Let θ = π 5 and A = cosθ sinθ - sinθ cosθ . If B = A + A 4 , then det B :

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

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

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

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

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

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

$=2+2\left(\cos 4\theta ·\cos \theta +\sin 4\theta ·\sin \theta \right)$

$=2+2\cos \left(4\theta -\theta \right)$

$=2+2\cos 3\theta$

$\left|B\right|=2+2\cos \frac{3\mathrm{\pi }}{5}$

$=2-\left(\frac{\sqrt{5}-1}{2}\right)=\frac{5-\sqrt{5}}{2}\in \left(1,2\right)$