Question
If \[A = \left[ {\begin{array}{*{20}{c}}
{\cos \theta }&{\sin \theta }\\
{ - \sin \theta }&{\cos \theta }
\end{array}} \right]\] then $$\mathop {\lim }\limits_{x \to \infty } \frac{1}{n}{A^n}$$ is
A.
a null matrix
B.
an identity matrix
C.
\[\left[ {\begin{array}{*{20}{c}}
0&1\\
{ - 1}&0
\end{array}} \right]\]
D.
None of these
Answer :
a null matrix
Solution :
\[\begin{array}{l}
{A^n} = \left[ {\begin{array}{*{20}{c}}
{\cos \,n\theta }&{\sin \,n\theta }\\
{ - \sin \,n\theta }&{\cos \,n\theta }
\end{array}} \right]\\
\frac{1}{n}{A^n} = \left[ {\begin{array}{*{20}{c}}
{\frac{{\cos \,n\theta }}{n}}&{\frac{{\sin \,n\theta }}{n}}\\
{ - \frac{{\sin \,n\theta }}{n}}&{\frac{{\cos \,n\theta }}{n}}
\end{array}} \right]
\end{array}\]
But $$ - 1 \leqslant \cos \,n\theta \leqslant 1{\text{ and }} - 1 \leqslant \sin \,n\theta \leqslant 1$$
$$\mathop {\lim }\limits_{n \to \infty } \frac{{\sin \,n\theta }}{n} = 0,\mathop {\lim }\limits_{n \to \infty } \frac{{\cos \,n\theta }}{n} = 0$$
\[\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}{A^n} = \left[ {\begin{array}{*{20}{c}}
0&0\\
0&0
\end{array}} \right].\]