{72}{81} - \frac{4}{9} \times \frac{36}{81} + \frac{1}{9} \times \left(-\frac{9}{81}\right) \)

\( = \frac{8 \times 8}{81} - \frac{4 \times 4}{81} - \frac{1}{81} \)

\( = \frac{64 - 16 - 1}{81} = -1 \)

Now, we need to find adj A, and for that we must find cofactors:

\( a_{11} = \begin{vmatrix} \frac{4}{9} & \frac{7}{9} \\ -\frac{8}{9} & \frac{4}{9} \end{vmatrix} = \frac{4}{9} \times \frac{9}{9} - \left(\frac{7}{9} \times \left(-\frac{8}{9}\right)\right) = \frac{16 + 56}{81} = \frac{72}{81} = \frac{8}{9} \)

\( a_{12} = -\begin{vmatrix} \frac{4}{9} & \frac{7}{9} \\ \frac{1}{9} & \frac{4}{9} \end{vmatrix} = -\left[\frac{4}{9} \times \frac{9}{9} - \left(\frac{7}{9} \times \frac{1}{9}\right)\right] = -\left[\frac{16 - 7}{81}\right] = -\frac{9}{81} = -\frac{1}{9} \)

\( a_{13} = \begin{vmatrix} \frac{4}{9} & \frac{4}{9} \\ \frac{1}{9} & -\frac{8}{9} \end{vmatrix} = \frac{4}{9} \times \left(-\frac{8}{9}\right) - \left(\frac{4}{9} \times \frac{1}{9}\right) = \left[-\frac{32 + 4}{81}\right] = -\frac{36}{81} = -\frac{4}{9} \)

\( a_{21} = -\begin{vmatrix} \frac{1}{9} & \frac{4}{9} \\ -\frac{8}{9} & \frac{4}{9} \end{vmatrix} = -\left[\frac{1}{9} \times \frac{9}{9} - \left(\frac{4}{9} \times \left(-\frac{8}{9}\right)\right)\right] = -\left[\frac{4 + 32}{81}\right] = -\frac{36}{81} = -\frac{4}{9} \)

\( a_{22} = \begin{vmatrix} -\frac{8}{9} & \frac{4}{9} \\ \frac{1}{9} & \frac{4}{9} \end{vmatrix} = \left(-\frac{8}{9}\right) \times \frac{9}{9} - \left(\frac{4}{9} \times \frac{1}{9}\right) = \left[-\frac{32 - 4}{81}\right] = -\frac{36}{81} = -\frac{4}{9} \)

\( a_{23} = -\begin{vmatrix} -\frac{8}{9} & \frac{1}{9} \\ \frac{1}{9} & -\frac{8}{9} \end{vmatrix} = -\left[\left(-\frac{8}{9}\right) \times \left(-\frac{8}{9}\right) - \left(\frac{1}{9} \times \frac{1}{9}\right)\right] = -\left[\frac{64 - 1}{81}\right] = -\frac{63}{81} = -\frac{7}{9} \)

\( a_{31} = \begin{vmatrix} \frac{1}{9} & \frac{4}{9} \\ \frac{4}{9} & \frac{7}{9} \end{vmatrix} = \frac{1}{9} \times \frac{9}{9} - \left(\frac{4}{9} \times \frac{4}{9}\right) = \left[\frac{7 - 16}{81}\right] = -\frac{9}{81} = -\frac{1}{9} \)

\( a_{32} = -\begin{vmatrix} -\frac{8}{9} & \frac{4}{9} \\ \frac{4}{9} & \frac{7}{9} \end{vmatrix} = -\left[\left(-\frac{8}{9}\right) \times \frac{9}{9} - \left(\frac{4}{9} \times \frac{4}{9}\right)\right] = -\left[-\frac{56 + 16}{81}\right] = \frac{72}{81} = \frac{8}{9} \)

\( a_{33} = \begin{vmatrix} -\frac{8}{9} & \frac{1}{9} \\ \frac{4}{9} & \frac{4}{9} \end{vmatrix} = \left(-\frac{8}{9}\right) \times \frac{9}{9} - \left(\frac{1}{9} \times \frac{4}{9}\right) = \left[-\frac{32 - 4}{81}\right] = -\frac{36}{81} = -\frac{4}{9} \)

\( \therefore \text{adj } A = \begin{bmatrix} \frac{8}{9} & -\frac{1}{9} & -\frac{4}{9} \\ -\frac{4}{9} & -\frac{4}{9} & -\frac{7}{9} \\ -\frac{1}{9} & \frac{8}{9} & -\frac{4}{9} \end{bmatrix}^T = \begin{bmatrix} \frac{8}{9} & -\frac{4}{9} & -\frac{1}{9} \\ -\frac{1}{9} & -\frac{4}{9} & \frac{8}{9} \\ -\frac{4}{9} & -\frac{7}{9} & -\frac{4}{9} \end{bmatrix} \)

\( \therefore A^{-1} = \frac{\text{adj } A}{|A|} = \frac{\begin{bmatrix} \frac{8}{9} & -\frac{4}{9} & -\frac{1}{9} \\ -\frac{1}{9} & -\frac{4}{9} & \frac{8}{9} \\ -\frac{4}{9} & -\frac{7}{9} & -\frac{4}{9} \end{bmatrix}}{-1} = \begin{bmatrix} -\frac{8}{9} & \frac{4}{9} & \frac{1}{9} \\ \frac{1}{9} & \frac{4}{9} & -\frac{8}{9} \\ \frac{4}{9} & \frac{7}{9} & \frac{4}{9} \end{bmatrix} \)
\( = A' \) [from eq. (i)]

Hence, A^-1 = A'
In simple words: The inverse of this matrix is simply its transpose - a special property that occurs when the matrix preserves distances and angles, making it an orthogonal matrix (scaled appropriately).

Exam Tip: Recognize that when A^-1 = A', the matrix has special geometric properties - always compute both to verify this elegant relationship.

Question 26. Let D = diag [d₁, d₂, d₃], where none of d₁, d₂, d₃ is 0; prove that D^-1 = diag [d₁^-1, d₂^-1, d₃^-1].
Answer: Given: D = diag [d₁, d₂, d₃]

It is also given that d₁ ≠ 0, d₂ ≠ 0, d₃ ≠ 0

\( D = \begin{bmatrix} d_1 & 0 & 0 \\ 0 & d_2 & 0 \\ 0 & 0 & d_3 \end{bmatrix} \)

A diagonal matrix D = diag(d₁, d₂, ..., dₙ) has an inverse if and only if all its diagonal entries are nonzero, that is, dᵢ ≠ 0 for 1 ≤ i ≤ n. When D does have an inverse, then D^-1 = diag(d₁^-1, ..., dₙ^-1).

According to the Inverting Diagonal Matrices Theorem, which states that the inverse of a diagonal matrix is obtained by taking reciprocals of its diagonal entries.

Here, it is given that d₁ ≠ 0, d₂ ≠ 0, d₃ ≠ 0

∴ D is invertible

\( \implies D^{-1} = \text{diag} \left[d_1^{-1}, d_2^{-1}, d_3^{-1}\right] \)

Hence Proved.
In simple words: For a diagonal matrix where no entry on the diagonal is zero, the inverse is simply another diagonal matrix whose entries are the reciprocals of the original diagonal elements.

Exam Tip: Remember this shortcut - inverting a diagonal matrix is straightforward: just flip each nonzero diagonal entry to its reciprocal, keeping everything else zero.

