Question

If $n$ is a positive integer and if $A=\left\{\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right\}$ , then $A^n=$

• A. $\left\{\begin{array}{cc}cos\theta & -sin\theta \\ sin\theta & cos\theta \end{array}\right\}$
• B. $\left\{\begin{array}{cc}cos\theta & -s\dot{m}\theta \\ -s\dot{m}\theta & cos\theta \end{array}\right\}$
• C. $\left\{\begin{array}{cc}cosn\theta & sinn\theta \\ -sinn\theta & cosn\theta \end{array}\right\}$
• D. $\left\{\begin{array}{cc}cosn\theta & cosn\theta \\ s\dot{m}\theta & sinn\theta \end{array}\right\}$

Answer 1

Answer: (C)

$\left\{\begin{array}{cc}cosn\theta & sinn\theta \\ -sinn\theta & cosn\theta \end{array}\right\}$

$A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
$A^2=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
$=\left[\begin{array}{cc}{\cos }^2\theta -{\sin }^2\theta & 2\sin \theta \cos \theta \\ -2\sin \theta \cos \theta & {\cos }^2\theta -{\sin }^2\theta \end{array}\right]$
$=\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right]$
Assume its true For n-1 $\therefore$ for n
$A^n=\left[\begin{array}{cc}\cos (n-1)\theta & \sin (n-1)\theta \\ -\sin (n-1)\theta & \cos (n-1)\theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
$=\left[\begin{array}{cc}\cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{array}\right]$
So by induction ans is C.