A) A null matrix
B) An identity matrix
C) \[\left[ \begin{matrix} 0 & 1 \\ -1 & 0 \\ \end{matrix} \right]\]
D) None of these
Correct Answer: A
Solution :
[a] \[{{A}^{n}}=\left[ \begin{matrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \\ \end{matrix} \right]\] \[\frac{1}{n}{{A}^{n}}=\left[ \begin{matrix} \frac{\cos n\theta }{n} & \frac{\sin n\theta }{n} \\ -\frac{\sin n\theta }{n} & \frac{\cos n\theta }{n} \\ \end{matrix} \right]\] But \[-1\le \cos n\theta \le 1\] and \[-1\le sinn\theta \le 1\] \[\underset{n\to \infty }{\mathop{\lim }}\,\frac{\sin n\theta }{n}=0,\underset{n\to \infty }{\mathop{\lim }}\,\frac{\cos n\theta }{n}=0\] \[\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}{{A}^{n}}=\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]\]
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