Question 27. If A = \( \begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix} \) and B = \( \begin{bmatrix} 6 & 7 \\ 8 & 9 \end{bmatrix} \), verify that (AB)^-1 = B^-1A^-1.
Answer: Given: \( A = \begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix} \) & \( B = \begin{bmatrix} 6 & 7 \\ 8 & 9 \end{bmatrix} \)

To Verify: (AB)^-1 = B^-1A^-1

First, we find (AB)^-1.

Calculating AB

\( AB = \begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix}\begin{bmatrix} 6 & 7 \\ 8 & 9 \end{bmatrix} \)
\( = \begin{bmatrix} 18 + 16 & 21 + 18 \\ 42 + 40 & 49 + 45 \end{bmatrix} \)
\( = \begin{bmatrix} 34 & 39 \\ 82 & 94 \end{bmatrix} \)

We need to find (AB)^-1 and \( (AB)^{-1} = \frac{\text{adj } (AB)}{|AB|} \)

First, we find adj AB and to do that we must find cofactors:

\( a_{11} \text{ (cofactor of 34)} = (-1)^{1+1}(94) = (-1)^2(94) = 94 \)

\( a_{12} \text{ (cofactor of 39)} = (-1)^{1+2}(82) = (-1)^3(82) = -82 \)

\( a_{21} \text{ (cofactor of 82)} = (-1)^{2+1}(39) = (-1)^3(39) = -39 \)

\( a_{22} \text{ (cofactor of 94)} = (-1)^{2+2}(34) = (-1)^4(34) = 34 \)

∴ The cofactor matrix = \( \begin{bmatrix} 94 & -82 \\ -39 & 34 \end{bmatrix} \)

Now, adj AB = Transpose of cofactor matrix

∴ \( \text{adj } AB = \begin{bmatrix} 94 & -82 \\ -39 & 34 \end{bmatrix}^T = \begin{bmatrix} 94 & -39 \\ -82 & 34 \end{bmatrix} \)

Calculating |AB|

\( |AB| = \begin{vmatrix} 34 & 39 \\ 82 & 94 \end{vmatrix} \)

If \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), then determinant of A is given by

\( |A| = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \)

\( = [34 \times 94 - (82) \times (39)] \)
\( = (3196 - 3198) \)
\( = -2 \)

\( \therefore (AB)^{-1} = \frac{\text{adj } A}{|A|} = \frac{\begin{bmatrix} 94 & -39 \\ -82 & 34 \end{bmatrix}}{-2} = -\frac{1}{2}\begin{bmatrix} 94 & -39 \\ -82 & 34 \end{bmatrix} \)

Now, we find B^-1A^-1

Calculating B^-1

Here, \( B = \begin{bmatrix} 6 & 7 \\ 8 & 9 \end{bmatrix} \)

We need to find A^-1 and \( B^{-1} = \frac{\text{adj } B}{|B|} \)

First, we find adj B and to do that we must find cofactors:

\( a_{11} \text{ (cofactor of 6)} = (-1)^{1+1}(9) = (-1)^2(9) = 9 \)

\( a_{12} \text{ (cofactor of 7)} = (-1)^{1+2}(8) = (-1)^3(8) = -8 \)

\( a_{21} \text{ (cofactor of 8)} = (-1)^{2+1}(7) = (-1)^3(7) = -7 \)

\( a_{22} \text{ (cofactor of 9)} = (-1)^{2+2}(6) = (-1)^4(6) = 6 \)

Now, adj B = Transpose of cofactor matrix

\( \therefore \text{adj } B = \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}^T = \begin{bmatrix} 9 & -7 \\ -8 & 6 \end{bmatrix} \)

Calculating |B|

\( |B| = \begin{vmatrix} 6 & 7 \\ 8 & 9 \end{vmatrix} \)

If \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), then determinant of A is given by

\( |A| = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \)

\( = [6 \times 9 - 7 \times 8] \)
\( = (54 - 56) \)
\( = -2 \)

\( \therefore B^{-1} = \frac{\text{adj } B}{|B|} = \frac{\begin{bmatrix} 9 & -7 \\ -8 & 6 \end{bmatrix}}{-2} = -\frac{1}{2}\begin{bmatrix} 9 & -7 \\ -8 & 6 \end{bmatrix} \)

Calculating A^-1

Here, \( A = \begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix} \)

We need to find A^-1 and \( A^{-1} = \frac{\text{adj } A}{|A|} \)

First, we find adj A and to do that we must find cofactors:

\( a_{11} \text{ (cofactor of 3)} = (-1)^{1+1}(5) = (-1)^2(5) = 5 \)

\( a_{12} \text{ (cofactor of 2)} = (-1)^{1+2}(7) = (-1)^3(7) = -7 \)

\( a_{21} \text{ (cofactor of 7)} = (-1)^{2+1}(2) = (-1)^3(2) = -2 \)

\( a_{22} \text{ (cofactor of 5)} = (-1)^{2+2}(3) = (-1)^4(3) = 3 \)

∴ The cofactor matrix = \( \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} \)

Now, adj A = Transpose of cofactor matrix

\( \therefore \text{adj } A = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}^T = \begin{bmatrix} 5 & -2 \\ -7 & 3 \end{bmatrix} \)

Calculating |A|

\( |A| = \begin{vmatrix} 3 & 2 \\ 7 & 5 \end{vmatrix} \)

If \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), then determinant of A is given by

\( |A| = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \)

\( = [3 \times 5 - 2 \times 7] \)
\( = (15 - 14) \)
\( = 1 \)

\( \therefore A^{-1} = \frac{\text{adj } A}{|A|} = \frac{\begin{bmatrix} 5 & -2 \\ -7 & 3 \end{bmatrix}}{1} = \begin{bmatrix} 5 & -2 \\ -7 & 3 \end{bmatrix} \)

Calculating B^-1A^-1

Here, \( B^{-1} = -\frac{1}{2}\begin{bmatrix} 9 & -7 \\ -8 & 6 \end{bmatrix} \) & \( A^{-1} = \begin{bmatrix} 5 & -2 \\ -7 & 3 \end{bmatrix} \)

So,

\( B^{-1}A^{-1} = \left(-\frac{1}{2}\begin{bmatrix} 9 & -7 \\ -8 & 6 \end{bmatrix}\right)\left(\begin{bmatrix} 5 & -2 \\ -7 & 3 \end{bmatrix}\right) \)
\( = -\frac{1}{2}\begin{bmatrix} 45 + 49 & -18 - 21 \\ -40 - 42 & 16 + 18 \end{bmatrix} \)
\( = -\frac{1}{2}\begin{bmatrix} 94 & -39 \\ -82 & 34 \end{bmatrix} \)

So, we get
\( (AB)^{-1} = -\frac{1}{2}\begin{bmatrix} 94 & -39 \\ -82 & 34 \end{bmatrix} \) and \( B^{-1}A^{-1} = -\frac{1}{2}\begin{bmatrix} 94 & -39 \\ -82 & 34 \end{bmatrix} \)

∴ (AB)^-1 = B^-1A^-1
In simple words: The inverse of a product of two matrices equals the product of their inverses in reverse order - a fundamental property that simplifies many matrix computations.

Exam Tip: Always calculate both sides (AB)^-1 and B^-1A^-1 separately to verify - mismatched dimensions will immediately alert you to computational errors.

Question 28. If A = \( \begin{bmatrix} 9 & -1 \\ 6 & -2 \end{bmatrix} \) and B = \( \begin{bmatrix} -4 & 3 \\ 5 & -4 \end{bmatrix} \), verify that (AB)^-1 = B^-1A^-1.
Answer: Given: \( A = \begin{bmatrix} 9 & -1 \\ 6 & -2 \end{bmatrix} \) & \( B = \begin{bmatrix} -4 & 3 \\ 5 & -4 \end{bmatrix} \)

To Verify: (AB)^-1 = B^-1A^-1

First, we find (AB)^-1.

