Question

$A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$, then $A^n=\dots$

• A. $\left[\begin{array}{cc}ncosn\theta & sinn\theta \\ -sinn\theta & cosn\theta \end{array}\right]$
• B. $\left[\begin{array}{cc}cosn\theta & cos\theta \\ -sin\theta & sin\theta \end{array}\right]$
• C. $\left[\begin{array}{cc}cosn\theta & sinn\theta \\ -sinn\theta & cosn\theta \end{array}\right]$
• D. $\left[\begin{array}{cc}cos(n+1)\theta & sin(n+1)\theta \\ -sin(n+l)\theta & sin(n+1)\theta \end{array}\right]$

Answer 1

The correct option is C $\left[\begin{array}{cc}cosn\theta & sinn\theta \\ -sinn\theta & cosn\theta \end{array}\right]$

Given , $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
$A^2=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
$=\left[\begin{array}{cc}{\cos }^2\theta -{\sin }^2\theta & 2\sin \theta \cos \theta \\ -2\sin \theta \cos \theta & {\cos }^2\theta -{\sin }^2\theta \end{array}\right]$
$=\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right]$
$A^3=A^{2}A=\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
$=\left[\begin{array}{cc}\cos 2\theta \cos \theta -\sin 2\theta \sin \theta & \cos 2\theta \sin \theta +\sin 2\theta \cos \theta \\ -\cos 2\theta \sin \theta -\sin 2\theta \cos \theta & \cos 2\theta \cos \theta -\sin 2\theta \sin \theta \end{array}\right]$
$=\left[\begin{array}{cc}\cos 3\theta & \sin 3\theta \\ -\sin 3\theta & \cos 3\theta \end{array}\right]$
Hence, $A^n=\left[\begin{array}{cc}\cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{array}\right]$
Alternative Method:
We will prove by Mathematical induction method
Given , $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
Hence, the result is true for n=1.
Now, let the result is true for n=k
Then $A^k=\left[\begin{array}{cc}\cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{array}\right]$
So, we will prove for n=k+1.
$A^{k+1}=A^{k}A$
$=\left[\begin{array}{cc}\cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
$=\left[\begin{array}{cc}\cos k\theta \cos \theta -\sin k\theta \sin \theta & \cos k\theta \sin \theta +\sin k\theta \cos \theta \\ -\cos k\theta \sin \theta -\sin k\theta \cos \theta & \cos k\theta \cos \theta -\sin k\theta \sin \theta \end{array}\right]=\left[\begin{array}{cc}\cos (k+1)\theta & \sin (k+1)\theta \\ -\sin (k+1)\theta & \cos (k+1)\theta \end{array}\right]$
Hence, the result is true for n=k+1.
So, by principle of mathematical induction , the result is true for all $n\in N$
$A^n=\left[\begin{array}{cc}\cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{array}\right]$