Solve this following
Question:

Let $\theta=\frac{\pi}{5}$ and $\mathrm{A}=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right] .$ If $\mathrm{B}=\mathrm{A}$

$+\mathrm{A}^{4}$, then $\operatorname{det}(\mathrm{B}):$

1. is one

2. lies in $(1,2)$

3. is zero

4. lies in $(2,3)$

Correct Option: , 2

Solution:

$A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$

$A^{2}=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$

$A^{2}=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$

$B=A+A^{4}$

$=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]+\left[\begin{array}{cc}\cos 4 \theta & \sin 4 \theta \\ -\sin 4 \theta & \cos 4 \theta\end{array}\right]$

$B=\left[\begin{array}{cc}(\cos \theta+\cos 4 \theta) & (\sin \theta+\sin 4 \theta) \\ -(\sin \theta+\sin 4 \theta) & (\cos \theta+\cos 4 \theta)\end{array}\right]$

$|\mathrm{B}|=(\cos \theta+\cos 4 \theta)^{2}+(\sin \theta+\sin 4 \theta)^{2}$

$|\mathrm{B}|=2+2 \cos 3 \theta$ when $\theta=\frac{\pi}{5}$

$|B|=2+2 \cos \frac{3 \pi}{5}=2(1-\sin 18)$

$|\mathrm{B}|=2\left(1-\frac{\sqrt{5}-1}{4}\right)=2\left(\frac{5-\sqrt{5}}{4}\right)=\frac{5-\sqrt{5}}{2}$