Let θ = (π)/(5) and A = [ cases cos θ & sin θ - sin θ & cos θ cases...

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

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

⇒ {A}^{4} = \left[\begin{cases}\cos 4\theta & \sin 4\theta \\ - \sin 4\theta & \cos 4\theta\end{cases}\right]

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

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

\left|B\right| = {\left(\cos 4\theta + cos\theta\right)}^{2} + {\left(\sin 4\theta + sin\theta\right)}^{2}$</p><p>$= 2 + 2\left(\cos 4\theta \cdot cos\theta + \sin 4\theta \cdot sin\theta\right)$</p><p>$= 2 + 2\cos \left(4\theta - \theta\right)$</p><p>$= 2 + 2\cos 3\theta$</p><p>\left|B\right| = 2 + 2cos\dfrac{3\pi}{5}$

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