Question

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

• A. True
• B. False

Answer 1

The correct option is A True

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

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

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

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

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

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

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

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

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