Question

Let $\alpha =\pi /5$ and $A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\text{and}B=A+A^2+A^3+A^4$, then which of the following is/are true for $B$ :

• A. $B$ is symmetric matrix
• B. $B$ is null matrix
• C. $B$ is Skew-symmetric
• D. $Tr\left(B\right)=0$

Answer 1

Answer: (C), (D)

$Tr\left(B\right)=0$

Given $A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$ and $\alpha =\pi /5$

$A^2=\left[\begin{array}{cc}\cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & \cos 2\alpha \end{array}\right]$

$A^3=\left[\begin{array}{cc}\cos 3\alpha & \sin 3\alpha \\ -\sin 3\alpha & \cos 3\alpha \end{array}\right]$

and $A^4=\left[\begin{array}{cc}\cos 4\alpha & \sin 4\alpha \\ -\sin 4\alpha & \cos 4\alpha \end{array}\right]$

We have $\cos \alpha +\cos 2\alpha +\cos 3\alpha +\cos 4\alpha$
$=\cos \alpha +\cos 2\alpha +\cos (\pi -2\alpha )+\cos (\pi -\alpha )[\because 5\alpha =\pi ]$
$=\cos \alpha +\cos 2\alpha -\cos 2\alpha -\cos \alpha =0$
and $\sin \alpha +\sin 2\alpha +\sin 3\alpha +\sin 4\alpha$
$=\sin \alpha +\sin 2\alpha +\sin (\pi -2\alpha )+\sin (\pi -\alpha )$
$=2[\sin \alpha +\sin 2\alpha ]$

$=2\left\{2\sin \left[\frac{3\alpha }{2}\right]\cos \frac{\alpha }{2}\right\}=4\sin \left[\frac{3\pi }{10}\right]\cos \frac{\pi }{10}$
$=4\sin [\frac{\pi }{2}-\frac{\pi }{5}]\cos \frac{\pi }{10}$
$=4\cos \frac{\pi }{5}\cos \frac{\pi }{10}=a\left(say\right)$

Thus, $B=\left[\begin{array}{cc}0& a\\ -a& 0\end{array}\right]$
$\therefore B$ is skew-symmetric.
$Tr\left(B\right)=0+0=0$