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  4. If A=(1/2)[1 √3 -√3 1], then :

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Q. If $A=\frac{1}{2}\begin{bmatrix}1 & \sqrt{3} \\ -\sqrt{3} & 1\end{bmatrix}$, then :

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Solution:

$A =\frac{1}{2}\begin{bmatrix}1 & \sqrt{3} \\ -\sqrt{3} & 1\end{bmatrix}$
$A=\begin{bmatrix} \cos 60^{\circ} & \sin 60^{\circ} \\ -\sin 60^{\circ} & \cos 60^{\circ} \end{bmatrix}$
If $A=\begin{bmatrix}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{bmatrix}$ Here $\alpha=\frac{\pi}{3}$
$A^2=\begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}\begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$
$=\begin{bmatrix} \cos 2 \alpha & \sin 2 \alpha \\ -\sin 2 \alpha & \cos 2 \alpha \end{bmatrix}$
$A^{30}=\begin{bmatrix} \cos 30 \alpha & \sin 30 \alpha \\ -\sin 30 \alpha & \cos 30 \alpha \end{bmatrix}$
$ A ^{30}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}= I$
$A^{25}=\begin{bmatrix}\cos 25 \alpha & \sin 25 \alpha \\ -\sin 25 \alpha & \cos 25 \alpha\end{bmatrix}=\begin{bmatrix}\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{-\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}$
$ A ^{25}= A $
$ A ^{25}- A =0$