Matrices and Determinants 1 Question 3
3. Let $A=\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array},(\alpha \in R)$ such that $A^{32}=\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}$. Then, a value of $\alpha$ is
(a) $\frac{\pi}{32}$
(b) 0
(c) $\frac{\pi}{64}$
(d) $\frac{\pi}{16}$
(2019 Main, 8 April I)
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Answer:
Correct Answer: 3. (c)
Solution:
- Given, matrix $A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
$ \begin{aligned} & \left.\therefore A^2=\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] \begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] \\ & =\left[\begin{array}{cc} \cos ^2 \alpha-\sin ^2 \alpha & -\cos \alpha \sin \alpha-\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha+\cos \alpha \sin \alpha & -\sin ^2 \alpha+\cos ^2 \alpha \end{array}\right] \\ & =\left[\begin{array}{cc} \cos 2 \alpha & -\sin 2 \alpha \\ \sin 2 \alpha & \cos 2 \alpha \end{array}\right] \end{aligned} $
Similarly,
$ \begin{aligned} A^n & =\left[\begin{array}{cc} \cos (n \alpha) & -\sin (n \alpha) \\ \sin (n \alpha) & \cos (n \alpha) \end{array}\right], n \in N \\ \Rightarrow A^{32} & =\left[\begin{array}{cc} \cos (32 \alpha) & -\sin (32 \alpha) \\ \sin (32 \alpha) & \cos (32 \alpha) \end{array}\right]=\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] \text { (given) } \end{aligned} $
So, $\cos (32 \alpha)=0$ and $\sin (32 \alpha)=1$
$ \Rightarrow \quad 32 \alpha=\frac{\pi}{2} \Rightarrow \alpha=\frac{\pi}{64} $
