If A = begin{bmatrix} 4 & 1 3 & 2 end{bmatri | vedclass

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(C) Given $A = \begin{bmatrix} 4 & 1 \ 3 & 2 \end{bmatrix}$.First,calculate $A^2 = A 	imes A = \begin{bmatrix} 4 & 1 \ 3 & 2 \end{bmatrix} \begin{b

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$-5I$

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(C) Given $A = \begin{bmatrix} 4 & 1 \ 3 & 2 \end{bmatrix}$.First,calculate $A^2 = A 	imes A = \begin{bmatrix} 4 & 1 \ 3 & 2 \end{bmatrix} \begin{bmatrix} 4 & 1 \ 3 & 2 \end{bmatrix}$.$A^2 = \begin{bmatrix} (4 	imes 4 + 1 	imes 3) & (4 	imes 1 + 1 	imes 2) \ (3 	imes 4 + 2 	imes 3) & (3 	imes 1 + 2 	imes 2) \end{bmatrix} = \begin{bmatrix} 19 & 6 \ 18 & 7 \end{bmatrix}$.Now,calculate $6A = 6 \begin{bmatrix} 4 & 1 \ 3 & 2 \end{bmatrix} = \begin{bmatrix} 24 & 6 \ 18 & 12 \end{bmatrix}$.Finally,compute $A^2 - 6A = \begin{bmatrix} 19 & 6 \ 18 & 7 \end{bmatrix} - \begin{bmatrix} 24 & 6 \ 18 & 12 \end{bmatrix} = \begin{bmatrix} -5 & 0 \ 0 & -5 \end{bmatrix}$.Since $\begin{bmatrix} -5 & 0 \ 0 & -5 \end{bmatrix} = -5 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = -5I$,the correct option is $C$.

If $A = \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,then ${A^2} - 6A = $

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