Let θ = π 5 and A=[ … ]. If B=A+A^4, then det (B) :

$${A}^{2}=[\begin{matrix}\mathrm{cos}\theta & \mathrm{sin}\theta \\ -\mathrm{sin}\theta & \mathrm{cos}\theta \end{matrix}][\begin{matrix}\mathrm{cos}\theta & \mathrm{sin}\theta \\ -\mathrm{sin}\theta & \mathrm{cos}\theta \end{matrix}]$$A2=[cosθ−sinθ​sinθcosθ​][cosθ−sinθ​sinθcosθ​]

$${A}^{2}=[\begin{matrix}\mathrm{cos}2\theta & \mathrm{sin}2\theta \\ -\mathrm{sin}2\theta & \mathrm{cos}2\theta \end{matrix}]$$A2=[cos2θ−sin2θ​sin2θcos2θ​]

$$\Rightarrow {A}^{4}=[\begin{matrix}\mathrm{cos}4\theta & \mathrm{sin}4\theta \\ -\mathrm{sin}4\theta & \mathrm{cos}4\theta \end{matrix}]$$⇒A4=[cos4θ−sin4θ​sin4θcos4θ​]

$$B=[\begin{matrix}\mathrm{cos}4\theta & \mathrm{sin}4\theta \\ -\mathrm{sin}4\theta & \mathrm{cos}4\theta \end{matrix}]+[\begin{matrix}\mathrm{cos}\theta & \mathrm{sin}\theta \\ -\mathrm{sin}\theta & \mathrm{cos}\theta \end{matrix}]$$B=[cos4θ−sin4θ​sin4θcos4θ​]+[cosθ−sinθ​sinθcosθ​]

$$=[\begin{matrix}\mathrm{cos}4\theta +\mathrm{cos}\theta & \mathrm{sin}4\theta +\mathrm{sin}\theta \\ -(\mathrm{sin}4\theta +\mathrm{sin}\theta ) & \mathrm{cos}4\theta +\mathrm{cos}\theta \end{matrix}]$$=[cos4θ+cosθ−(sin4θ+sinθ)​sin4θ+sinθcos4θ+cosθ​]

$$|B|={(\mathrm{cos}4\theta +\mathrm{cos}\theta )}^{2}+{(\mathrm{sin}4\theta +\mathrm{sin}\theta )}^{2}$$∣B∣=(cos4θ+cosθ)2+(sin4θ+sinθ)2

$$=2+2(\mathrm{cos}4\theta \cdot \mathrm{cos}\theta +\mathrm{sin}4\theta \cdot \mathrm{sin}\theta )$$=2+2(cos4θ⋅cosθ+sin4θ⋅sinθ)

$$=2+2\mathrm{cos}(4\theta -\theta )$$=2+2cos(4θ−θ)

$$=2+2\mathrm{cos}3\theta$$=2+2cos3θ

$$|B|=2+2\mathrm{cos}\frac{3\pi }{5}$$∣B∣=2+2cos53π​

$$=2-(\frac{\sqrt{5}-1}{2})=\frac{5-\sqrt{5}}{2}\in (1,2)$$=2−(25​−1​)=25−5​​∈(1,2)