Calculating AB

\( AB = \begin{bmatrix} 9 & -1 \\ 6 & -2 \end{bmatrix}\begin{bmatrix} -4 & 3 \\ 5 & -4 \end{bmatrix} \)
\( = \begin{bmatrix} -36 - 5 & 27 + 4 \\ -24 - 10 & 18 + 8 \end{bmatrix} \)
\( = \begin{bmatrix} -41 & 31 \\ -34 & 26 \end{bmatrix} \)

We need to find (AB)^-1 and \( (AB)^{-1} = \frac{\text{adj } (AB)}{|AB|} \)

First, we find adj AB and to do that we must find cofactors:

\( a_{11} \text{ (cofactor of -41)} = (-1)^{1+1}(26) = (-1)^2(26) = 26 \)

\( a_{12} \text{ (cofactor of 31)} = (-1)^{1+2}(-34) = (-1)^3(-34) = 34 \)

\( a_{21} \text{ (cofactor of -34)} = (-1)^{2+1}(31) = (-1)^3(31) = -31 \)

\( a_{22} \text{ (cofactor of 26)} = (-1)^{2+2}(-41) = (-1)^4(-41) = -41 \)

∴ The cofactor matrix = \( \begin{bmatrix} 26 & 34 \\ -31 & -41 \end{bmatrix} \)

Now, adj AB = Transpose of cofactor matrix

\( \therefore \text{adj } AB = \begin{bmatrix} 26 & 34 \\ -31 & -41 \end{bmatrix}^T = \begin{bmatrix} 26 & -31 \\ 34 & -41 \end{bmatrix} \)

Calculating |AB|

\( |AB| = \begin{vmatrix} -41 & 31 \\ -34 & 26 \end{vmatrix} \)

If \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), then determinant of A is given by

\( |A| = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \)

\( = [-41 \times 26 - (-34) \times (31)] \)
\( = (-1066 + 1054) \)
\( = -12 \)

\( \therefore (AB)^{-1} = \frac{\text{adj } A}{|A|} = \frac{\begin{bmatrix} 26 & -31 \\ 34 & -41 \end{bmatrix}}{-12} = \frac{1}{12}\begin{bmatrix} 26 & -31 \\ 34 & -41 \end{bmatrix} \)

Now, we find B^-1A^-1

Calculating B^-1

Here, \( B = \begin{bmatrix} -4 & 3 \\ 5 & -4 \end{bmatrix} \)

We need to find A^-1 and \( B^{-1} = \frac{\text{adj } B}{|B|} \)

First, we find adj B and to do that we must find cofactors:

\( a_{11} \text{ (cofactor of -4)} = (-1)^{1+1}(-4) = (-1)^2(-4) = -4 \)

\( a_{12} \text{ (cofactor of 3)} = (-1)^{1+2}(5) = (-1)^3(5) = -5 \)

\( a_{21} \text{ (cofactor of 5)} = (-1)^{2+1}(3) = (-1)^3(3) = -3 \)

\( a_{22} \text{ (cofactor of -4)} = (-1)^{2+2}(-4) = (-1)^4(-4) = -4 \)

∴ The cofactor matrix = \( \begin{bmatrix} -4 & -5 \\ -3 & -4 \end{bmatrix} \)

Now, adj B = Transpose of cofactor matrix

\( \therefore \text{adj } B = \begin{bmatrix} -4 & -5 \\ -3 & -4 \end{bmatrix}^T = \begin{bmatrix} -4 & -3 \\ -5 & -4 \end{bmatrix} \)

Calculating |B|

\( |B| = \begin{vmatrix} -4 & 3 \\ 5 & -4 \end{vmatrix} \)

If \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), then determinant of A is given by

\( |A| = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \)

\( = [(-4) \times (-4) - 3 \times 5] \)
\( = (16 - 15) \)
\( = 1 \)

\( \therefore B^{-1} = \frac{\text{adj } B}{|B|} = \frac{\begin{bmatrix} -4 & -3 \\ -5 & -4 \end{bmatrix}}{1} = \begin{bmatrix} -4 & -3 \\ -5 & -4 \end{bmatrix} \)

Calculating A^-1

Here, \( A = \begin{bmatrix} 9 & -1 \\ 6 & -2 \end{bmatrix} \)

We need to find A^-1 and \( A^{-1} = \frac{\text{adj } A}{|A|} \)

First, we find adj A and to do that we must find cofactors:

\( a_{11} \text{ (cofactor of 9)} = (-1)^{1+1}(-2) = (-1)^2(-2) = -2 \)

\( a_{12} \text{ (cofactor of -1)} = (-1)^{1+2}(6) = (-1)^3(6) = -6 \)

\( a_{21} \text{ (cofactor of 6)} = (-1)^{2+1}(-1) = (-1)^3(-1) = 1 \)

\( a_{22} \text{ (cofactor of -2)} = (-1)^{2+2}(9) = (-1)^4(9) = 9 \)

∴ The cofactor matrix = \( \begin{bmatrix} -2 & -6 \\ 1 & 9 \end{bmatrix} \)

Now, adj A = Transpose of cofactor matrix

\( \therefore \text{adj } A = \begin{bmatrix} -2 & -6 \\ 1 & 9 \end{bmatrix}^T = \begin{bmatrix} -2 & 1 \\ -6 & 9 \end{bmatrix} \)

Calculating |A|

\( |A| = \begin{vmatrix} 9 & -1 \\ 6 & -2 \end{vmatrix} \)

If \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), then determinant of A is given by

\( |A| = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \)

\( = [9 \times (-2) - (-1) \times 6] \)
\( = (-18 + 6) \)
\( = -12 \)

\( \therefore A^{-1} = \frac{\text{adj } A}{|A|} = \frac{\begin{bmatrix} -2 & 1 \\ -6 & 9 \end{bmatrix}}{-12} = -\frac{1}{12}\begin{bmatrix} -2 & 1 \\ -6 & 9 \end{bmatrix} \)

Calculating B^-1A^-1

Here, \( B^{-1} = \begin{bmatrix} -4 & -3 \\ -5 & -4 \end{bmatrix} \) & \( A^{-1} = -\frac{1}{12}\begin{bmatrix} -2 & 1 \\ -6 & 9 \end{bmatrix} \)

So,

\( B^{-1}A^{-1} = \left(\begin{bmatrix} -4 & -3 \\ -5 & -4 \end{bmatrix}\right)\left(-\frac{1}{12}\begin{bmatrix} -2 & 1 \\ -6 & 9 \end{bmatrix}\right) \)
\( = -\frac{1}{12}\begin{bmatrix} 8 + 18 & -4 - 27 \\ 10 + 24 & -5 - 36 \end{bmatrix} \)
\( = -\frac{1}{12}\begin{bmatrix} 26 & -31 \\ 34 & -41 \end{bmatrix} \)

So, we get

\( (AB)^{-1} = \frac{1}{12}\begin{bmatrix} 26 & -31 \\ 34 & -41 \end{bmatrix} \) and \( B^{-1}A^{-1} = -\frac{1}{12}\begin{bmatrix} 26 & -31 \\ 34 & -41 \end{bmatrix} \)

Hmm, let me recalculate. Actually:

\( (AB)^{-1} = \frac{1}{-12}\begin{bmatrix} 26 & -31 \\ 34 & -41 \end{bmatrix} = -\frac{1}{12}\begin{bmatrix} 26 & -31 \\ 34 & -41 \end{bmatrix} \)

∴ (AB)^-1 = B^-1A^-1
In simple words: The inverse of a product of matrices always equals the product of their inverses computed in the reverse order - this property makes complex matrix manipulations much more manageable.

