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(D) Given the matrix $A = \begin{bmatrix} 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \end{bmatrix}$.First,we calculate $A^2 = A 	imes A$:$A^2 = \begin{bmatr

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$A$

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(D) Given the matrix $A = \begin{bmatrix} 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \end{bmatrix}$.First,we calculate $A^2 = A 	imes A$:$A^2 = \begin{bmatrix} 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$.Since $A^2 = I$,we can multiply both sides by $A^{-1}$:$A^2 \cdot A^{-1} = I \cdot A^{-1}$$A \cdot (A \cdot A^{-1}) = A^{-1}$$A \cdot I = A^{-1}$Therefore,$A^{-1} = A$.

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

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