Let $A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$. Verify that $A(\text{adj } A) = (\text{adj } A) A = |A| I$.
(A) $A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$
$|A| = 1(0 - 0) - (-1)(9 - (-2)) + 2(0 - 0) = 1(0) + 1(11) + 2(0) = 11$
$\therefore |A| I = 11 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$
Now,calculating the cofactors $A_{ij}$:
$A_{11} = (0 - 0) = 0, A_{12} = -(9 - (-2)) = -11, A_{13} = (0 - 0) = 0$
$A_{21} = -(-3 - 0) = 3, A_{22} = (3 - 2) = 1, A_{23} = -(0 - (-1)) = -1$
$A_{31} = (2 - 0) = 2, A_{32} = -(-2 - 6) = 8, A_{33} = (0 - (-3)) = 3$
$\therefore \text{adj } A = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix}$
Now,$A(\text{adj } A) = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$
Also,$(\text{adj } A) A = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$
Hence,$A(\text{adj } A) = (\text{adj } A) A = |A| I$ is verified.
$|A| = 1(0 - 0) - (-1)(9 - (-2)) + 2(0 - 0) = 1(0) + 1(11) + 2(0) = 11$
$\therefore |A| I = 11 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$
Now,calculating the cofactors $A_{ij}$:
$A_{11} = (0 - 0) = 0, A_{12} = -(9 - (-2)) = -11, A_{13} = (0 - 0) = 0$
$A_{21} = -(-3 - 0) = 3, A_{22} = (3 - 2) = 1, A_{23} = -(0 - (-1)) = -1$
$A_{31} = (2 - 0) = 2, A_{32} = -(-2 - 6) = 8, A_{33} = (0 - (-3)) = 3$
$\therefore \text{adj } A = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix}$
Now,$A(\text{adj } A) = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$
Also,$(\text{adj } A) A = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$
Hence,$A(\text{adj } A) = (\text{adj } A) A = |A| I$ is verified.
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