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

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(C) First,we calculate the determinant of $A$:$|A| = 3(-3 + 4) - (-3)(2 - 0) + 4(-2 - 0) = 3(1) + 3(2) + 4(-2) = 3 + 6 - 8 = 1$.Next,we find the cofac

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$A^3$

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(C) First,we calculate the determinant of $A$:$|A| = 3(-3 + 4) - (-3)(2 - 0) + 4(-2 - 0) = 3(1) + 3(2) + 4(-2) = 3 + 6 - 8 = 1$.Next,we find the cofactor matrix $C$:$C_{11} = (-3+4) = 1, C_{12} = -(2-0) = -2, C_{13} = (-2-0) = -2$$C_{21} = -(-3+4) = -1, C_{22} = (3-0) = 3, C_{23} = -(-3-0) = 3$$C_{31} = (-12+12) = 0, C_{32} = -(12-8) = -4, C_{33} = (-9+6) = -3$Wait,let us recompute $A^{-1}$ using the formula $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$:$A^{-1} = \begin{bmatrix} 1 & -1 & 0 \ -2 & 3 & -4 \ -2 & 3 & -3 \end{bmatrix}$.Now,calculate $A^2 = \begin{bmatrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -4 & 4 \ 0 & -1 & 0 \ -2 & 2 & -3 \end{bmatrix}$.Now,calculate $A^3 = A^2 \cdot A = \begin{bmatrix} 3 & -4 & 4 \ 0 & -1 & 0 \ -2 & 2 & -3 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \ -2 & 3 & -4 \ -2 & 3 & -3 \end{bmatrix}$.Since $A^{-1} = A^3$,the correct option is $C$.

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

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