JEE Main 2020 — Matrices Question with Solution

$A^2=\left[\begin{array}{l}\cos \theta i\sin \theta \\ i\sin \theta \cos \theta \end{array}\right]\left[\begin{array}{l}\cos \theta i\sin \theta \\ i\sin \theta \cos \theta \end{array}\right]$

$=\left[\begin{array}{l}{\cos }^2\theta -{\sin }^2\theta 2i\sin \theta \cos \theta \\ 2i\sin \theta \cos \theta {\cos }^2\theta -{\sin }^2\theta \end{array}\right]$

$=\left[\begin{array}{l}\cos 2\theta i\sin 2\theta \\ i\sin 2\theta \cos 2\theta \end{array}\right]$

$$$A^3=\left[\begin{array}{l}\cos 2\theta i\sin 2\theta \\ i\sin 2\theta \cos 2\theta \end{array}\right]\left[\begin{array}{l}\cos \theta i\sin \theta \\ i\sin \theta \cos \theta \end{array}\right]$

$=\left[\begin{array}{l}\cos 3\theta i\sin 3\theta \\ i\sin 3\theta \cos 3\theta \end{array}\right]$

Similarly, $A^5=\left[\begin{array}{l}\cos 5\theta i\sin 5\theta \\ i\sin 5\theta \cos 5\theta \end{array}\right]=\left[\begin{array}{l}a b\\ c d\end{array}\right]$

$a=\cos 5\theta$, $b=i\sin 5\theta$, $c=i\sin 5\theta$

$d=\cos 5\theta$

$a=d, b=c$

$\left(1\right)$ $a^2+b^2={\cos }^{2}5\theta -{\sin }^{2}5\theta =\cos 10\theta \in \left(0,1\right)$ as $\theta =\frac{\pi }{24}$

$\left(2\right)$ $a^2-d^2=0$

$\left(3\right)$ $a^2-b^2={\cos }^{2}5\theta +{\sin }^{2}5\theta =1$

$\left(4\right)$ $a^2-c^2=a^2-b^2=1$