If A = begin{bmatrix} 3 & -3 & 4 2 & -3 & 4 | vedclass

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(A) Given $A = \begin{bmatrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{bmatrix}$.First, calculate $A^2 = A 	imes A$:$A^2 = \begin{bmatrix} 3 & -3

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$A^2$

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(A) Given $A = \begin{bmatrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{bmatrix}$.First, calculate $A^2 = A 	imes A$:$A^2 = \begin{bmatrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -4 & 4 \ 0 & -1 & 0 \ -2 & 2 & -3 \end{bmatrix}$.Now, find the determinant $|A^2|$:$|A^2| = 3((-1)(-3) - (0)(2)) - (-4)((0)(-3) - (0)(-2)) + 4((0)(2) - (-1)(-2)) = 3(3) + 4(0) + 4(-2) = 9 - 8 = 1$.Since $|A^2| = 1$, $(A^2)^{-1} = \frac{1}{|A^2|} 	ext{adj}(A^2) = 	ext{adj}(A^2)$.The cofactor matrix of $A^2$ is calculated as:$C_{11} = 3, C_{12} = 0, C_{13} = -2$$C_{21} = 4, C_{22} = -1, C_{23} = 2$$C_{31} = 4, C_{32} = 0, C_{33} = -3$Thus, $	ext{adj}(A^2) = \begin{bmatrix} 3 & 4 & 4 \ 0 & -1 & 0 \ -2 & 2 & -3 \end{bmatrix}$.Wait, re-evaluating $A^2$: $A^2 = \begin{bmatrix} 3 & -4 & 4 \ 0 & -1 & 0 \ -2 & 2 & -3 \end{bmatrix}$.The inverse $(A^2)^{-1}$ is indeed $A^2$ because $A^2 	imes A^2 = I$.

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

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