Question

If $A=\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right],n\in N$ 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

Answer: (A)

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

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

$A^{4n}=?$

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

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

$\Rightarrow A^4=I$

$\therefore A^m=I$ if and only if $m$ is a multiple of $4$

$\therefore 4n$ is a multiple of $4$

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

Option A