If A = (1)/(2)[ cases 1 & 3 - 3 & 1 cases ] then,

Given A = \dfrac{1}{2}\left[\begin{cases}1 & \sqrt{3} \\ - \sqrt{3} & 1\end{cases}\right]

{A}^{2} = \dfrac{1}{4}\left[\begin{cases}1 & \sqrt{3} \\ - \sqrt{3} & 1\end{cases}\right]\left[\begin{cases}1 & \sqrt{3} \\ - \sqrt{3} & 1\end{cases}\right]

= \dfrac{1}{4}\left[\begin{cases}1 - 3 & 2\sqrt{3} \\ - 2\sqrt{3} & - 3 + 1\end{cases}\right]

{A}^{2} = \dfrac{1}{2}\left[\begin{cases}- 1 & \sqrt{3} \\ - \sqrt{3} & - 1\end{cases}\right]

{A}^{3} = \dfrac{1}{4}\left[\begin{cases}- 1 & \sqrt{3} \\ - \sqrt{3} & - 1\end{cases}\right]\left[\begin{cases}- 1 & \sqrt{3} \\ - \sqrt{3} & - 1\end{cases}\right]

⇒ {A}^{3} = \dfrac{1}{4}\left[\begin{cases}- 4 & 0 \\ 0 & - 4\end{cases}\right] = - \left[\begin{cases}1 & 0 \\ 0 & 1\end{cases}\right]

$⇒ {A}^{3} = - I . . . . . ( 1 )$

Now from equation \left(1\right)$,</p><p>${A}^{30} = ( - I{)}^{10} = I . . . . . ( 2 )$</p><p>Again from equation \left(1\right)$,

${A}^{25} = {\left({A}^{3}\right)}^{8} . A = {\left(- I\right)}$^{8} . A$</p><p>$⇒ {A}^{25} = A$</p><p>$⇒ {A}^{25} - A = 0 . . . . . . ( 3 )$</p><p>On adding equations \left(2\right) and \left(3\right)$ we get,

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