Question

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

Answer 1

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

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

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

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

.

.

Aⁿ = 2^{2n - 2}A

∴ k = 2n - 2 = 2(n-1)