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 matrix of $2\times 2\&A^T$ 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

Answer: (A)

We have, $A=\left[\begin{array}{cc}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]$
$\therefore A^T=\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right]$
Now, $A+A^T=\left[\begin{array}{cc}\cos 2\theta +\cos 2\theta & -\sin 2\theta +\sin 2\theta \\ \sin 2\theta -\sin 2\theta & \cos 2\theta +\cos 2\theta \end{array}\right]$
$=\left[\begin{array}{cc}2\cos 2\theta & 0\\ 0& 2\cos 2\theta \end{array}\right]$
$\therefore \left[\begin{array}{cc}2\cos 2\theta & 0\\ 0& 2\cos 2\theta \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\because A+A^T=I\right]$
$\Rightarrow 2\cos 2\theta =1$
$\Rightarrow \cos 2\theta =\frac{1}{2}$
$\Rightarrow 2\theta =\frac{\pi }{3}$
$\therefore \theta =\frac{\pi }{6}$