If θ=π5 and A=(cosθsinθ−sinθcosθ) and B=A+A4 then detB

A=(cosθsinθ−sinθcosθ) ⇒An=(cosθsinnθsinnθcosnθ)B=A+A4 =(cosθsinθ−sinθcosθ) +(cos4θsin4θ−sin4θcos4θ)detB=4sin2θ|0+1−10|=4sin2(π5)=10−234=1.35∈(1,2)

If θ=π5 and A=(cosθsinθ−sinθcosθ) and B=A+A4 then detB

A=(cosθsinθ−sinθcosθ) ⇒An=(cosθsinnθsinnθcosnθ)B=A+A4 =(cosθsinθ−sinθcosθ) +(cos4θsin4θ−sin4θcos4θ)detB=4sin2θ|0+1−10|=4sin2(π5)=10−234=1.35∈(1,2)

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If $θ = \frac{π}{5} a n d A = (\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}) a n d B = A + A^{4} t h e n \det B$

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$l i e s i n (1, 2)$

$l i e s i n (2, 3)$

answer is D.

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Detailed Solution

$A = (\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}) \Rightarrow A^{n} = (\begin{matrix} \cos θ & \sin n θ \\ \sin n θ & \cos n θ \end{matrix})$

$B = A + A^{4} = (\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}) + (\begin{matrix} \cos 4 θ & \sin 4 θ \\ - \sin 4 θ & \cos 4 θ \end{matrix})$