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

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

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

$B=A+A^{4}$

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

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

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

$|B|=2+2 \cos 3 \theta$

when $\theta=\frac{\pi}{5}$

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

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