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

• A. $I+A$
• B. $I-A$
• C. $-I-A$
• D. none of these

Answer 1

Answer: (A)

$I+A$

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