If $A=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ then $A^{2005}$ is 

Given $A=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ we have

$\begin{array}{l}{A}^2=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)=-I-\left(1\right)\\ \Rightarrow A^4=A^2{A}^2=(-I)(-I)\text{ [using (1) ] }-\left(2\right)\end{array}$

Now $\begin{array}{l}{\mathrm{A}}^{2005}={\mathrm{A}}^{2004}\cdot \mathrm{A}=\mathrm{I}\end{array}$

$={\left({\mathrm{A}}^4\right)}^{501}\cdot \mathrm{A}={\mathrm{I}}^{501}\cdot \mathrm{A}\left[\mathrm{Using}\right(2\left)\right]$

$=A$