Solve the 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 $\operatorname{det}(B):$

1. (1) is one

2. (2) lies in $(2,3)$

3. (3) is zero

4. (4) lies in $(1,2)$

Correct Option: , 4

Solution:

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

$\therefore A^{n}=\left[\begin{array}{cc}\cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta\end{array}\right], n \in \mathbf{N}$

$\therefore B=A+A^{4}$

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

$\therefore B=\left[\begin{array}{ll}\cos \frac{\pi}{5}+\cos \frac{4 \pi}{5} & \sin \frac{\pi}{5}+\sin \frac{4 \pi}{5} \\ -\sin \frac{\pi}{5}-\sin \frac{4 \pi}{5} & \cos \frac{\pi}{5}+\cos \frac{4 \pi}{5}\end{array}\right]$

Then, $\operatorname{det}(B)=2 \sin \left(\frac{\pi}{5}\right) \cdot\left|\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right|$

$=\frac{\sqrt{10-2 \sqrt{5}}}{2} \approx \frac{2.35}{2} \approx 1.175$

$\therefore \operatorname{det} B \in(1,2)$