Question

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

Answer 1

L.H.S.
$=I+A$
$=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}0& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 0\end{array}\right]$
$=\left[\begin{array}{cc}1& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 1\end{array}\right]...\left(1\right)$
and R.H.S.
$=(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$
$=(\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}0& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 0\end{array}\right])\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$
$=\left[\begin{array}{cc}1& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 1\end{array}\right]\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$
$=\left[\begin{array}{cc}\cos \alpha +\sin \alpha \tan \frac{\alpha }{2}& -\sin \alpha +\cos \alpha \tan \frac{\alpha }{2}\\ -\cos \alpha \tan \frac{\alpha }{2}+\sin \alpha & \sin \alpha \tan \frac{\alpha }{2}+\cos \alpha \end{array}\right]$
$=\left[\begin{array}{cc}1-2{sin}^2\frac{\alpha }{2}+2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\tan \frac{\alpha }{2}& -2\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2}+(2{cos}^2\frac{\alpha }{2}-1)\tan \frac{\alpha }{2}\\ -(2{cos}^2\frac{\alpha }{2}-1)\tan \frac{\alpha }{2}+2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}& 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\tan \frac{\alpha }{2}+1-2{sin}^2\frac{\alpha }{2}\end{array}\right]$
$=\left[\begin{array}{cc}1-2{sin}^2\frac{\alpha }{2}+2{sin}^2\frac{\alpha }{2}& -2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}+2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}-\tan \frac{\alpha }{2}\\ -2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\tan \frac{\alpha }{2}+2\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2}& 2{sin}^2\frac{\alpha }{2}+1-2{sin}^2\frac{\alpha }{2}\end{array}\right]$
$=\left[\begin{array}{cc}1& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 1\end{array}\right]$
Clearly L.H.S. $=$ R.H.S. Hence proved