Question

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

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

Answer 1

The correct option is D $2(n-1)$

$A=\left[\begin{array}{cc}2& -2\\ -2& 2\end{array}\right]$

$\text{Then}{A}^2=\left[\begin{array}{cc}2& -2\\ -2& 2\end{array}\right]\left[\begin{array}{cc}2& -2\\ -2& 2\end{array}\right]$

$=\left[\begin{array}{cc}8& -8\\ -8& 8\end{array}\right]=4\left[\begin{array}{cc}2& -2\\ -2& 2\end{array}\right]=4A\cdots \left(i\right)$

$A^3=A^{2}A=4A\cdot A\left[\text{From}\right(i\left)\right]$
$=4A^2=4\cdot 4A=16A=2^{4}A=2^{2(3-1)}A$

$A^4=A^{3}A=16A.A$
$=16A^2=64A=2^{6}A=2^{2(4-1)}A$
$⋮⋮$
$A^n=2^{2(n-1)}A$
$\therefore k=2(n-1)$