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(D) Given $A = \begin{bmatrix} 1 & -1 & 1 \ 2 & -1 & 0 \ 1 & 0 & 0 \end{bmatrix}$.First, calculate $A^2 = A \cdot A = \begin{bmatrix} 1 & -1 & 1 \

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$A^{-1}$

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(D) Given $A = \begin{bmatrix} 1 & -1 & 1 \ 2 & -1 & 0 \ 1 & 0 & 0 \end{bmatrix}$.First, calculate $A^2 = A \cdot A = \begin{bmatrix} 1 & -1 & 1 \ 2 & -1 & 0 \ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \ 2 & -1 & 0 \ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \ 0 & -1 & 2 \ 1 & -1 & 1 \end{bmatrix}$.Next, calculate $A^3 = A^2 \cdot A = \begin{bmatrix} 0 & 0 & 1 \ 0 & -1 & 2 \ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \ 2 & -1 & 0 \ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$.Since $A^3 = I$, we can find $A^5$ as follows:$A^5 = A^3 \cdot A^2 = I \cdot A^2 = A^2$.Calculating $A^2$ again: $A^2 = \begin{bmatrix} 0 & 0 & 1 \ 0 & -1 & 2 \ 1 & -1 & 1 \end{bmatrix}$.However, checking the options provided, we re-evaluate $A^3$.$A^3 = \begin{bmatrix} 0 & 0 & 1 \ 0 & -1 & 2 \ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \ 2 & -1 & 0 \ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$.Thus $A^5 = A^2$. Since $A^2$ is not listed, let us check $A^{-1}$.$A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$. $|A| = 1(0-0) - (-1)(0-0) + 1(0 - (-1)) = 1$.$	ext{adj}(A) = \begin{bmatrix} 0 & 0 & 1 \ 0 & -1 & 2 \ 1 & -1 & 1 \end{bmatrix}^T = \begin{bmatrix} 0 & 0 & 1 \ 0 & -1 & -1 \ 1 & 2 & 1 \end{bmatrix}$.Since $A^3 = I$, $A^2 = A^{-1}$. Therefore $A^5 = A^2 = A^{-1}$.

If $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$, then $A^5 =$

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