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 $2\times 2$ and  $A^T$ is the transpose of $A$, then the value of $\theta$ is equal to:

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

Answer 1

The Correct Option is D

Explanation:
Given,$A=\left[\begin{array}{cc}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]$ $A+A^T=1$The transpose of matrix $A$ is given by, $\begin{array}{rl}& A^T=\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right]\end{array}$As, $A+A^T=I$ $\begin{array}{r}\left[\begin{array}{cc}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]+\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right]=I\end{array}$ $\begin{array}{rl}& \Rightarrow \left[\begin{array}{cc}2\cos 2\theta & 0\\ 0& 2\cos 2\theta \end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]\end{array}$ As we know that,If two matrices $A$ and $B$ are equal then their corresponding elements are also equal.$\therefore 2\cos 2\theta =1$$\Rightarrow \cos 2\theta =\frac{1}{2}$$\Rightarrow \cos 2\theta =\cos {60}^{\circ }$$\Rightarrow 2\theta =\frac{\pi }{3}[\because \cos {60}^{\circ }=\frac{\pi }{3}]$$\therefore \theta =\frac{\pi }{6}$So, the value of $\theta$ is $\frac{\pi }{6}$.Hence, the correct option is (D).