Exam Tip: Verify both computations independently and match them exactly - any small arithmetic error will cause a mismatch, immediately revealing calculation mistakes.

Question 29. Compute (AB)^-1 when A = \( \begin{bmatrix} 1 & 1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \) and B^-1 = \( \begin{bmatrix} 1 & 2 & 0 \\ 0 & 3 & -1 \\ 1 & 0 & 2 \end{bmatrix} \).
Answer: We have, \( A = \begin{bmatrix} 1 & 1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \)

To find: (AB)^-1

We know that,
(AB)^-1 = B^-1A^-1

and here, B^-1 is given but we need to find A^-1 and \( A^{-1} = \frac{\text{adj } A}{|A|} \)

First, we find |A|

Expanding |A| along C₁, we get
\( |A| = a_{11} (-1)^{1+1} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} + a_{21} (-1)^{2+1} \begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix} + a_{31} (-1)^{3+1} \begin{vmatrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{vmatrix} \)

\( |A| = (1) \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} - (0) \begin{vmatrix} 1 & 2 \\ -2 & 4 \end{vmatrix} + (3) \begin{vmatrix} 1 & 2 \\ 2 & -3 \end{vmatrix} \)
\( = 1(8 - 6) - 0 + 3(-3 - 4) \)
\( = 1(2) + 3(-7) \)
\( = 2 - 21 \)
\( = -19 \)

Now, we need to find adj A and to do that we must find cofactors:

\( a_{11} = \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} = 8 - 6 = 2 \)

\( a_{12} = -\begin{vmatrix} 0 & -3 \\ 3 & 4 \end{vmatrix} = -(0 + 9) = -9 \)

\( a_{13} = \begin{vmatrix} 0 & 2 \\ 3 & -2 \end{vmatrix} = 0 - 6 = -6 \)

\( a_{21} = -\begin{vmatrix} 1 & 2 \\ -2 & 4 \end{vmatrix} = -(4 + 4) = -8 \)

\( a_{22} = \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = 4 - 6 = -2 \)

\( a_{23} = -\begin{vmatrix} 1 & 1 \\ 3 & -2 \end{vmatrix} = -(-2 - 3) = 5 \)

\( a_{31} = \begin{vmatrix} 1 & 2 \\ 2 & -3 \end{vmatrix} = -3 - 4 = -7 \)

\( a_{32} = -\begin{vmatrix} 1 & 2 \\ 0 & -3 \end{vmatrix} = -(-3 - 0) = 3 \)

\( a_{33} = \begin{vmatrix} 1 & 1 \\ 0 & 2 \end{vmatrix} = 2 - 0 = 2 \)

\( \therefore \text{adj } A = \begin{bmatrix} 2 & -9 & -6 \\ -8 & -2 & 5 \\ -7 & 3 & 2 \end{bmatrix}^T = \begin{bmatrix} 2 & -8 & -7 \\ -9 & -2 & 3 \\ -6 & 5 & 2 \end{bmatrix} \)

\( \therefore A^{-1} = \frac{\text{adj } A}{|A|} = \frac{\begin{bmatrix} 2 & -8 & -7 \\ -9 & -2 & 3 \\ -6 & 5 & 2 \end{bmatrix}}{-19} = \frac{1}{19}\begin{bmatrix} -2 & 8 & 7 \\ 9 & 2 & -3 \\ 6 & -5 & -2 \end{bmatrix} \)

Now, we have
\( B^{-1} = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 3 & -1 \\ 1 & 0 & 2 \end{bmatrix} \) & \( A^{-1} = \frac{1}{19}\begin{bmatrix} -2 & 8 & 7 \\ 9 & 2 & -3 \\ 6 & -5 & -2 \end{bmatrix} \)

So,
\( (AB)^{-1} = B^{-1}A^{-1} = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 3 & -1 \\ 1 & 0 & 2 \end{bmatrix}\left(\frac{1}{19}\begin{bmatrix} -2 & 8 & 7 \\ 9 & 2 & -3 \\ 6 & -5 & -2 \end{bmatrix}\right) \)

\( = \frac{1}{19}\begin{bmatrix} -2 + 18 + 0 & 8 + 4 + 0 & 7 - 6 + 0 \\ 0 + 27 - 6 & 0 + 6 + 5 & 0 - 9 + 2 \\ -2 + 0 + 12 & 8 + 0 - 10 & 7 + 0 - 4 \end{bmatrix} \)

\( = \frac{1}{19}\begin{bmatrix} 16 & 12 & 1 \\ 21 & 11 & -7 \\ 10 & -2 & 3 \end{bmatrix} \)

Ans. \( (AB)^{-1} = \frac{1}{19}\begin{bmatrix} 16 & 12 & 1 \\ 21 & 11 & -7 \\ 10 & -2 & 3 \end{bmatrix} \)
In simple words: Since B^-1 is already provided, we only computed A^-1 and then multiplied them in the correct order to get the final result - this saved considerable computation time.

Exam Tip: When B^-1 is directly provided, always use it immediately - never waste effort computing B then inverting it, which would be redundant and time-consuming.

Question 30. Obtain the inverses of the matrices \( \begin{bmatrix} 1 & p & 0 \\ 0 & 1 & p \\ 0 & 0 & 1 \end{bmatrix} \) and \( \begin{bmatrix} 1 & 0 & 0 \\ q & 1 & 0 \\ 0 & q & 1 \end{bmatrix} \). And, hence find the inverse of the matrix \( \begin{bmatrix} 1 + pq & p & 0 \\ q & 1 + pq & p \\ 0 & q & 1 \end{bmatrix} \).
Answer: Let \( A = \begin{bmatrix} 1 & p & 0 \\ 0 & 1 & p \\ 0 & 0 & 1 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 0 & 0 \\ q & 1 & 0 \\ 0 & q & 1 \end{bmatrix} \) & \( C = \begin{bmatrix} 1 + pq & p & 0 \\ q & 1 + pq & p \\ 0 & q & 1 \end{bmatrix} \)

To find: A^-1, B^-1 and C^-1

Calculating A^-1

We have, \( A = \begin{bmatrix} 1 & p & 0 \\ 0 & 1 & p \\ 0 & 0 & 1 \end{bmatrix} \)

We need to find A^-1 and \( A^{-1} = \frac{\text{adj } A}{|A|} \)

First, we find |A|

Expanding |A| along C₁, we get
\( |A| = a_{11} (-1)^{1+1} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} + a_{21} (-1)^{2+1} \begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix} + a_{31} (-1)^{3+1} \begin{vmatrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{vmatrix} \)

\( |A| = (1) \begin{vmatrix} 1 & p \\ 0 & 1 \end{vmatrix} - (0) + (0) \)
\( = 1(1 - 0) \)
\( = 1 \)

Now, we need to find adj A and to do that we must find cofactors:

\( a_{11} = \begin{vmatrix} 1 & p \\ 0 & 1 \end{vmatrix} = 1 - 0 = 1 \)

\( a_{12} = -\begin{vmatrix} 0 & p \\ 0 & 1 \end{vmatrix} = 0 \)

\( a_{13} = \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} = 0 \)

\( a_{21} = -\begin{vmatrix} p & 0 \\ 0 & 1 \end{vmatrix} = -(p - 0) = -p \)

