Determinants Matrices Question 143
Question: If $ A= \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{bmatrix} $ then $ \underset{n\to \infty }{\mathop{\lim }}\frac{1}{n}A^{n} $ is
Options:
A) A null matrix
B) An identity matrix
C) $ \begin{bmatrix} 0 & 1 \\ -1 & 0 \\ \end{bmatrix} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] $ A^{n}= \begin{bmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \\ \end{bmatrix} $
$ \frac{1}{n}A^{n}= \begin{bmatrix} \frac{\cos n\theta }{n} & \frac{\sin n\theta }{n} \\ -\frac{\sin n\theta }{n} & \frac{\cos n\theta }{n} \\ \end{bmatrix} $ But $ -1\le \cos n\theta \le 1 $ and $ -1\le sinn\theta \le 1 $
$ \underset{n\to \infty }{\mathop{\lim }}\frac{\sin n\theta }{n}=0,\underset{n\to \infty }{\mathop{\lim }}\frac{\cos n\theta }{n}=0 $
$ \underset{n\to \infty }{\mathop{\lim }}\frac{1}{n}A^{n}= \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix} $
BETA
