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(B) Given the matrix equation $A^2-4A-5I=0$.Multiply both sides by $A^{-1}$:$A^2 A^{-1} - 4A A^{-1} - 5I A^{-1} = 0 A^{-1}$$A - 4I - 5A^{-1} = 0$$5A^{

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$\frac{1}{5}\begin{bmatrix} -3 & 2 & 2 \ 2 & -3 & 2 \ 2 & 2 & -3 \end{bmatrix}$

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(B) Given the matrix equation $A^2-4A-5I=0$.Multiply both sides by $A^{-1}$:$A^2 A^{-1} - 4A A^{-1} - 5I A^{-1} = 0 A^{-1}$$A - 4I - 5A^{-1} = 0$$5A^{-1} = A - 4I$$A^{-1} = \frac{1}{5}(A - 4I)$Now, substitute the matrix $A$ and $I$:$A^{-1} = \frac{1}{5} \left( \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix} - 4 \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} ight)$$A^{-1} = \frac{1}{5} \left( \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix} - \begin{bmatrix} 4 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 4 \end{bmatrix} ight)$$A^{-1} = \frac{1}{5} \begin{bmatrix} 1-4 & 2-0 & 2-0 \ 2-0 & 1-4 & 2-0 \ 2-0 & 2-0 & 1-4 \end{bmatrix}$$A^{-1} = \frac{1}{5} \begin{bmatrix} -3 & 2 & 2 \ 2 & -3 & 2 \ 2 & 2 & -3 \end{bmatrix}$

If the matrix $A=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$ satisfies the matrix equation $A^2-4A-5I=0$, then $A^{-1}=$

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