Question

If $A=\left[\begin{array}{cc}0& -\tan \alpha /2\\ \tan \alpha /2& 0\end{array}\right]$ and $I$ is a $2\times 2$ unit matrix then is

$I+A=(I-A)\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$ If true enter 1 else enter 0.

Answer 1

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

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

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

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

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

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

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

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

$=\left[\begin{array}{cc}\frac{(1+{\tan }^2\alpha /2)}{(1+{\tan }^2\alpha /2)}& -\frac{\tan \alpha /2(1+{\tan }^2\alpha /2)}{(1+{\tan }^2\alpha /2)}\\ \frac{\tan \alpha /2(1+{\tan }^2\alpha /2)}{(1+{\tan }^2\alpha /2)}& \frac{(1+{\tan }^2\alpha /2)}{(1+{\tan }^2\alpha /2)}\end{array}\right]$
$=\left[\begin{array}{cc}1& -\tan \alpha /2\\ \tan \alpha /2& 1\end{array}\right]$$=I+A$$=L.H.S$