Class 12 Maths NCERT Solutions Chapter 4 Ex 4.4 – Adjoint and Inverse of a Matrix

For a 2\times2 matrix \begin{bmatrix} a & b \\ c & d \end{bmatrix}, the adjoint is \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} (swap the diagonal entries, negate the off-diagonal entries).

The adjoint is the transpose of the cofactor matrix. Computing all nine cofactors:

A_{11}=\begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix}=3,\quad A_{12}=-\begin{vmatrix} 2 & 5 \\ -2 & 1 \end{vmatrix}=-12,\quad A_{13}=\begin{vmatrix} 2 & 3 \\ -2 & 0 \end{vmatrix}=6

A_{21}=-\begin{vmatrix} -1 & 2 \\ 0 & 1 \end{vmatrix}=1,\quad A_{22}=\begin{vmatrix} 1 & 2 \\ -2 & 1 \end{vmatrix}=5,\quad A_{23}=-\begin{vmatrix} 1 & -1 \\ -2 & 0 \end{vmatrix}=2

A_{31}=\begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix}=-11,\quad A_{32}=-\begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix}=-1,\quad A_{33}=\begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix}=5

The cofactor matrix is \begin{bmatrix} 3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5 \end{bmatrix}; taking its transpose gives the adjoint.

|A| = 2(-6)-3(-4) = -12+12 = 0, and \text{adj }A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix}.

A(\text{adj }A) = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}\begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} -12+12 & -6+6 \\ 24-24 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

(\text{adj }A)A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} = \begin{bmatrix} -12+12 & -18+18 \\ 8-8 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

Expanding along R_1: |A| = 1(0-0)-(-1)(9+2)+2(0-0) = 0+11+0 = 11

Computing all cofactors gives the cofactor matrix \begin{bmatrix} 0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3 \end{bmatrix}, so \text{adj }A = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} (its transpose).

Multiplying out A(\text{adj }A) row by row gives \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}, and multiplying (\text{adj }A)A gives the same result.

|A| = 2(3)-(-2)(4) = 6+8 = 14, and \text{adj }A = \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}.

|A| = -1(2)-5(-3) = -2+15 = 13, and \text{adj }A = \begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}.

Since A is upper triangular, |A| = 1\times2\times5 = 10.

Computing the cofactor matrix gives \begin{bmatrix} 10 & 0 & 0 \\ -10 & 5 & 0 \\ 2 & -4 & 2 \end{bmatrix}, so \text{adj }A = \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix}.

Since A is lower triangular, |A| = 1\times3\times(-1) = -3.

Computing the cofactor matrix gives \begin{bmatrix} -3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3 \end{bmatrix}, so \text{adj }A = \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}.

Expanding along R_1: |A| = 2(-1-0)-1(4-0)+3(8-7) = -2-4+3 = -3

Computing all nine cofactors gives the cofactor matrix \begin{bmatrix} -1 & -4 & 1 \\ 5 & 23 & -11 \\ 3 & 12 & -6 \end{bmatrix}, so \text{adj }A = \begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix}.

Expanding along R_1: |A| = 1(8-6)-(-1)(0+9)+2(0-6) = 2+9-12 = -1

Computing all nine cofactors gives the cofactor matrix \begin{bmatrix} 2 & -9 & -6 \\ 0 & -2 & -1 \\ -1 & 3 & 2 \end{bmatrix}, so \text{adj }A = \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix}.

Expanding along R_1: |A| = 1(-\cos^2\alpha-\sin^2\alpha) = -1

Computing the cofactors gives the cofactor matrix \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix}, which happens to be symmetric, so \text{adj }A equals this same matrix.

A^{-1} = \dfrac{1}{-1}\begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix}

|A| = 15-14 = 1 \Rightarrow A^{-1} = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}

|B| = 54-56 = -2 \Rightarrow B^{-1} = \dfrac{1}{-2}\begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} = \begin{bmatrix} -\tfrac{9}{2} & 4 \\ \tfrac{7}{2} & -3 \end{bmatrix}

AB = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}, with |AB| = 67(61)-87(47) = -2

(AB)^{-1} = \dfrac{1}{-2}\begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} = \begin{bmatrix} -\tfrac{61}{2} & \tfrac{87}{2} \\ \tfrac{47}{2} & -\tfrac{67}{2} \end{bmatrix}

Now compute B^{-1}A^{-1} = \begin{bmatrix} -\tfrac{9}{2} & 4 \\ \tfrac{7}{2} & -3 \end{bmatrix}\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} -\tfrac{61}{2} & \tfrac{87}{2} \\ \tfrac{47}{2} & -\tfrac{67}{2} \end{bmatrix}

A^2 = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}, 5A = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}, 7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}

A^2-5A+7I = \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O

Multiplying the identity A^2-5A+7I=O by A^{-1} throughout: A-5I+7A^{-1}=O \Rightarrow A^{-1} = \dfrac{1}{7}(5I-A)

5I-A = \begin{bmatrix} 5-3 & 0-1 \\ 0+1 & 5-2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}

A^2 = \begin{bmatrix} 9+2 & 6+2 \\ 3+1 & 2+1 \end{bmatrix} = \begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix}

A^2+aA+bI = \begin{bmatrix} 11+3a+b & 8+2a \\ 4+a & 3+a+b \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

From the (1,2) entry: 8+2a=0 \Rightarrow a=-4. Substituting into the (1,1) entry: 11+3(-4)+b=0 \Rightarrow b=1.

Checking against the (2,1) and (2,2) entries: 4+(-4)=0 ✓ and 3+(-4)+1=0 ✓ — both consistent.

Computing A^2 = A\cdot A = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} and then A^3 = A\cdot A^2 = \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix}.

Substituting into A^3-6A^2+5A+11I, every entry cancels to 0, confirming the identity.

Multiplying A^3-6A^2+5A+11I=O by A^{-1}: A^2-6A+5I+11A^{-1}=O \Rightarrow A^{-1} = \dfrac{1}{11}(6A-A^2-5I)

6A-A^2-5I = \begin{bmatrix} 6-4-5 & 6-2+0 & 6-1+0 \\ 6+3+0 & 12-8-5 & -18+14+0 \\ 12-7+0 & -6+3+0 & 18-14-5 \end{bmatrix} = \begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix}

Computing A^2 = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} and then A^3 = \begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix}.

Substituting into A^3-6A^2+9A-4I, every entry cancels to 0, confirming the identity.

Multiplying by A^{-1}: A^2-6A+9I-4A^{-1}=O \Rightarrow A^{-1} = \dfrac{1}{4}(A^2-6A+9I)

A^2-6A+9I = \begin{bmatrix} 6-12+9 & -5+6+0 & 5-6+0 \\ -5+6+0 & 6-12+9 & -5+6+0 \\ 5-6+0 & -5+6+0 & 6-12+9 \end{bmatrix} = \begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}

Since AA^{-1}=I, taking determinants of both sides: \det(A)\det(A^{-1}) = \det(I) = 1

So \det(A^{-1}) = \dfrac{1}{\det(A)}, which holds for an invertible matrix of any order, not just order 2.

For any square matrix, the transpose has the same determinant as the original matrix: \det(A') = \det(A).

Also, for an invertible matrix, \det(A^{-1}) = \dfrac{1}{\det(A)} (as shown in Question 17).

Using the product rule for determinants: \det(A'A^{-1}) = \det(A')\cdot\det(A^{-1}) = \det(A)\cdot\dfrac{1}{\det(A)} = 1