Question

If $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$ then $\underset{n\to \infty }{Lt}\frac{1}{n}|A^n|=?$

• A. $1$
• B. $0$
• C. A
• D. $\frac{1}{n}A$

Answer 1

The correct option is B $0$

$A^2=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]=\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right]$

Thus we can say that $A^n=\left[\begin{array}{cc}\cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{array}\right]$

Also, $\left|A^n\right|=1$

So, ${lim}_{n\to \infty }\frac{1}{n}\left|A^n\right|={lim}_{n\to \infty }\left(\frac{1}{n}\right)=0$