Question

If $A=\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right],nϵN$, then $A^{4n}$ equals

• A. $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right]$
• C. $\left[\begin{array}{cc}0& i\\ i& 0\end{array}\right]$
• D. $\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$A=\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right]$

$A^2=\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right]\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$

$A^4=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$A^{4n}={\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]}^n=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$