If $$A = {1 \over 2}\left[ {\matrix{ 1 & {\sqrt 3 } \cr { - \sqrt 3 } & 1 \cr } } \right]$$, then :

$$A = {1 \over 2}\left[ {\matrix{ 1 & {\sqrt 3 } \cr { - \sqrt 3 } & 1 \cr } } \right]$$ <br/><br/>Let $\theta=\frac{\pi}{3}$ <br/><br/>$$ \begin{aligned} A^2 & =\left[\begin{array}{cc} \cos \theta & \sin \theta \\\\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc} \cos \theta & \sin \theta \\\\ -\sin \theta & \cos \theta \end{array}\right] \\\\ & =\left[\begin{array}{cc} \cos 2 \theta & \sin 2 \theta \\\\ -\sin 2 \theta & \cos 2 \theta \end{array}\right] \\\\ A^3 & =\left[\begin{array}{cc} \cos 2 \theta & \sin 2 \theta \\\\ -\sin 2 \theta & \cos 2 \theta \end{array}\right]\left[\begin{array}{cc} \cos \theta & \sin \theta \\\\ -\sin \theta & \cos \theta \end{array}\right] \\\\ & =\left[\begin{array}{cc} \cos 3 \theta & \sin 3 \theta \\\\ -\sin 3 \theta & \cos 3 \theta \end{array}\right] \end{aligned} $$ <br/><br/>$\begin{aligned} & \therefore \quad A^{30}=\left[\begin{array}{cc}\cos 30 \theta & \sin 30 \theta \\\\ -\sin 30 \theta & \cos 30 \theta\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\\\ & A^{25}=\left[\begin{array}{cc}\cos 25 \theta & \sin 25 \theta \\\\ -\sin 25 \theta & \cos 25 \theta\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}1 & \sqrt{3} \\ -\sqrt{3} & 1\end{array}\right]=A \\\\ & \therefore \quad A^{30}+A^{25}-A=I\end{aligned}$