Question

Let $\alpha =\frac{\pi }{5}$ and $A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$, then $B=A+A^2+A^3+A^4$ is

• A. Symmetric and Singular
• B. Symmetric and Non-singular
• C. Skew-Symmetric and Non-Singular
• D. Skew-Symmetric and Singular

Answer 1

Answer: (C)

Skew-Symmetric and Non-Singular

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

$B=A+A^2+A^3+A^4$

$B=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]+\left[\begin{array}{cc}\cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & \cos 2\alpha \end{array}\right]+\left[\begin{array}{cc}\cos 3\alpha & \sin 3\alpha \\ -\sin 3\alpha & \cos 3\alpha \end{array}\right]+\left[\begin{array}{cc}\cos 4\alpha & \sin 4\alpha \\ -\sin 4\alpha & \cos 4\alpha \end{array}\right]$

$=\left[\begin{array}{cc}\cos \alpha +\cos 2\alpha +\cos 3\alpha +\cos 4\alpha & \sin \alpha +\sin 2\alpha +\sin 3\alpha +\sin 4\alpha \\ -(\sin \alpha +\sin 2\alpha +\sin 3\alpha +\sin 4\alpha )& \cos \alpha +\cos 2\alpha +\cos 3\alpha +\cos 4\alpha \end{array}\right]$

We know

$\cos \alpha +\cos 2\alpha +\cos 3\alpha +\cos 4\alpha =\frac{\sin \frac{4\beta }{2}}{\sin \frac{\beta }{2}}\cos \left(\frac{\alpha +4\alpha }{2}\right)$

$\therefore \cos \frac{\pi }{5}+\cos \frac{2\pi }{5}+\cos \frac{3\pi }{5}+\cos \frac{4\pi }{5}=\frac{\sin \frac{2\pi }{5}}{\sin \frac{\pi }{5}\times \frac{1}{2}}\cos (\frac{5}{2}\times \frac{\pi }{5})$

$=0$

$\sin \frac{\pi }{5}+\sin \frac{2\pi }{5}+\sin \frac{3\pi }{5}+\sin \frac{4\pi }{5}=\frac{\sin \frac{2\pi }{5}}{\sin \frac{\pi }{5}\times \frac{1}{2}}\sin \frac{\pi }{2}=\frac{\sin {72}^o}{\sin {18}^o}$

$B=\left[\begin{array}{cc}0& \frac{\sin {72}^o}{\sin {18}^o}\\ \frac{-\sin {72}^o}{\sin {18}^o}& 0\end{array}\right]$

$\therefore B$ is a skew symmetric matrices and non-singular