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

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$7$

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(D) Given $A = \begin{bmatrix} 2 & 3 \ 4 & 5 \end{bmatrix}$.First, calculate $A^2 = A 	imes A = \begin{bmatrix} 2 & 3 \ 4 & 5 \end{bmatrix} \begin{bmatrix} 2 & 3 \ 4 & 5 \end{bmatrix} = \begin{bmatrix} (2 	imes 2 + 3 	imes 4) & (2 	imes 3 + 3 	imes 5) \ (4 	imes 2 + 5 	imes 4) & (4 	imes 3 + 5 	imes 5) \end{bmatrix} = \begin{bmatrix} 16 & 21 \ 28 & 37 \end{bmatrix}$.Now, calculate $A^2 - 2I = \begin{bmatrix} 16 & 21 \ 28 & 37 \end{bmatrix} - \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} 14 & 21 \ 28 & 35 \end{bmatrix}$.We can factor out $7$ from the resulting matrix: $\begin{bmatrix} 14 & 21 \ 28 & 35 \end{bmatrix} = 7 \begin{bmatrix} 2 & 3 \ 4 & 5 \end{bmatrix} = 7A$.Comparing this with $A^2 - 2I = KA$, we get $K = 7$.

For matrix $A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$, if $A^2 - 2I = KA$, then $K = \dots$

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