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

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(A) Given that,$A = \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix}$.First,we calculate $A^2$:$A^2 = A \cdot A = \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bm

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$2A$

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(A) Given that,$A = \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix}$.First,we calculate $A^2$:$A^2 = A \cdot A = \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix} \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} (-1)(-1) + (0)(0) & (-1)(0) + (0)(2) \ (0)(-1) + (2)(0) & (0)(0) + (2)(2) \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 4 \end{bmatrix}$.Next,we calculate $A^3$:$A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 0 \ 0 & 4 \end{bmatrix} \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} (1)(-1) + (0)(0) & (1)(0) + (0)(2) \ (0)(-1) + (4)(0) & (0)(0) + (4)(2) \end{bmatrix} = \begin{bmatrix} -1 & 0 \ 0 & 8 \end{bmatrix}$.Now,calculate $A^3 - A^2$:$A^3 - A^2 = \begin{bmatrix} -1 & 0 \ 0 & 8 \end{bmatrix} - \begin{bmatrix} 1 & 0 \ 0 & 4 \end{bmatrix} = \begin{bmatrix} -1 - 1 & 0 - 0 \ 0 - 0 & 8 - 4 \end{bmatrix} = \begin{bmatrix} -2 & 0 \ 0 & 4 \end{bmatrix}$.We observe that $2A = 2 \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 0 \ 0 & 4 \end{bmatrix}$.Therefore,$A^3 - A^2 = 2A$.

If $A = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}$,then $A^3 - A^2$ is equal to

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