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

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(C) To find $A^{-1}$,we first calculate the determinant $|A|$.$|A| = 3((-3)(1) - (4)(-1)) - (-3)((2)(1) - (4)(0)) + 4((2)(-1) - (-3)(0))$$|A| = 3(-3 +

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$A^3$

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(C) To find $A^{-1}$,we first calculate the determinant $|A|$.$|A| = 3((-3)(1) - (4)(-1)) - (-3)((2)(1) - (4)(0)) + 4((2)(-1) - (-3)(0))$$|A| = 3(-3 + 4) + 3(2) + 4(-2) = 3(1) + 6 - 8 = 3 + 6 - 8 = 1$.Since $|A| = 1 
eq 0$,$A^{-1}$ exists.Next,we find the matrix of cofactors $C_{ij}$.$C_{11} = +((-3)(1) - (4)(-1)) = 1$,$C_{12} = -((2)(1) - (4)(0)) = -2$,$C_{13} = +((2)(-1) - (-3)(0)) = -2$.$C_{21} = -((-3)(1) - (4)(-1)) = -1$,$C_{22} = +((3)(1) - (4)(0)) = 3$,$C_{23} = -((3)(-1) - (-3)(0)) = 3$.$C_{31} = +((-3)(4) - (-3)(4)) = 0$,$C_{32} = -((3)(4) - (2)(4)) = -4$,$C_{33} = +((3)(-3) - (2)(-3)) = -3$.Thus,$adj(A) = \begin{bmatrix} 1 & -1 & 0 \ -2 & 3 & -4 \ -2 & 3 & -3 \end{bmatrix}$.Calculating $A^2 = \begin{bmatrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -4 & 4 \ 0 & -1 & 0 \ -2 & 2 & -3 \end{bmatrix}$.Calculating $A^3 = A^2 \cdot A = \begin{bmatrix} 3 & -4 & 4 \ 0 & -1 & 0 \ -2 & 2 & -3 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \ -2 & 3 & -4 \ -2 & 3 & -3 \end{bmatrix}$.Since $A^{-1} = \frac{1}{|A|} adj(A) = 1 \cdot adj(A) = adj(A)$,and we found $A^3 = adj(A)$,therefore $A^{-1} = A^3$.

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

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