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(A) Given $A \begin{bmatrix} 1 & 2 & 3 \ 0 & 2 & 3 \ 0 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \end{bmatrix}$.Le

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$\begin{bmatrix} 3 & 1 & 2 \ 3 & 0 & 2 \ 1 & 0 & 1 \end{bmatrix}$

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(A) Given $A \begin{bmatrix} 1 & 2 & 3 \ 0 & 2 & 3 \ 0 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \end{bmatrix}$.Let $B = \begin{bmatrix} 1 & 2 & 3 \ 0 & 2 & 3 \ 0 & 1 & 1 \end{bmatrix}$ and $C = \begin{bmatrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \end{bmatrix}$. So,$AB = C$.We know that $A^{-1} = B C^{-1}$.First,find $C^{-1}$. Since $C$ is a permutation matrix,$C^{-1} = C^T = \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{bmatrix}$.Now,$A^{-1} = \begin{bmatrix} 1 & 2 & 3 \ 0 & 2 & 3 \ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{bmatrix}$.Performing matrix multiplication:$A^{-1} = \begin{bmatrix} (1)(0)+(2)(0)+(3)(1) & (1)(1)+(2)(0)+(3)(0) & (1)(0)+(2)(1)+(3)(0) \ (0)(0)+(2)(0)+(3)(1) & (0)(1)+(2)(0)+(3)(0) & (0)(0)+(2)(1)+(3)(0) \ (0)(0)+(1)(0)+(1)(1) & (0)(1)+(1)(0)+(1)(0) & (0)(0)+(1)(1)+(1)(0) \end{bmatrix} = \begin{bmatrix} 3 & 1 & 2 \ 3 & 0 & 2 \ 1 & 0 & 1 \end{bmatrix}$.

Let $A$ be a $3 \times 3$ matrix such that $A \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$. Then $A^{-1}$ is

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