Question

If $A=\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right]$, then $A^{27}=$

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

Answer 1

The correct option is C $\left[\begin{array}{cc}-i& 0\\ 0& -i\end{array}\right]$

$A^2=\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right]\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right]=\left[\begin{array}{cc}{i}^2& 0\\ 0& i^2\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$
$A^3=A^2\times A=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\left[\begin{array}{cc}i& 0\\ 0& i\end{array}\right]=\left[\begin{array}{cc}-i& 0\\ 0& -i\end{array}\right]$
$A^4=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$
$\Rightarrow A^{4n}=I^n=I\therefore A^{24}=I$
$\therefore A^{27}=A^{24}\times A^3=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}-i& 0\\ 0& -i\end{array}\right]=\left[\begin{array}{cc}-i& 0\\ 0& -i\end{array}\right]$