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If \( A=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right] \), then \( A^{n}=2^{k} A \), where \( k= \) 

(a) \( 2^{n-1} \) 

(b) \( n+1 \) 

(c) \( n-1 \) 

(d) 2

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1 Answer

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Correct option is (A) 2(n - 1)

\(A = \begin{bmatrix}2&-2\\-2&2\end{bmatrix}\) = 20A = 22-2A = 22x1 - 2A

\(A^2=\)\(\begin{bmatrix}8&-8\\-8&8\end{bmatrix}\) = \(4\begin{bmatrix}2&-2\\-2&2\end{bmatrix}\) = 22 A = 22x2-2A

\(A^3 = \begin{bmatrix}32&-32\\-2&2\end{bmatrix}\) = 24A = 22x3-2A

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An = 22n - 2A

∴ k = 2n - 2 = 2(n-1)

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