# Solve this

Question:

If $A=\left[\begin{array}{rr}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$, then $A^{T}+A=I_{2}$, if

(a) $\theta=n \pi, n \in \mathrm{Z}$

(b) $\theta=(2 n+1) \frac{\pi}{2}, n \in Z$

(c) $\theta=2 n \pi+\frac{\pi}{3}, n \in Z$

(d) none of these

Solution:

(c) $\theta=2 n \pi+\frac{\pi}{3}, n \in Z$

Here,

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

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

Now,

$A^{T}+A=I_{2}$

$\Rightarrow\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]+\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

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

$\Rightarrow 2 \cos \theta=1$

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

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

$\Rightarrow \theta=2 n \pi \pm \frac{\pi}{3} \quad(n \in Z)$