If A = begin{bmatrix} 3 & 1 -1 & 2 end{bmatr | vedclass

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(D) Given that,$A = \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix}$.First,we calculate $A^{2} = A 	imes A$:$A^{2} = \begin{bmatrix} 3 & 1 \ -1 & 2 \e

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$-7I$

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(D) Given that,$A = \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix}$.First,we calculate $A^{2} = A 	imes A$:$A^{2} = \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} = \begin{bmatrix} (3)(3) + (1)(-1) & (3)(1) + (1)(2) \ (-1)(3) + (2)(-1) & (-1)(1) + (2)(2) \end{bmatrix} = \begin{bmatrix} 9-1 & 3+2 \ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \ -5 & 3 \end{bmatrix}$.Next,we calculate $5A$:$5A = 5 \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} = \begin{bmatrix} 15 & 5 \ -5 & 10 \end{bmatrix}$.Finally,we compute $A^{2} - 5A$:$A^{2} - 5A = \begin{bmatrix} 8 & 5 \ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \ -5 & 10 \end{bmatrix} = \begin{bmatrix} 8-15 & 5-5 \ -5-(-5) & 3-10 \end{bmatrix} = \begin{bmatrix} -7 & 0 \ 0 & -7 \end{bmatrix}$.This can be written as $-7 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = -7I$.

If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$,then $A^{2} - 5A$ is equal to:

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