Question

For $\theta =\frac{2\pi }{5},letB=\left[b_{ij}\right]$ be a square matrix of order 2 such that
$b_{ij}$ $=\left\{\begin{array}{cc}\cos \theta ,& i=j\\ \cos (\frac{j\pi }{2}+\theta ),& i>j\\ \cos (\frac{i\pi }{2}-\theta ),& i<j\end{array}\right\}$
The trace of matrix $B^5$ is equal to

• A. -2
• B. -1
• C. 1
• D. 2

Answer 1

The correct option is D 2

$B=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
$B^2=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]=\left[\begin{array}{cc}{\cos }^2\theta -{\sin }^2\theta & \cos \theta \sin \theta +\sin \theta \cos \theta \\ -\cos \theta \sin \theta -\sin \theta \cos \theta & -{\sin }^2\theta +{\cos }^2\theta \end{array}\right]=\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right],\text{Similarly},B^5=\left[\begin{array}{cc}\cos 5\theta & \sin 5\theta \\ -\sin 5\theta & \cos 5\theta \end{array}\right]$
So, $Tr\left(B^5\right)=2\cos 5\theta$
at $\theta =\frac{2\pi }{5}=2\times 1=2$