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

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

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$2A$

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(A) Given that, $A = \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix} \dots (i)$ 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 $A^3 - A^2$: $A^3 - A^2 = \begin{bmatrix} -1 & 0 \ 0 & 8 \end{bmatrix} - \begin{bmatrix} 1 & 0 \ 0 & 4 \end{bmatrix} = \begin{bmatrix} -1-1 & 0-0 \ 0-0 & 8-4 \end{bmatrix} = \begin{bmatrix} -2 & 0 \ 0 & 4 \end{bmatrix}$ Since $A = \begin{bmatrix} -1 & 0 \ 0 & 2 \end{bmatrix}$, we can see 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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