Question

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

• A. $2^{n -1}$
• B. n + 1
• C. n - 1
• D. 2(n -1)

Answer 1

The Correct Option is D

$A^{2} = \begin{bmatrix}2&-2\\ -2&2\end{bmatrix}\begin{bmatrix}2&-2\\ -2&2\end{bmatrix} =\begin{bmatrix}8&-8\\ -8&8\end{bmatrix} =4A = 2^{2}A $
$ A^{3} =A^{2} .A = 4A .A = 4 \left(4A\right) = 16A = 2^{4} A$
$ A^{4}= A^{3} .A = 16A.A= 16\left(4A\right) =64A=2^{6}A $
$ \therefore $ By inspection $k = 2(n - 1)$