If A = begin{bmatrix} 1 & -2 & 2 0 & 2 & -3 | vedclass

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(C) Given $A = \begin{bmatrix} 1 & -2 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4 \end{bmatrix}$.First,calculate the determinant $|A| = 1(8 - 6) - (-2)(0 - (-9))

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$\begin{bmatrix} 9 & -2 & 2 \ 0 & 10 & -3 \ 3 & -2 & 12 \end{bmatrix}$

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(C) Given $A = \begin{bmatrix} 1 & -2 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4 \end{bmatrix}$.First,calculate the determinant $|A| = 1(8 - 6) - (-2)(0 - (-9)) + 2(0 - 6) = 1(2) + 2(9) + 2(-6) = 2 + 18 - 12 = 8$.Next,find $\operatorname{adj} A$. The matrix of cofactors is:$C_{11} = +(8-6) = 2, C_{12} = -(0 - (-9)) = -9, C_{13} = +(0-6) = -6$$C_{21} = -(-8 - (-4)) = 4, C_{22} = +(4-6) = -2, C_{23} = -(-2 - (-6)) = -4$$C_{31} = +(6-4) = 2, C_{32} = -(-3-0) = 3, C_{33} = +(2-0) = 2$So,$\operatorname{adj} A = \begin{bmatrix} 2 & 4 & 2 \ -9 & -2 & 3 \ -6 & -4 & 2 \end{bmatrix}$.Then $A(I + \operatorname{adj} A) = A + A(\operatorname{adj} A) = A + |A|I$.$A + 8I = \begin{bmatrix} 1 & -2 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4 \end{bmatrix} + \begin{bmatrix} 8 & 0 & 0 \ 0 & 8 & 0 \ 0 & 0 & 8 \end{bmatrix} = \begin{bmatrix} 9 & -2 & 2 \ 0 & 10 & -3 \ 3 & -2 & 12 \end{bmatrix}$.Thus,the correct option is $C$.

If $A = \begin{bmatrix} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$,then $A(I + \operatorname{adj} A) = $

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