Question

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

Answer 1

Since $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ and given

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

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

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

R.H.S$=(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \sin \alpha \end{array}\right]$

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

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

Let $\tan \left(\frac{\alpha }{2}\right)=\lambda$, then

R.H.S$=\left[\begin{array}{cc}1& \lambda \\ -\lambda & 1\end{array}\right]\left[\begin{array}{cc}\frac{1-{\lambda }^2}{1+{\lambda }^2}& \frac{-2\lambda }{1+{\lambda }^2}\\ \frac{2\lambda }{1+{\lambda }^2}& \frac{1-{\lambda }^2}{1+{\lambda }^2}\end{array}\right]$

$=\left[\begin{array}{cc}\frac{1-{\lambda }^2+2{\lambda }^2}{1+{\lambda }^2}& \frac{-2\lambda +\lambda (1-{\lambda }^2)}{1+{\lambda }^2}\\ \frac{-\lambda (1-{\lambda }^2)+2\lambda }{1+{\lambda }^2}& \frac{2{\lambda }^2+1-{\lambda }^2}{1+{\lambda }^2}\end{array}\right]$

$=\left[\begin{array}{cc}\frac{1+{\lambda }^2}{1+{\lambda }^2}& \frac{-\lambda (1+{\lambda }^2)}{1+{\lambda }^2}\\ \frac{\lambda (1+{\lambda }^2)}{1+{\lambda }^2}& \frac{2{\lambda }^2+1-{\lambda }^2}{1+{\lambda }^2}\end{array}\right]$

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

$=I+A$

$=$LHS

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

Hence Proved