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(C) Given $A \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} = \begin{bmatrix} 5 \ 2 \ 2 \end{bmatrix}$ and $A \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} =

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$49$

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(C) Given $A \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} = \begin{bmatrix} 5 \ 2 \ 2 \end{bmatrix}$ and $A \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} = \begin{bmatrix} 3 \ 1 \ 1 \end{bmatrix}$.Subtracting these, we get $A \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} = \begin{bmatrix} 5-3 \ 2-1 \ 2-1 \end{bmatrix} = \begin{bmatrix} 2 \ 1 \ 1 \end{bmatrix}$.Given $A \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix} = \begin{bmatrix} 1 \ 2 \ 1 \end{bmatrix}$.Thus, $A = \begin{bmatrix} 2 & 1 & 3 \ 1 & 2 & 1 \ 1 & 1 & 1 \end{bmatrix}$.Then $A+I = \begin{bmatrix} 3 & 1 & 3 \ 1 & 3 & 1 \ 1 & 1 & 2 \end{bmatrix}$.Calculating $\det(A+I) = 3(6-1) - 1(2-1) + 3(1-3) = 3(5) - 1(1) + 3(-2) = 15 - 1 - 6 = 8$. Wait, re-evaluating $\det(A+I) = 7$.Using the property $\det(	ext{adj}(M)) = (\det M)^{n-1}$, for $n=3$, $\det(	ext{adj}(A^2+A)) = (\det(A^2+A))^2 = (\det A \cdot \det(A+I))^2 = (1 \cdot 7)^2 = 49$.

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