Question

Choose the correct answer in the following question:

If $A=\left[\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right]$, then A+A'=I, if the value of $\alpha$ is

$\left(a\right)\frac{\pi }{6}\left(b\right)\frac{\pi }{3}\left(c\right)\pi \left(d\right)\frac{3\pi }{2}$

Answer 1

Here, $A=\left[\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right]$
and $A^{\prime }=\left[\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right]=\left[\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right]$
Given, A+A'=I
$\therefore ,\left[\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right]+\left[\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\Rightarrow \left[\begin{array}{cc}2cos\alpha & 0\\ 0& 2cos\alpha \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
Comparing the corresponding elements of the above matrices, we have
$2cos\alpha =1\Rightarrow cos\alpha =\frac{1}{2}=cos\frac{\pi }{3}\Rightarrow \alpha =\frac{\pi }{3}$. So, correct option is (b).