\( a_{22} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1 - 0 = 1 \)

\( a_{23} = -\begin{vmatrix} 1 & p \\ 0 & 0 \end{vmatrix} = -(0) = 0 \)

\( a_{31} = \begin{vmatrix} p & 0 \\ 1 & p \end{vmatrix} = p^2 - 0 = p^2 \)

\( a_{32} = -\begin{vmatrix} 1 & 0 \\ 0 & p \end{vmatrix} = -(p - 0) = -p \)

\( a_{33} = \begin{vmatrix} 1 & p \\ 0 & 1 \end{vmatrix} = 1 \)

\( \therefore \text{adj } A = \begin{bmatrix} 1 & 0 & 0 \\ -p & 1 & 0 \\ p^2 & -p & 1 \end{bmatrix}^T = \begin{bmatrix} 1 & -p & p^2 \\ 0 & 1 & -p \\ 0 & 0 & 1 \end{bmatrix} \)

\( \therefore A^{-1} = \frac{\text{adj } A}{|A|} = \frac{\begin{bmatrix} 1 & -p & p^2 \\ 0 & 1 & -p \\ 0 & 0 & 1 \end{bmatrix}}{1} = \begin{bmatrix} 1 & -p & p^2 \\ 0 & 1 & -p \\ 0 & 0 & 1 \end{bmatrix} \) ... (i)

Calculating B^-1

We have, \( B = \begin{bmatrix} 1 & 0 & 0 \\ q & 1 & 0 \\ 0 & q & 1 \end{bmatrix} \)

We need to find B^-1 and \( B^{-1} = \frac{\text{adj } B}{|B|} \)

First, we find |B|

Expanding |B| along C₁, we get
\( |B| = a_{11} (-1)^{1+1} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} + a_{21} (-1)^{2+1} \begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix} + a_{31} (-1)^{3+1} \begin{vmatrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{vmatrix} \)

\( |B| = (1) \begin{vmatrix} 1 & 0 \\ q & 1 \end{vmatrix} - (q) \begin{vmatrix} 0 & 0 \\ q & 1 \end{vmatrix} + (0) \)
\( = 1(1 - 0) - q(0) + 0 \)
\( = 1(1 - 0) \)
\( = 1 \)

Now, we need to find adj B and to do that we must find cofactors:

\( a_{11} = \begin{vmatrix} 1 & 0 \\ q & 1 \end{vmatrix} = 1 - 0 = 1 \)

\( a_{12} = -\begin{vmatrix} q & 0 \\ 0 & 1 \end{vmatrix} = 0 \)

\( a_{13} = \begin{vmatrix} q & 1 \\ 0 & q \end{vmatrix} = 0 \)

\( a_{21} = -\begin{vmatrix} 0 & 0 \\ q & 1 \end{vmatrix} = -(0 - 0) = 0 \)

\( a_{22} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1 - 0 = 1 \)

\( a_{23} = -\begin{vmatrix} 1 & 0 \\ 0 & q \end{vmatrix} = -(0) = 0 \)

\( a_{31} = \begin{vmatrix} 0 & 0 \\ 1 & 0 \end{vmatrix} = 0 - 0 = 0 \)

\( a_{32} = -\begin{vmatrix} 1 & 0 \\ q & 0 \end{vmatrix} = -(0 - 0) = 0 \)

\( a_{33} = \begin{vmatrix} 1 & 0 \\ q & 1 \end{vmatrix} = 1 - 0 = 1 \)

\( \therefore \text{adj } B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}^T = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

\( \therefore B^{-1} = \frac{\text{adj } B}{|B|} = \frac{\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}}{1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

Wait, this is incorrect. Let me recalculate B^-1 more carefully. Actually, the cofactors are:

\( a_{21} = -\begin{vmatrix} 0 & 0 \\ q & 1 \end{vmatrix} = -(0 \cdot 1 - 0 \cdot q) = 0 \)

Hmm, I need to be more careful. Let me redo this. For a lower triangular matrix, we can see the pattern more directly. But let me compute the cofactors correctly by using the standard formula. Actually, the inverse of the lower triangular matrix is:

\( B^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -q & 1 & 0 \\ q^2 & -q & 1 \end{bmatrix} \)

Actually, let me verify more carefully. We have \( B = \begin{bmatrix} 1 & 0 & 0 \\ q & 1 & 0 \\ 0 & q & 1 \end{bmatrix} \)

The cofactors should be (recalculating):
\( a_{21} = -\begin{vmatrix} 0 & 0 \\ q & 1 \end{vmatrix} = -(0 - 0) = 0 \)
\( a_{31} = -\begin{vmatrix} 0 & 0 \\ 1 & 0 \end{vmatrix} = -(0 - 0) = 0 \)
\( a_{32} = -\begin{vmatrix} 1 & 0 \\ q & 0 \end{vmatrix} = -(0 - 0) = 0 \)
\( a_{12} = -\begin{vmatrix} q & 0 \\ 0 & 1 \end{vmatrix} = -(q \cdot 1 - 0 \cdot 0) = -q \)

So the cofactor matrix should be:\br />\( \begin{bmatrix} 1 & -q & 0 \\ 0 & 1 & -q \\ 0 & 0 & 1 \end{bmatrix} \)

Transposing this:\br />\( \text{adj } B = \begin{bmatrix} 1 & 0 & 0 \\ -q & 1 & 0 \\ 0 & -q & 1 \end{bmatrix} \)

\( \therefore B^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -q & 1 & 0 \\ 0 & -q & 1 \end{bmatrix} \)

Now, we have
\( B^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -q & 1 & 0 \\ 0 & -q & 1 \end{bmatrix} \) & \( A^{-1} = \begin{bmatrix} 1 & -p & p^2 \\ 0 & 1 & -p \\ 0 & 0 & 1 \end{bmatrix} \)

So,
\( (AB)^{-1} = B^{-1}A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -q & 1 & 0 \\ 0 & -q & 1 \end{bmatrix}\left(\begin{bmatrix} 1 & -p & p^2 \\ 0 & 1 & -p \\ 0 & 0 & 1 \end{bmatrix}\right) \)

\( = \begin{bmatrix} 1 & -p & p^2 \\ -q + 0 + 0 & q \cdot p + 1 + 0 & -q \cdot p^2 - p + 0 \\ 0 + 0 + 0 & 0 + q \cdot 0 + 0 & 0 - q \cdot (-p) + 1 \end{bmatrix} \)

\( = \begin{bmatrix} 1 & -p & p^2 \\ -q & qp + 1 & -qp^2 - p \\ 0 & 0 & qp + 1 \end{bmatrix} \)

Wait, I need to be more careful. Let me compute this product more carefully:

\( B^{-1}A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -q & 1 & 0 \\ 0 & -q & 1 \end{bmatrix}\begin{bmatrix} 1 & -p & p^2 \\ 0 & 1 & -p \\ 0 & 0 & 1 \end{bmatrix} \)

Row 1: \( [1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0, 1 \cdot (-p) + 0 \cdot 1 + 0 \cdot 0, 1 \cdot p^2 + 0 \cdot (-p) + 0 \cdot 1] = [1, -p, p^2] \)
Row 2: \( [-q \cdot 1 + 1 \cdot 0 + 0 \cdot 0, -q \cdot (-p) + 1 \cdot 1 + 0 \cdot 0, -q \cdot p^2 + 1 \cdot (-p) + 0 \cdot 1] = [-q, qp + 1, -qp^2 - p] \)
Row 3: \( [0 \cdot 1 + (-q) \cdot 0 + 1 \cdot 0, 0 \cdot (-p) + (-q) \cdot 1 + 1 \cdot 0, 0 \cdot p^2 + (-q) \cdot (-p) + 1 \cdot 1] = [0, -q, qp + 1] \)

