If $A = \begin{bmatrix} \cos heta & {sin\ } heta \ - \sin heta & \cos heta \end{bmatrix}$, then for $n \in N,\ A^{n}$ is equal to

Question

If $A = \begin{bmatrix} \cos	heta & {sin\ }	heta \ - \sin	heta & \cos	heta \end{bmatrix}$, then for $n \in N,\ A^{n}$ is equal to

Dr. Abubacker · Accepted Answer

(b)
Given, $A = \begin{bmatrix}
\cos	heta & {sin\ }	heta \
 - \sin	heta & \cos	heta
\end{bmatrix}$
Now, $A^{2} = \begin{bmatrix}
\cos	heta & {sin\ }	heta \
 - \sin	heta & \cos	heta
\end{bmatrix}$ $\begin{bmatrix}
\cos	heta & {sin\ }	heta \
 - \sin	heta & \cos	heta
\end{bmatrix}$
$= \begin{bmatrix}
\cos^{2}	heta - \sin^{2}	heta & 2\sin{	heta\cos	heta} \
 - 2\sin{	heta\cos	heta} & \cos^{2}{	heta - \sin^{2}	heta}
\end{bmatrix}$
$= \begin{bmatrix}
\cos{2	heta} & \sin{2	heta} \
 - \sin{2	heta} & \cos{2	heta}
\end{bmatrix}$
$	herefore\$ By induction, $A^{n} = \begin{bmatrix}
\cos{n	heta} & \sin{n	heta} \
 - \sin{n	heta} & \cos n	heta
\end{bmatrix}$

Q. If $A = \begin{bmatrix} \cos\theta & {sin\ }\theta \\ - \sin\theta & \cos\theta \end{bmatrix}$, then for $n \in N,\ A^{n}$ is equal to

Explanation

Reviewed by Dr. Abubacker

Want more practice?