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 order $2\times 2$ 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 $\frac{\pi }{6}$

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

$\text{Given,}A+A^T=I$

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

On comparing, we get
$2\cos 2\theta =1$
$\Rightarrow \cos 2\theta =\frac{1}{2}$
$\Rightarrow \theta =\frac{\pi }{6}$