Question

If $A=\left[\begin{array}{cc}0& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 0\end{array}\right]$ and $I$ is the unit matrix, show that $I+A=(I-A)\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$.

Answer 1

$A=\left[\begin{array}{cc}0& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 0\end{array}\right]$

$\therefore I+A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}0& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 0\end{array}\right]$

$=\left[\begin{array}{cc}1& \tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 1\end{array}\right]$

$\therefore I-A=\left[\begin{array}{cc}1& \tan \frac{\alpha }{2}\\ -\tan \frac{\alpha }{2}& 1\end{array}\right]$

$\therefore (I-A)\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]=\left[\begin{array}{cc}1& \tan \frac{\alpha }{2}\\ -\tan \frac{\alpha }{2}& 1\end{array}\right]\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$

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

$=\left[\begin{array}{cc}1& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 1\end{array}\right]$

Hence proved.