Question

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

• A. $A^{30}-A^{25}=2I$
• B. $A^{30}=A^{25}$
• C. $A^{30}+A^{25}-A=I$
• D. $A^{30}+A^{25}+A=I$

Answer 1

Answer: (C)

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