Question

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