HSC 12th Mathematics: Matrices (Exercise 2.2, Q4)
Question 4
If $A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$, verify that $A(\text{adj } A) = (\text{adj } A)A = |A|I$.
Solution:
Step 1: Find the determinant of matrix $A$
$$|A| = \begin{vmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{vmatrix}$$
$$|A| = 1(0 - 0) - (-1)(9 - (-2)) + 2(0 - 0)$$
$$|A| = 0 + 1(11) + 0 = 11$$
Step 2: Find the cofactors of all elements
The cofactor is given by $A_{ij} = (-1)^{i+j} M_{ij}$:
$A_{11} = (-1)^{1+1}(0 - 0) = 0$
$A_{12} = (-1)^{1+2}(9 - (-2)) = -11$
$A_{13} = (-1)^{1+3}(0 - 0) = 0$
$A_{21} = (-1)^{2+1}(-3 - 0) = 3$
$A_{22} = (-1)^{2+2}(3 - 2) = 1$
$A_{23} = (-1)^{2+3}(0 - (-1)) = -1$
$A_{31} = (-1)^{3+1}(2 - 0) = 2$
$A_{32} = (-1)^{3+2}(-2 - 6) = 8$
$A_{33} = (-1)^{3+3}(0 - (-3)) = 3$
The cofactor matrix $C$ is:
$$C = \begin{bmatrix} 0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3 \end{bmatrix}$$
Step 3: Find the Adjoint of $A$
The adjoint of $A$ is the transpose of the cofactor matrix ($C^T$):
$$\text{adj } A = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix}$$
Step 4: Calculate $A(\text{adj } A)$
$$A(\text{adj } A) = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix}$$
$$= \begin{bmatrix} (0+11+0) & (3-1-2) & (2-8+6) \\ (0+0+0) & (9+0+2) & (6+0-6) \\ (0+0+0) & (3+0-3) & (2+0+9) \end{bmatrix}$$
$$= \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} \quad \text{--- (Equation 1)}$$
Step 5: Calculate $(\text{adj } A)A$
$$(\text{adj } A)A = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$$
$$= \begin{bmatrix} (0+9+2) & (0+0+0) & (0-6+6) \\ (-11+3+8) & (11+0+0) & (-22-2+24) \\ (0-3+3) & (0+0+0) & (0+2+9) \end{bmatrix}$$
$$= \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} \quad \text{--- (Equation 2)}$$
Step 6: Calculate $|A|I$
We know $|A| = 11$, and $I$ is the $3 \times 3$ identity matrix:
$$|A|I = 11 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} \quad \text{--- (Equation 3)}$$
Conclusion:
From Equations 1, 2, and 3, it is clearly verified that:
$$A(\text{adj } A) = (\text{adj } A)A = |A|I$$
