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

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$A^3$

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(C) Given,$A = \begin{bmatrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{bmatrix}$.First,we calculate the determinant $|A|$:$|A| = 3((-3)(1) - (4)(-1)) - (-3)((2)(1) - (4)(0)) + 4((2)(-1) - (-3)(0))$$|A| = 3(-3 + 4) + 3(2 - 0) + 4(-2 - 0) = 3(1) + 3(2) + 4(-2) = 3 + 6 - 8 = 1$.Since $|A| = 1 
eq 0$,$A^{-1}$ exists.Next,we find the adjoint of $A$ by calculating the cofactor matrix and transposing it.The cofactor matrix $C$ is:$C_{11} = (-1)^{1+1} \begin{vmatrix} -3 & 4 \ -1 & 1 \end{vmatrix} = 1, C_{12} = (-1)^{1+2} \begin{vmatrix} 2 & 4 \ 0 & 1 \end{vmatrix} = -2, C_{13} = (-1)^{1+3} \begin{vmatrix} 2 & -3 \ 0 & -1 \end{vmatrix} = -2$$C_{21} = (-1)^{2+1} \begin{vmatrix} -3 & 4 \ -1 & 1 \end{vmatrix} = -1, C_{22} = (-1)^{2+2} \begin{vmatrix} 3 & 4 \ 0 & 1 \end{vmatrix} = 3, C_{23} = (-1)^{2+3} \begin{vmatrix} 3 & -3 \ 0 & -1 \end{vmatrix} = 3$$C_{31} = (-1)^{3+1} \begin{vmatrix} -3 & 4 \ -3 & 4 \end{vmatrix} = 0, C_{32} = (-1)^{3+2} \begin{vmatrix} 3 & 4 \ 2 & 4 \end{vmatrix} = -4, C_{33} = (-1)^{3+3} \begin{vmatrix} 3 & -3 \ 2 & -3 \end{vmatrix} = -3$Thus,$	ext{adj}(A) = C^T = \begin{bmatrix} 1 & -1 & 0 \ -2 & 3 & -4 \ -2 & 3 & -3 \end{bmatrix}$.Since $|A| = 1$,$A^{-1} = \frac{	ext{adj}(A)}{|A|} = \begin{bmatrix} 1 & -1 & 0 \ -2 & 3 & -4 \ -2 & 3 & -3 \end{bmatrix}$.Now,calculate $A^2 = A \cdot A = \begin{bmatrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -4 & 4 \ 0 & -1 & 0 \ -2 & 2 & -3 \end{bmatrix}$.Then,$A^3 = A^2 \cdot A = \begin{bmatrix} 3 & -4 & 4 \ 0 & -1 & 0 \ -2 & 2 & -3 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \ -2 & 3 & -4 \ -2 & 3 & -3 \end{bmatrix}$.Comparing $A^{-1}$ and $A^3$,we see that $A^{-1} = A^3$.

For the matrix $A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$,find $A^{-1}$.

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