A=left[begin{array}{lll}1 & 0 & 1 0 & 1 & 1 | vedclass

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(B) Given,$A=\left[\begin{array}{lll}1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 1 & 0\end{array}ight]$.First,calculate $A^2$:$A^2 = \left[\begin{array}{lll}1 & 0

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$-A^{-1}$

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(B) Given,$A=\left[\begin{array}{lll}1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 1 & 0\end{array}ight]$.First,calculate $A^2$:$A^2 = \left[\begin{array}{lll}1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 1 & 0\end{array}ight] \left[\begin{array}{lll}1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 1 & 0\end{array}ight] = \left[\begin{array}{lll}1 & 1 & 1 \ 0 & 2 & 1 \ 0 & 1 & 1\end{array}ight]$.Now,calculate $A^2 - 2A$:$A^2 - 2A = \left[\begin{array}{lll}1 & 1 & 1 \ 0 & 2 & 1 \ 0 & 1 & 1\end{array}ight] - 2\left[\begin{array}{lll}1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 1 & 0\end{array}ight] = \left[\begin{array}{lll}1 & 1 & 1 \ 0 & 2 & 1 \ 0 & 1 & 1\end{array}ight] - \left[\begin{array}{lll}2 & 0 & 2 \ 0 & 2 & 2 \ 0 & 2 & 0\end{array}ight] = \left[\begin{array}{rrr}-1 & 1 & -1 \ 0 & 0 & -1 \ 0 & -1 & 1\end{array}ight] \dots (i)$.Next,find $A^{-1}$. The determinant $|A| = 1(0-1) - 0 + 1(0) = -1$.The cofactor matrix $C$ is:$C_{11} = -1, C_{12} = 0, C_{13} = 0$$C_{21} = 1, C_{22} = 0, C_{23} = -1$$C_{31} = -1, C_{32} = -1, C_{33} = 1$$Adj(A) = C^T = \left[\begin{array}{rrr}-1 & 1 & -1 \ 0 & 0 & -1 \ 0 & -1 & 1\end{array}ight]$.Since $A^{-1} = \frac{1}{|A|} Adj(A) = \frac{1}{-1} \left[\begin{array}{rrr}-1 & 1 & -1 \ 0 & 0 & -1 \ 0 & -1 & 1\end{array}ight] = -\left[\begin{array}{rrr}-1 & 1 & -1 \ 0 & 0 & -1 \ 0 & -1 & 1\end{array}ight]$.Comparing this with $(i)$,we see that $A^2 - 2A = -A^{-1}$.

$A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{array}\right] \Rightarrow A^2-2 A=$

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