If A = 1 2 1 3 - 3 1 then,

Given $\mathrm{A}=\frac{1}{2}\left[\begin{array}{cc}1& \sqrt{3}\\ -\sqrt{3}& 1\end{array}\right]$

$A^2=\frac{1}{4}\left[\begin{array}{cc}1& \sqrt{3}\\ -\sqrt{3}& 1\end{array}\right]\left[\begin{array}{cc}1& \sqrt{3}\\ -\sqrt{3}& 1\end{array}\right]$

$=\frac{1}{4}\left[\begin{array}{cc}1-3 & 2\sqrt{3}\\ -2\sqrt{3}& -3+1\end{array}\right]$

$A^2=\frac{1}{2}\left[\begin{array}{cc}-1 & \sqrt{3}\\ -\sqrt{3}& -1\end{array}\right]$

$A^3=\frac{1}{4}\left[\begin{array}{cc}-1 & \sqrt{3}\\ -\sqrt{3}& -1\end{array}\right]\left[\begin{array}{cc}-1 & \sqrt{3}\\ -\sqrt{3}& -1\end{array}\right]$

$\Rightarrow A^3=\frac{1}{4}\left[\begin{array}{cc}-4 & 0\\ 0& -4\end{array}\right]=-\left[\begin{array}{cc}1 & 0\\ 0& 1\end{array}\right]$

$\Rightarrow \boxed{A^3=-I} .....\left(1\right)$

Now from equation $\left(1\right)$,

$A^{30}=(-I{)}^{10}=I .....(2)$

Again from equation $\left(1\right)$,

$A^{25}={\left(A^3\right)}^8.A={\left(-I\right)}^8.A$

$\Rightarrow A^{25}=A$

$\Rightarrow A^{25}-A=0 ......\left(3\right)$

On adding equations $\left(2\right)$ and $\left(3\right)$ we get,

$A^{30}+A^{25}-A=I$