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

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(B) Given $A = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{bmatrix}$.To find $A^2$,we multiply $A$ by itself:$A^2 = A 	imes A = \begi

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(B) Given $A = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{bmatrix}$.To find $A^2$,we multiply $A$ by itself:$A^2 = A 	imes A = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{bmatrix}$Performing matrix multiplication:$A^2 = \begin{bmatrix} (-1)(-1) + 0 + 0 & 0 + 0 + 0 & 0 + 0 + 0 \ 0 + 0 + 0 & 0 + (-1)(-1) + 0 & 0 + 0 + 0 \ 0 + 0 + 0 & 0 + 0 + 0 & 0 + 0 + (-1)(-1) \end{bmatrix}$$A^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$Thus,$A^2$ is the unit matrix (identity matrix) $I$.

If $A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$,then $A^2$ is

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