Question

If $A=\left[\begin{array}{cc}cos\theta & -sin\theta \\ sin\theta & cos\theta \end{array}\right]$, then $A^T+A=I_2$, if

• A. $\theta =n\pi ,nϵZ$
• B. $\theta =(2n+1)\frac{\pi }{2},nϵZ$
• C. $\theta =2n\pi +\frac{\pi }{3},nϵZ$
• D. none of these

Answer 1

Answer: (C)

$\theta =2n\pi +\frac{\pi }{3},nϵZ$

$A=\left[\begin{array}{cc}cos\theta & -sin\theta \\ sin\theta & cos\theta \end{array}\right]$
$A^T+A=\left[\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right]+\left[\begin{array}{cc}cos\theta & -sin\theta \\ sin\theta & cos\theta \end{array}\right]$
$\Rightarrow A^T+A=\left[\begin{array}{cc}2cos\theta & 0\\ 0& 2cos\theta \end{array}\right]$
Given, $A^T+A=I_2$
$\Rightarrow \left[\begin{array}{cc}2cos\theta & 0\\ 0& 2cos\theta \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$\Rightarrow 2\cos \theta =1$
$\Rightarrow \cos \theta =\cos \frac{\pi }{3}$

$\Rightarrow \theta =2n\pi +\frac{\pi }{3},n\in Z$