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

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(A) Given $A = \begin{bmatrix} 3 & -2 \ 4 & -2 \end{bmatrix}$.First,calculate $A^{2} = A \cdot A = \begin{bmatrix} 3 & -2 \ 4 & -2 \end{bmatrix} \be

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$k = 1$

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(A) Given $A = \begin{bmatrix} 3 & -2 \ 4 & -2 \end{bmatrix}$.First,calculate $A^{2} = A \cdot A = \begin{bmatrix} 3 & -2 \ 4 & -2 \end{bmatrix} \begin{bmatrix} 3 & -2 \ 4 & -2 \end{bmatrix}$.$A^{2} = \begin{bmatrix} 3(3) + (-2)(4) & 3(-2) + (-2)(-2) \ 4(3) + (-2)(4) & 4(-2) + (-2)(-2) \end{bmatrix} = \begin{bmatrix} 9-8 & -6+4 \ 12-8 & -8+4 \end{bmatrix} = \begin{bmatrix} 1 & -2 \ 4 & -4 \end{bmatrix}$.Now,substitute into the equation $A^{2} = kA - 2I$:$\begin{bmatrix} 1 & -2 \ 4 & -4 \end{bmatrix} = k \begin{bmatrix} 3 & -2 \ 4 & -2 \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.$\begin{bmatrix} 1 & -2 \ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k & -2k \ 4k & -2k \end{bmatrix} - \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix}$.$\begin{bmatrix} 1 & -2 \ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k-2 & -2k \ 4k & -2k-2 \end{bmatrix}$.Comparing the corresponding elements,we get $3k - 2 = 1$.$3k = 3 \Rightarrow k = 1$.

If $A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,find $k$ such that $A^{2} = kA - 2I$.

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