Question

Let $A=\left[\begin{array}{cc}0& \sqrt{-1}\\ -\sqrt{-1}& 0\end{array}\right]$ then prove $A^{10}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Answer 1

$A=\left[\begin{array}{cc}0& \sqrt{-1}\\ -\sqrt{-1}& 0\end{array}\right]A^2=\left[\begin{array}{cc}0& \sqrt{-1}\\ -\sqrt{-1}& 0\end{array}\right]\times \left[\begin{array}{cc}0& \sqrt{-1}\\ -\sqrt{-1}& 0\end{array}\right]=\left[\begin{array}{cc}-(-1)& 0\\ 0& -(-1)\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I{A}^5=A^2\cdot A^2\cdot A=I\cdot I\cdot A=A{A}^{10}=A^5\cdot A^5=A\cdot A=A^2=I\therefore A^{10}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Proved