Mathematics - Matrices Question with Solution | TestHub

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

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

$\Rightarrow I-A=\left[\begin{array}{cc}1& \tan \frac{\theta }{2}\\ -\tan \frac{\theta }{2}& 1\end{array}\right] \left\{\therefore |I-A|={\sec }^2\theta /2\right\}$

$\Rightarrow {\left(I-A\right)}^{-1}=\frac{1}{{\sec }^2\frac{\theta }{2}}\left[\begin{array}{cc}1& -\tan \frac{\theta }{2}\\ \tan \frac{\theta }{2}& 1\end{array}\right]$

$\Rightarrow \left(I+A\right){\left(I-A\right)}^{-1}=\frac{1}{{\sec }^2\frac{\theta }{2}}\left[\begin{array}{cc}1& -\tan \frac{\theta }{2}\\ \tan \frac{\theta }{2}& 1\end{array}\right] \left[\begin{array}{cc}1& -\tan \frac{\theta }{2}\\ \tan \frac{\theta }{2}& 1\end{array}\right]$

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

$a=\frac{1-{\tan }^2\frac{\theta }{2}}{{\sec }^2\frac{\theta }{2}}$

$b=\frac{2 \tan \frac{\theta }{2}}{{\sec }^2\frac{\theta }{2}}$

$\therefore a^2+b^2=1$

$\therefore 13\left(a^2+b^2\right)=13$