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(B) Given that $5 A^{-1}=\begin{bmatrix} -3 & 2 & 2 \ 2 & -3 & 2 \ 2 & 2 & -3 \end{bmatrix}$.Post-multiplying both sides by $A$, we get:$5 I = \begi

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$5 I$

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(B) Given that $5 A^{-1}=\begin{bmatrix} -3 & 2 & 2 \ 2 & -3 & 2 \ 2 & 2 & -3 \end{bmatrix}$.Post-multiplying both sides by $A$, we get:$5 I = \begin{bmatrix} -3 & 2 & 2 \ 2 & -3 & 2 \ 2 & 2 & -3 \end{bmatrix} \begin{bmatrix} x & y & y \ y & x & y \ y & y & x \end{bmatrix}$$\begin{bmatrix} 5 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 5 \end{bmatrix} = \begin{bmatrix} -3x+4y & 2x-y & 2x-y \ 2x-y & -3x+4y & 2x-y \ 2x-y & 2x-y & -3x+4y \end{bmatrix}$Comparing elements, we get $-3x+4y=5$ and $2x-y=0$.From $2x-y=0$, we have $y=2x$. Substituting into the first equation: $-3x+4(2x)=5 \Rightarrow 5x=5 \Rightarrow x=1$.Then $y=2(1)=2$.Thus, $A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}$.From the given equation $5 A^{-1} = \begin{bmatrix} -3 & 2 & 2 \ 2 & -3 & 2 \ 2 & 2 & -3 \end{bmatrix}$, we can write this as:$5 A^{-1} = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix} - \begin{bmatrix} 4 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 4 \end{bmatrix} = A - 4 I$.Multiplying by $A$ on the right: $5 I = A^2 - 4 A$.

If $A=\begin{bmatrix} x & y & y \\ y & x & y \\ y & y & x \end{bmatrix}$ is a matrix such that $5 A^{-1}=\begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}$, then $A^2-4 A=$

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