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(n -1)

Answer 1

Answer: (D)

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