Question

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

• A. A null matrix
• B. An identity matrix
• C. $\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$
• D. None of these

Answer 1

Answer: (A)

A null matrix

$A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]{lim}_{n\to \infty }\frac{1}{n}{A}^n$

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

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

Now,

${lim}_{n\to \infty }\frac{1}{n}{A}^n={lim}_{n\to \infty }\frac{1}{n}\left[\begin{array}{cc}\cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{array}\right]$

$=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=$Null matrix

Proof for $A^n=\left[\begin{array}{cc}\cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{array}\right]=P\left(n\right)$

$P\left(n\right)$is true for $n=1$

For $n=k,k\ge 1$

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

$A^{k+1}=A^{k}A$

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

$\Rightarrow \left[\begin{array}{cc}\cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{array}\right]$

$\therefore P\left(n\right)$ is true for $n=k+1(k\ge 1)$