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 , then $\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$ is equal to

• A. $I+A$
• B. $I-A$
• C. $A-I$
• D. $A$

Answer 1

Answer: (A)

Here, $A=\left[\begin{array}{cc}0& -t\\ t& 0\end{array}\right]$, where $t=\tan \left(\frac{\alpha }{2}\right)$
Now, $\cos \alpha =\frac{1-{\tan }^2\left(\frac{\alpha }{2}\right)}{1+{\tan }^2\left(\frac{\alpha }{2}\right)}=\frac{1-t^2}{1+t^2}$
and $\sin \alpha =\frac{2\tan \left(\frac{\alpha }{2}\right)}{1+{\tan }^2\left(\frac{\alpha }{2}\right)}=\frac{2t}{1+t^2}$
$=(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$
$=\left(\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}0& -t\\ +t& 0\end{array}\right]\right)\left[\begin{array}{cc}\frac{1-t^2}{1+t^2}& \frac{-2t}{1+t^2}\\ \frac{2t}{1+t^2}& \frac{1-t^2}{1+t^2}\end{array}\right]$
$=\left[\begin{array}{cc}1& t\\ -t& 1\end{array}\right]\left[\begin{array}{cc}\frac{1-t^2}{1+t^2}& \frac{-2t}{1+t^2}\\ \frac{2t}{1+t^2}& \frac{1-t^2}{1+t^2}\end{array}\right]$
$=\left[\begin{array}{cc}\frac{1-t^2+2t^2}{1+t^2}& \frac{-2t+t\left(1-t^2\right)}{1+t^2}\\ \frac{-t\left(1-t^2\right)+2t}{1+t^2}& \frac{2t^2+1-t^2}{1+t^2}\end{array}\right]$
$=\left[\begin{array}{cc}\frac{1+t^2}{1+t^2}& \frac{-2t+t-t^3}{1+t^2}\\ \frac{-t+t^3+2t}{1+t^2}& \frac{2t^2+1-t^2}{1+t^2}\end{array}\right]$
$=\left[\begin{array}{cc}\frac{1+t^2}{1+t^2}& \frac{-t\left(1+t^2\right)}{1+t^2}\\ \frac{t\left(1+t^2\right)}{1+t^2}& \frac{1+t^2}{1+t^2}\end{array}\right]=\left[\begin{array}{cc}1& -t\\ t& 1\end{array}\right]$
Also, $I+A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}0& -t\\ t& 0\end{array}\right]$
$=\left[\begin{array}{cc}0+1& -t+0\\ t+0& 0+1\end{array}\right]=\left[\begin{array}{cc}1& -t\\ t& 1\end{array}\right]$
$=(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$