\( B^{-1}A^{-1} = \begin{bmatrix} 1 & -p & p^2 \\ -q & 1 + pq & -p - qp^2 \\ 0 & -q & 1 + qp \end{bmatrix} \)

Since C = AB and \( C^{-1} = B^{-1}A^{-1} \), we have

\( C^{-1} = \begin{bmatrix} 1 & -p & p^2 \\ -q & 1 + pq & -p - qp^2 \\ 0 & -q & 1 + qp \end{bmatrix} \)

Ans. \( C^{-1} = \begin{bmatrix} 1 & -p & p^2 \\ -q & 1 + pq & -p(1 + qp) \\ 0 & -q & 1 + pq \end{bmatrix} \) or \( C^{-1} = \begin{bmatrix} 1 & -p & p^2 \\ -q & 1 + pq & -(p + pqp) \\ 0 & -q & 1 + pq \end{bmatrix} \)

Note that \( -p - qp^2 = -p(1 + qp) \)
In simple words: By computing the inverses of the two simpler matrices and then multiplying them in reverse order, we obtaine the inverse of their product - this approach decomposes a complicated inversion problem into manageable pieces.

Exam Tip: When a matrix can be expressed as a product of simpler matrices, always compute their individual inverses first - the result is often cleaner and faster than direct computation.

Question 31. If A = \( \begin{bmatrix} 3 & 2 \\ 2 & 1 \end{bmatrix} \), verify that A² - 4A - I = O, and hence find A⁻¹.
Answer: Given: \( A = \begin{bmatrix} 3 & 2 \\ 2 & 1 \end{bmatrix} \)
To verify: A² - 4A - I = 0
First, we calculate A²
\( A^2 = A.A = \begin{bmatrix} 3 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 9+4 & 6+2 \\ 6+2 & 4+1 \end{bmatrix} = \begin{bmatrix} 13 & 8 \\ 8 & 5 \end{bmatrix} \)
Taking the LHS of the given equation, which is A² - 4A - I
\( \Rightarrow \begin{bmatrix} 13 & 8 \\ 8 & 5 \end{bmatrix} - 4\begin{bmatrix} 3 & 2 \\ 2 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 13 & 8 \\ 8 & 5 \end{bmatrix} - \begin{bmatrix} 12 & 8 \\ 8 & 4 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 13-12-1 & 8-8-0 \\ 8-8-0 & 5-4-1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
= 0 = RHS
∴ LHS = RHS, hence verified.
Now, we need to find A⁻¹
Finding A⁻¹ using the given equation
A² - 4A - I = O
Post multiplying both sides by A⁻¹, we get
(A² - 4A - I)A⁻¹ = OA⁻¹
\( \Rightarrow A^2.A^{-1} - 4A.A^{-1} - I.A^{-1} = O [OA^{-1} = O] \)
\( \Rightarrow A.(AA^{-1}) - 4I - A^{-1} = O [AA^{-1} = I] \)
\( \Rightarrow A(I) - 4I - A^{-1} = O \)
\( \Rightarrow A - 4I - A^{-1} = O \)
\( \Rightarrow A - 4I - O = A^{-1} \)
\( \Rightarrow A - 4I = A^{-1} \)
\( \Rightarrow A^{-1} = \begin{bmatrix} 3 & 2 \\ 2 & 1 \end{bmatrix} - 4\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( \Rightarrow A^{-1} = \begin{bmatrix} 3 & 2 \\ 2 & 1 \end{bmatrix} - \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} \)
\( \Rightarrow A^{-1} = \begin{bmatrix} 3-4 & 2-0 \\ 2-0 & 1-4 \end{bmatrix} = \begin{bmatrix} -1 & 2 \\ 2 & -3 \end{bmatrix} \)
In simple words: We first squared the given matrix and then substituted it into the given equation to confirm both sides equal zero. Once verified, we rearranged the equation to isolate the inverse by multiplying both sides by A⁻¹, which gave us a simple formula to calculate it.

Exam Tip: Always verify the given equation completely before deriving the inverse. The algebraic rearrangement to isolate A⁻¹ is straightforward once the equation is confirmed.

Question 32. Show that the matrix A = \( \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} \) satisfies the equation x² + 4x - 42 = 0 and hence find A⁻¹.
Answer: Given: \( A = \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} \)
To show: Matrix A satisfies the equation x² + 4x - 42 = 0
If matrix A satisfies the given equation, then A² + 4A - 42 = 0
First, we calculate A²
\( A^2 = A.A = \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} 64+10 & -40+20 \\ -16+8 & 10+16 \end{bmatrix} = \begin{bmatrix} 74 & -20 \\ -8 & 26 \end{bmatrix} \)
Taking the LHS of the given equation, which is A² + 4A - 42
\( \Rightarrow \begin{bmatrix} 74 & -20 \\ -8 & 26 \end{bmatrix} + 4\begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} - 42\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 74 & -20 \\ -8 & 26 \end{bmatrix} + \begin{bmatrix} -32 & 20 \\ 8 & 16 \end{bmatrix} - \begin{bmatrix} 42 & 0 \\ 0 & 42 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 74-32-42 & -20+20-0 \\ -8+8-0 & 26+16-42 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
= 0 = RHS
∴ LHS = RHS, hence matrix A satisfies the equation x² + 4x - 42 = 0
Now, we need to find A⁻¹
Finding A⁻¹ using the given equation
A² + 4A - 42 = O
Post multiplying both sides by A⁻¹, we get
(A² + 4A - 42)A⁻¹ = OA⁻¹
\( \Rightarrow A^2.A^{-1} + 4A.A^{-1} - 42.A^{-1} = O [OA^{-1} = O] \)
\( \Rightarrow A.(AA^{-1}) + 4I - 42A^{-1} = O [AA^{-1} = I] \)
\( \Rightarrow A(I) + 4I - 42A^{-1} = O \)
\( \Rightarrow A + 4I - 42A^{-1} = O \)
\( \Rightarrow A + 4I - O = 42A^{-1} \)
\( \Rightarrow A^{-1} = \frac{1}{42}(A + 4I) \)
\( \Rightarrow A^{-1} = \frac{1}{42}\left(\begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} + 4\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\right) \)
\( \Rightarrow A^{-1} = \frac{1}{42}\left(\begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} + \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}\right) \)
\( \Rightarrow A^{-1} = \frac{1}{42}\begin{bmatrix} -8+4 & 5+0 \\ 2+0 & 4+4 \end{bmatrix} = \frac{1}{42}\begin{bmatrix} -4 & 5 \\ 2 & 8 \end{bmatrix} \)
\( \Rightarrow A^{-1} = \begin{bmatrix} -\frac{1}{21} & \frac{5}{42} \\ \frac{1}{21} & \frac{4}{21} \end{bmatrix} \)
In simple words: We calculated A squared and substituted it along with A into the given equation to confirm it equals zero. Then we rearranged the equation and multiplied by A⁻¹ to isolate the inverse term, which we computed using simple matrix arithmetic.

Exam Tip: When deriving the inverse from a matrix equation, isolate the inverse term carefully by grouping all other terms on one side before multiplying by the scalar coefficient.

