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(C) Given $A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}$.First, we calculate the determinant $|A|$:$|A| = 1(1 - 4) - 2(2 - 4)

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$\frac{1}{5}(A - 4I)$

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(C) Given $A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}$.First, we calculate the determinant $|A|$:$|A| = 1(1 - 4) - 2(2 - 4) + 2(4 - 2) = -3 + 4 + 4 = 5$.Next, we find the adjoint of $A$ $(	ext{Adj } A)$:The matrix of cofactors is:$C_{11} = (1-4) = -3, C_{12} = -(2-4) = 2, C_{13} = (4-2) = 2$$C_{21} = -(2-4) = 2, C_{22} = (1-4) = -3, C_{23} = -(2-4) = 2$$C_{31} = (4-2) = 2, C_{32} = -(2-4) = 2, C_{33} = (1-4) = -3$So, $	ext{Adj } A = \begin{bmatrix} -3 & 2 & 2 \ 2 & -3 & 2 \ 2 & 2 & -3 \end{bmatrix}$.Thus, $A^{-1} = \frac{1}{|A|} 	ext{Adj } A = \frac{1}{5} \begin{bmatrix} -3 & 2 & 2 \ 2 & -3 & 2 \ 2 & 2 & -3 \end{bmatrix}$.We can rewrite this as:$A^{-1} = \frac{1}{5} \left( \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix} - \begin{bmatrix} 4 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 4 \end{bmatrix} ight) = \frac{1}{5}(A - 4I)$.

If $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$, then $A^{-1} =$

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