Question

If $A=\left[\begin{array}{cc}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]$ and $A+A^T=I$, where $I$ is the unit of matrix of $2\times 2$ and $A^T$ is the transpose of $A$, then the value of $\theta$ is equal to

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\pi$
• D. $\frac{3\pi }{2}$

Answer 1

The correct option is A $\frac{\pi }{6}$

We have, $A=\left[\begin{array}{cc}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]$

$\Rightarrow A^T=\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right]$, swap the rows and column

$\Rightarrow A+A^T=\left[\begin{array}{cc}2\cos 2\theta & 0\\ 0& 2\cos 2\theta \end{array}\right]=I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow 2\cos 2\theta =1$, since if two matrix are same, then corresponding elements should be same

$\Rightarrow \cos 2\theta =\frac{1}{2}=\cos \frac{\pi }{3}$

$\Rightarrow 2\theta =\frac{\pi }{3}$

$\therefore \theta =\frac{\pi }{6}$