Question 33. If A = \( \begin{bmatrix} 1 & 1 \\ 5 & 7 \end{bmatrix} \), show that A² + 3A + 4I2 = O and hence find A⁻¹.
Answer: Given: \( A = \begin{bmatrix} 1 & 1 \\ 5 & 7 \end{bmatrix} \)
To verify: A² + 3A + 4I = 0
First, we calculate A²
\( A^2 = A.A = \begin{bmatrix} 1 & 1 \\ 5 & 7 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 5 & 7 \end{bmatrix} \)
\( = \begin{bmatrix} 1+5 & 1+7 \\ 5+35 & 5+49 \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ 40 & 54 \end{bmatrix} \)

Taking the LHS of the given equation
\( \Rightarrow \begin{bmatrix} -1 & 3 \\ -6 & 2 \end{bmatrix} + 3\begin{bmatrix} -1 & -1 \\ 2 & -2 \end{bmatrix} + 4\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} -1 & 3 \\ -6 & 2 \end{bmatrix} + \begin{bmatrix} -3 & -3 \\ 6 & -6 \end{bmatrix} + \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} -1-3+4 & 3-3+0 \\ -6+6+0 & 2-6+4 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
= 0 = RHS
∴ LHS = RHS, hence verified.
Now, we need to find A⁻¹
Finding A⁻¹ using the given equation
A² + 3A + 4I = O
Post multiplying both sides by A⁻¹, we get
(A² + 3A + 4I)A⁻¹ = OA⁻¹
\( \Rightarrow A^2.A^{-1} + 3A.A^{-1} + 4I.A^{-1} = O [OA^{-1} = O] \)
\( \Rightarrow A.(AA^{-1}) + 3I + 4A^{-1} = O [AA^{-1} = I] \)
\( \Rightarrow A(I) + 3I + 4A^{-1} = O \)
\( \Rightarrow A + 3I + 4A^{-1} = O \)
\( \Rightarrow 4A^{-1} = -A - 3I + O \)
\( \Rightarrow A^{-1} = \frac{1}{4}(-A - 3I) \)
\( \Rightarrow A^{-1} = \frac{1}{4}\left(-\begin{bmatrix} -1 & -1 \\ 2 & -2 \end{bmatrix} - 3\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\right) \)
\( \Rightarrow A^{-1} = \frac{1}{4}\left(\begin{bmatrix} 1 & 1 \\ -2 & 2 \end{bmatrix} + \begin{bmatrix} -3 & 0 \\ 0 & -3 \end{bmatrix}\right) \)
\( \Rightarrow A^{-1} = \frac{1}{4}\begin{bmatrix} 1-3 & 1+0 \\ -2+0 & 2-3 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} -2 & 1 \\ -2 & -1 \end{bmatrix} \)
\( \Rightarrow A^{-1} = \begin{bmatrix} -\frac{1}{2} & \frac{1}{4} \\ -\frac{1}{2} & -\frac{1}{4} \end{bmatrix} \)
In simple words: We computed A squared and verified that the given equation holds by substituting A² and A into it. After confirmation, we rearranged to isolate the inverse term and calculated it using basic matrix operations.

Exam Tip: Double-check matrix arithmetic during squaring and addition steps - a single calculation error will invalidate the entire verification. Keep track of which terms contribute to which matrix element.

Question 34. If A = \( \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \), find x and y such that A² + xI = yA. Hence, find A⁻¹. [CBSE 2005]
Answer: Given: \( A = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \)
To find: values of x and y
Given equation: A² + xI = yA
First, we calculate A²
\( A^2 = A.A = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \)
\( = \begin{bmatrix} 9+7 & 3+5 \\ 21+35 & 7+25 \end{bmatrix} = \begin{bmatrix} 16 & 8 \\ 56 & 32 \end{bmatrix} \)
Putting the values in the given equation
A² + xI = yA
\( \Rightarrow \begin{bmatrix} 16 & 8 \\ 56 & 32 \end{bmatrix} + x\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = y\begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 16+x & 8 \\ 56 & 32+x \end{bmatrix} = \begin{bmatrix} 3y & y \\ 7y & 5y \end{bmatrix} \)
On comparing, we get
16 + x = 3y ...(i)
y = 8 ...(ii)
56 = 7y ...(iii)
32 + x = 5y ...(iv)
Putting the value of y = 8 in equation (i)
16 + x = 3(8)
\( \Rightarrow \) 16 + x = 24
\( \Rightarrow \) x = 8
Hence, the values are x = 8 and y = 8
The given equation becomes A² + 8I = 8A
Now, we need to find A⁻¹
Finding A⁻¹ using the given equation
A² + 8I = 8A
Post multiplying both sides by A⁻¹, we get
(A² + 8I)A⁻¹ = 8AA⁻¹
\( \Rightarrow A^2.A^{-1} + 8I.A^{-1} = 8AA^{-1} \)
\( \Rightarrow A.(AA^{-1}) + 8A^{-1} = 8I [AA^{-1} = I] \)
\( \Rightarrow A(I) + 8A^{-1} = 8I \)
\( \Rightarrow A + 8A^{-1} = 8I \)
\( \Rightarrow 8A^{-1} = -A + 8I \)
\( \Rightarrow A^{-1} = \frac{1}{8}(-A + 8I) \)
\( \Rightarrow A^{-1} = \frac{1}{8}\left(-\begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} + 8\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\right) \)
\( \Rightarrow A^{-1} = \frac{1}{8}\left(\begin{bmatrix} -3 & -1 \\ -7 & -5 \end{bmatrix} + \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix}\right) \)
\( \Rightarrow A^{-1} = \frac{1}{8}\begin{bmatrix} -3+8 & -1+0 \\ -7+0 & -5+8 \end{bmatrix} = \frac{1}{8}\begin{bmatrix} 5 & -1 \\ -7 & 3 \end{bmatrix} \)
\( \Rightarrow A^{-1} = \begin{bmatrix} \frac{5}{8} & -\frac{1}{8} \\ -\frac{7}{8} & \frac{3}{8} \end{bmatrix} \)
In simple words: We squared the matrix and substituted A² and A into the given equation, then compared corresponding elements to solve for x and y. Once we had these values, we rearranged the equation to isolate A⁻¹ and calculated it.

Exam Tip: When comparing matrices element by element, ensure you list all four equations - some may appear redundant but verify your values satisfy all of them to catch calculation errors early.

Question 35. If A = \( \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \), find the value of λ so that A² = λA - 2I. Hence, find A⁻¹. [CBSE 2007]
Answer: Given: \( A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \)
To find: value of λ
Given equation: A² = λA - 2I
First, we calculate A²
\( A^2 = A.A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \)
\( = \begin{bmatrix} 9-8 & -6+4 \\ 12-8 & -8+4 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} \)
Putting the values in the given equation
A² = λA - 2I
\( \Rightarrow \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = λ\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} - 2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3λ & -2λ \\ 4λ & -2λ \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3λ-2 & -2λ \\ 4λ & -2λ-2 \end{bmatrix} \)
On comparing, we get
3λ - 2 = 1 ...(i)
-2λ = -2 ...(ii)
4λ = 4 ...(iii)
-2λ - 2 = -4 ...(iv)
Solving equation (iii)
4λ = 4
\( \Rightarrow \) λ = 1
Hence, the value of λ = 1
The given equation becomes A² = A - 2I
Now, we need to find A⁻¹
Finding A⁻¹ using the given equation
A² = A - 2I
Post multiplying both sides by A⁻¹, we get
(A²)A⁻¹ = (A - 2I)A⁻¹
\( \Rightarrow A^2.A^{-1} = AA^{-1} - 2IA^{-1} \)
\( \Rightarrow A.(AA^{-1}) = I - 2A^{-1} [AA^{-1} = I] \)
\( \Rightarrow A(I) = I - 2A^{-1} \)
\( \Rightarrow A + 2A^{-1} = I \)
\( \Rightarrow 2A^{-1} = -A + I \)
\( \Rightarrow A^{-1} = \frac{1}{2}(-A + I) \)
\( \Rightarrow A^{-1} = \frac{1}{2}\left(-\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\right) \)
\( \Rightarrow A^{-1} = \frac{1}{2}\begin{bmatrix} -3+1 & 2+0 \\ -4+0 & 2+1 \end{bmatrix} = \frac{1}{2}\begin{bmatrix} -2 & 2 \\ -4 & 3 \end{bmatrix} \)
\( \Rightarrow A^{-1} = \begin{bmatrix} -1 & 1 \\ -2 & \frac{3}{2} \end{bmatrix} \)
In simple words: We calculated A squared and compared it element by element with the expression λA - 2I to solve for λ. Once λ was found, we rearranged the equation by multiplying both sides by A⁻¹ to isolate the inverse and compute it using matrix subtraction and scalar division.

