Question

If $A=\left(\begin{array}{cc}i& -i\\ -i& i\end{array}\right)$ and $B=\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right),$ then $A^8$ equals

• A. $128B$
• B. $23B$
• C. $16B$
• D. $64B$

Answer 1

The correct option is A $128B$

$A=\left(\begin{array}{cc}i& -i\\ -i& i\end{array}\right)$
$A^2=\left(\begin{array}{cc}i& -i\\ -i& i\end{array}\right)\left(\begin{array}{cc}i& -i\\ -i& i\end{array}\right)$$\Rightarrow A^2=\left(\begin{array}{cc}-2& 2\\ 2& -2\end{array}\right)$

$\Rightarrow A^4=\left(\begin{array}{cc}8& -8\\ -8& 8\end{array}\right)$

$\Rightarrow A^8=\left(\begin{array}{cc}128& -128\\ -128& 128\end{array}\right)=128\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$
$\therefore A^8=128B$