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(D) First,calculate $A^2$: $A^2 = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 &

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(D) First,calculate $A^2$: $A^2 = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 8 & 8 \ 8 & 9 & 8 \ 8 & 8 & 9 \end{bmatrix}$ Now,evaluate $A^2 - 4A - 5I_3$: $A^2 - 4A - 5I_3 = \begin{bmatrix} 9 & 8 & 8 \ 8 & 9 & 8 \ 8 & 8 & 9 \end{bmatrix} - 4 \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix} - 5 \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix} = O$. Thus,$A^2 - 4A - 5I_3 = 0$ is correct. From $A^2 - 4A = 5I_3$,we have $A(A - 4I_3) = 5I_3$. Multiplying by $A^{-1}$,we get $A^{-1} = \frac{1}{5}(A - 4I_3)$,which is also correct. Finally,$|A| = 1(1-4) - 2(2-4) + 2(4-2) = -3 + 4 + 4 = 5 
eq 0$. Since $|A| 
eq 0$,$A$ is invertible,and consequently $A^2$ is also invertible. Therefore,all statements are correct.

Let $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$,then:

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