Exam Tip: When solving for a scalar parameter like λ, use the simplest equation available (the one with fewest variables in its solution) to avoid fractional arithmetic. Always verify your answer in all four comparison equations.

Question 36. Show that the matrix A = \( \begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix} \) satisfies the equation A³ - A² - 3A - I = O, and hence find A⁻¹.
Answer: Given: \( A = \begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix} \)
We need to show that matrix A satisfies the equation A³ - A² - 3A - I = O
First, we calculate A²
\( A^2 = A.A = \begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1+0-6 & 0+0-8 & -2+0-2 \\ -2+2+6 & 0+1+8 & 4-1+2 \\ 3-8+3 & 0-4+4 & -6+4+1 \end{bmatrix} = \begin{bmatrix} -5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3 \end{bmatrix} \)
Now, we calculate A³
\( A^3 = A^2.A = \begin{bmatrix} -5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} -5+16-12 & 0+8-16 & 10-16-4 \\ 6-18+12 & 0-9+16 & -12-18+4 \\ -2+0+9 & 0+0+12 & 4+0+3 \end{bmatrix} \)
\( = \begin{bmatrix} -1 & -8 & -10 \\ 0 & 7 & -10 \\ 7 & 12 & 7 \end{bmatrix} \)
Taking LHS of the given equation, which is A³ - A² - 3A - I
\( = \begin{bmatrix} -1 & -8 & -10 \\ 0 & 7 & -10 \\ 7 & 12 & 7 \end{bmatrix} - \begin{bmatrix} -5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3 \end{bmatrix} - 3\begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} -1+5 & -8+8 & -10+4 \\ 0-6 & 7-9 & -10-4 \\ 7+2 & 12-0 & 7-3 \end{bmatrix} - \begin{bmatrix} 3 & 0 & -6 \\ -6 & -3 & 6 \\ 9 & 12 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 4 & 0 & -6 \\ -6 & -2 & -14 \\ 9 & 12 & 4 \end{bmatrix} - \begin{bmatrix} 3 & 0 & -6 \\ -6 & -3 & 6 \\ 9 & 12 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 4-3-1 & 0-0-0 & -6-(-6)-0 \\ -6-(-6)-0 & -2-(-3)-1 & -14-6-0 \\ 9-9-0 & 12-12-0 & 4-3-1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)
= O = RHS
∴ LHS = RHS
Hence, the given matrix A satisfies the equation A³ - A² - 3A - I = O
Now, we need to find A⁻¹
Finding A⁻¹ using the given equation
A³ - A² - 3A - I = O
Post multiplying both sides by A⁻¹, we get
(A³ - A² - 3A - I)A⁻¹ = OA⁻¹
\( \Rightarrow A^3.A^{-1} - A^2.A^{-1} - 3A.A^{-1} - I.A^{-1} = O [OA^{-1} = O] \)
\( \Rightarrow A^2.(AA^{-1}) - A.(AA^{-1}) - 3I - A^{-1} = O \)
\( \Rightarrow A^2(I) - A(I) - 3I - A^{-1} = O [AA^{-1} = I] \)
\( \Rightarrow A^2 - A - 3I - A^{-1} = O \)
\( \Rightarrow O + A^{-1} = A^2 - A - 3I \)
\( \Rightarrow A^{-1} = A^2 - A - 3I \)
\( \Rightarrow A^{-1} = \begin{bmatrix} -5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix} - 3\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} -5-1-3 & -8-0-0 & -4-(-2)-0 \\ 6-(-2)-0 & 9-(-1)-3 & 4-2-0 \\ -2-3-0 & 0-4-0 & 3-1-3 \end{bmatrix} \)
\( = \begin{bmatrix} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{bmatrix} \)
In simple words: We computed the second and third powers of the matrix and substituted all of them into the given equation to verify both sides equal the zero matrix. Once confirmed, we rearranged the equation by multiplying by A⁻¹ to isolate it, then calculated the inverse as a simple combination of A², A, and I.

Exam Tip: For higher-order polynomial equations in matrices, carefully organize your calculation of successive powers. The verification step is lengthy but essential - save the rearrangement for the inverse only after confirmation.

Question 37. Prove that: (i) adj I = I (ii) adj O = O (iii) I⁻¹ = I.
Answer:
(i) To prove: adj I = I
We know that I denotes the identity matrix. Let I be a 2 × 2 matrix
\( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
To find adj I, we first calculate all co-factors
a₁₁ (co-factor of 1) = (-1)¹⁺¹(1) = (-1)²(1) = 1
a₁₂ (co-factor of 0) = (-1)¹⁺²(0) = (-1)³(0) = 0
a₂₁ (co-factor of 0) = (-1)²⁺¹(0) = (-1)³(0) = 0
a₂₂ (co-factor of 1) = (-1)²⁺²(1) = (-1)⁴(1) = 1
The co-factor matrix = \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
Now, adj I = Transpose of co-factor matrix
\( \therefore adj I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}^T = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \)
Thus, adj I = I
Hence proved

(ii) To prove: adj O = O
We know that O denotes the zero matrix where all elements equal 0. Let O be a 2 × 2 matrix
\( O = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
To find adj O, we calculate all co-factors
a₁₁ (co-factor of 0) = (-1)¹⁺¹(0) = 0
a₁₂ (co-factor of 0) = (-1)¹⁺²(0) = 0
a₂₁ (co-factor of 0) = (-1)²⁺¹(0) = 0
a₂₂ (co-factor of 0) = (-1)²⁺²(0) = 0
The co-factor matrix = \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
Now, adj O = Transpose of co-factor matrix
\( \therefore adj O = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}^T = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \)
Thus, adj O = O
Hence proved

(iii) To prove: I⁻¹ = I
We know that
\( I^{-1} = \frac{\text{adj } I}{|I|} \)
From part (i), we get adj I = I
Now we need to find |I|
Calculating |I|
\( |I| = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1 \times 1 - 0 \times 0 = 1 \)
Therefore
\( I^{-1} = \frac{\text{adj } I}{|I|} = \frac{\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}}{1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \)
Thus, I⁻¹ = I
Hence proved
In simple words: For the identity matrix, each co-factor works out to match the original element because of the alternating sign pattern and the fact that the minors of an identity matrix are always 1 or 0. The adjugate (transpose of co-factors) gives us back the identity, and since |I| = 1, the inverse is simply I divided by 1, which is I itself.

Exam Tip: These proofs are fundamental: remember that any matrix property involving the identity or zero matrix becomes simplified. For adj I = I, the key insight is that both the co-factor matrix and its transpose are identical to I.