If A = frac{1}{2} begin{bmatrix} 1 & sqrt{3} | vedclass

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(C) Given $A = \begin{bmatrix} \cos 60^{\circ} & \sin 60^{\circ} \ -\sin 60^{\circ} & \cos 60^{\circ} \end{bmatrix}$.Let $\alpha = 60^{\circ} = \frac

Vedclass · Accepted Answer

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

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(C) Given $A = \begin{bmatrix} \cos 60^{\circ} & \sin 60^{\circ} \ -\sin 60^{\circ} & \cos 60^{\circ} \end{bmatrix}$.Let $\alpha = 60^{\circ} = \frac{\pi}{3}$.Then $A = \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix}$.By the property of rotation matrices,$A^n = \begin{bmatrix} \cos(n\alpha) & \sin(n\alpha) \ -\sin(n\alpha) & \cos(n\alpha) \end{bmatrix}$.For $A^{30}$,$n = 30$,so $n\alpha = 30 	imes \frac{\pi}{3} = 10\pi$.$A^{30} = \begin{bmatrix} \cos(10\pi) & \sin(10\pi) \ -\sin(10\pi) & \cos(10\pi) \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I$.For $A^{25}$,$n = 25$,so $n\alpha = 25 	imes \frac{\pi}{3} = 8\pi + \frac{\pi}{3}$.$A^{25} = \begin{bmatrix} \cos(8\pi + \frac{\pi}{3}) & \sin(8\pi + \frac{\pi}{3}) \ -\sin(8\pi + \frac{\pi}{3}) & \cos(8\pi + \frac{\pi}{3}) \end{bmatrix} = \begin{bmatrix} \cos(\frac{\pi}{3}) & \sin(\frac{\pi}{3}) \ -\sin(\frac{\pi}{3}) & \cos(\frac{\pi}{3}) \end{bmatrix} = A$.Thus,$A^{30} = I$ and $A^{25} = A$.Checking the options: $A^{30} + A^{25} - A = I + A - A = I$. Therefore,option $C$ is correct.

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

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