Question

If $A=\left(\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \sin \alpha \end{array}\right)$, then $A^{10}=$

• A. $\left(\begin{array}{cc}\cos 10\alpha & -\sin 10\alpha \\ \sin 10\alpha & \cos 10\alpha \end{array}\right)$
• B. $\left(\begin{array}{cc}\cos 10\alpha & \sin 10\alpha \\ -\sin 10\alpha & \cos 10\alpha \end{array}\right)$
• C. $\left(\begin{array}{cc}\cos 10\alpha & -\sin 10\alpha \\ -\sin 10\alpha & -\cos 10\alpha \end{array}\right)$
• D. $\left(\begin{array}{cc}-\cos 10\alpha & -\sin 10\alpha \\ -\sin 10\alpha & -\cos 10\alpha \end{array}\right)$

Answer 1

The correct option is B $\left(\begin{array}{cc}\cos 10\alpha & \sin 10\alpha \\ -\sin 10\alpha & \cos 10\alpha \end{array}\right)$

$A=\left({}_{-\sin \alpha }^{\cos \alpha }{}_{\cos \alpha }^{\sin \alpha }\right)$

$A^2=\left({}_{-\sin \alpha }^{\cos \alpha }{}_{\cos \alpha }^{\sin \alpha }\right)\left({}_{-\sin \alpha }^{\cos \alpha }{}_{\cos \alpha }^{\sin \alpha }\right)$

$A^2=\left({}_{-\sin 2\alpha }^{\cos 2\alpha }{}_{\cos 2\alpha }^{\sin 2\alpha }\right)$

By using mathematical induction we've,

So$,$ $A^{10}=\left({}_{-\sin 10\alpha }^{\cos 10\alpha }{}_{\cos 10\alpha }^{\sin 10\alpha }\right)$.