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If \( A=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right] \), then \( A^{n}=2^{k} A \), where \( K= \)
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The correct answer is:
\( 2(n-1) \)
Given that \( A=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right] \)
and \( A^{n}=2^{k} A \)
So, \( A^{2}=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right] \)
\( A^{3}=\left[\begin{array}{cc}8 & -8 \\ -8 & 8\end{array}\right]=2^{2} A \)
\( =\left[\begin{array}{cc}32 & -32 \\ -32 & 32\end{array}\right]=2^{4} A \)
\( A^{4}=\left[\begin{array}{cc}32 & -32 \\ -32 & 32\end{array}\right]\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right] \)
\( =\left[\begin{array}{cc}128 & -128 \\ -128 & 128\end{array}\right]=2^{6} A \)
So, \( k=2(n-1) \)
and \( A^{n}=2^{k} A \)
So, \( A^{2}=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right] \)
\( A^{3}=\left[\begin{array}{cc}8 & -8 \\ -8 & 8\end{array}\right]=2^{2} A \)
\( =\left[\begin{array}{cc}32 & -32 \\ -32 & 32\end{array}\right]=2^{4} A \)
\( A^{4}=\left[\begin{array}{cc}32 & -32 \\ -32 & 32\end{array}\right]\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right] \)
\( =\left[\begin{array}{cc}128 & -128 \\ -128 & 128\end{array}\right]=2^{6} A \)
So, \( k=2(n-1) \)
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