If A = begin{bmatrix} 1 & 2 & -1 -1 & 0 & 2 | vedclass

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(C) Given $A = \begin{bmatrix} 1 & 2 & -1 \ -1 & 0 & 2 \ 1 & 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} -3 & -2 & 4 \ 2 & 2 & -1 \ -2 & 0 & 3 \

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$A+B$

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(C) Given $A = \begin{bmatrix} 1 & 2 & -1 \ -1 & 0 & 2 \ 1 & 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} -3 & -2 & 4 \ 2 & 2 & -1 \ -2 & 0 & 3 \end{bmatrix}$.First, calculate $A^2 = A 	imes A$:$A^2 = \begin{bmatrix} 1 & 2 & -1 \ -1 & 0 & 2 \ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 & -1 \ -1 & 0 & 2 \ 1 & 2 & 0 \end{bmatrix} = \begin{bmatrix} (1-2-1) & (2+0-2) & (-1+4+0) \ (-1+0+2) & (-2+0+4) & (1+0+0) \ (1-2+0) & (2+0+0) & (-1+4+0) \end{bmatrix} = \begin{bmatrix} -2 & 0 & 3 \ 1 & 2 & 1 \ -1 & 2 & 3 \end{bmatrix}$.Next, calculate $A+B$:$A+B = \begin{bmatrix} 1-3 & 2-2 & -1+4 \ -1+2 & 0+2 & 2-1 \ 1-2 & 2+0 & 0+3 \end{bmatrix} = \begin{bmatrix} -2 & 0 & 3 \ 1 & 2 & 1 \ -1 & 2 & 3 \end{bmatrix}$.Comparing both results, we see that $A^2 = A+B$.

If $A = \begin{bmatrix} 1 & 2 & -1 \\ -1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} -3 & -2 & 4 \\ 2 & 2 & -1 \\ -2 & 0 & 3 \end{bmatrix}$, then $A^2 = $

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