If A = begin{bmatrix} 3 & -5 -4 & 2 end{bmat | vedclass

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(B) Given $A = \begin{bmatrix} 3 & -5 \ -4 & 2 \end{bmatrix}$.First,calculate $A^2 = A 	imes A$:$A^2 = \begin{bmatrix} 3 & -5 \ -4 & 2 \end{bmatrix

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$14I$

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(B) Given $A = \begin{bmatrix} 3 & -5 \ -4 & 2 \end{bmatrix}$.First,calculate $A^2 = A 	imes A$:$A^2 = \begin{bmatrix} 3 & -5 \ -4 & 2 \end{bmatrix} \begin{bmatrix} 3 & -5 \ -4 & 2 \end{bmatrix} = \begin{bmatrix} (3)(3) + (-5)(-4) & (3)(-5) + (-5)(2) \ (-4)(3) + (2)(-4) & (-4)(-5) + (2)(2) \end{bmatrix} = \begin{bmatrix} 9 + 20 & -15 - 10 \ -12 - 8 & 20 + 4 \end{bmatrix} = \begin{bmatrix} 29 & -25 \ -20 & 24 \end{bmatrix}$.Next,calculate $5A$:$5A = 5 \begin{bmatrix} 3 & -5 \ -4 & 2 \end{bmatrix} = \begin{bmatrix} 15 & -25 \ -20 & 10 \end{bmatrix}$.Now,find $A^2 - 5A$:$A^2 - 5A = \begin{bmatrix} 29 & -25 \ -20 & 24 \end{bmatrix} - \begin{bmatrix} 15 & -25 \ -20 & 10 \end{bmatrix} = \begin{bmatrix} 29-15 & -25-(-25) \ -20-(-20) & 24-10 \end{bmatrix} = \begin{bmatrix} 14 & 0 \ 0 & 14 \end{bmatrix}$.This can be written as $14 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = 14I$.Thus,the correct option is $B$.

If $A = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix}$,then $A^2 - 5A = $

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