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

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(D) Given $A = \begin{bmatrix} 2 & -2 \ -2 & 2 \end{bmatrix}$.First,calculate $A^2$:$A^2 = \begin{bmatrix} 2 & -2 \ -2 & 2 \end{bmatrix} \begin{bmat

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$2^{n-1}$

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(D) Given $A = \begin{bmatrix} 2 & -2 \ -2 & 2 \end{bmatrix}$.First,calculate $A^2$:$A^2 = \begin{bmatrix} 2 & -2 \ -2 & 2 \end{bmatrix} \begin{bmatrix} 2 & -2 \ -2 & 2 \end{bmatrix} = \begin{bmatrix} 4+4 & -4-4 \ -4-4 & 4+4 \end{bmatrix} = \begin{bmatrix} 8 & -8 \ -8 & 8 \end{bmatrix} = 4 \begin{bmatrix} 2 & -2 \ -2 & 2 \end{bmatrix} = 2^2 A$.Next,calculate $A^3$:$A^3 = A^2 \cdot A = (2^2 A) \cdot A = 2^2 A^2 = 2^2 (2^2 A) = 2^4 A$.By observing the pattern:$A^1 = 2^0 A$$A^2 = 2^2 A$$A^3 = 2^4 A$$A^4 = 2^6 A$In general,$A^n = 2^{2(n-1)} A$.Comparing this with $A^n = 2^k A$,we get $k = 2(n-1)$.

If $A = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix}$,then $A^n = 2^k A$,where $k = $

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