If A=cos⁡θisin⁡θisin⁡θcos⁡θ, θ=π24 and A5=a bc d,where i=−1, then which one of the following is not true?

We find thatA2=cos⁡2θisin⁡2θisin⁡2θcos⁡2θ,A3=cos⁡3θisin⁡3θisin⁡3θcos⁡3θ etc. ∴A5=cos⁡5θisin⁡5θisin⁡5θcos⁡5θ⇒abcd=cos⁡5θisin⁡5θisin⁡5θcos⁡5θ⇒a=cos⁡5θ,b=isin⁡5θ,c=isin⁡5θ and d=cos⁡5θ⇒a=d=cos⁡5θ and b=c=isin⁡5θ⇒a2−d2=0,a2−c2=cos2⁡5θ+sin2⁡5θ=1 and a2+b2=cos2⁡5θ−sin2⁡5θ=cos⁡10θ⇒a2−c2=cos2⁡5θ+sin2⁡5θ=1

If A=cos⁡θisin⁡θisin⁡θcos⁡θ, θ=π24 and A5=a bc d,where i=−1, then which one of the following is not true?

We find thatA2=cos⁡2θisin⁡2θisin⁡2θcos⁡2θ,A3=cos⁡3θisin⁡3θisin⁡3θcos⁡3θ etc. ∴A5=cos⁡5θisin⁡5θisin⁡5θcos⁡5θ⇒abcd=cos⁡5θisin⁡5θisin⁡5θcos⁡5θ⇒a=cos⁡5θ,b=isin⁡5θ,c=isin⁡5θ and d=cos⁡5θ⇒a=d=cos⁡5θ and b=c=isin⁡5θ⇒a2−d2=0,a2−c2=cos2⁡5θ+sin2⁡5θ=1 and a2+b2=cos2⁡5θ−sin2⁡5θ=cos⁡10θ⇒a2−c2=cos2⁡5θ+sin2⁡5θ=1

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If $A = [\begin{array}{cc} \cos θ & i \sin θ \\ i \sin θ & \cos θ \end{array}], θ = \frac{π}{24}$ and $A^{5} = [\begin{array}{ll} a b \\ c d \end{array}]$ ,where $i = \sqrt{- 1},$ then which one of the following is not true?

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$a^{2} - c^{2} = 1$

$a^{2} - d^{2} = 0$

$a^{2} - b^{2} = \frac{1}{2}$

$0 \leq a^{2} + b^{2} \leq 1$

answer is D.

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Detailed Solution

We find that

$\begin{array}{l} A^{2} = [\begin{array}{cc} \cos 2 θ & i \sin 2 θ \\ i \sin 2 θ & \cos 2 θ \end{array}], A^{3} = [\begin{array}{cc} \cos 3 θ & i \sin 3 θ \\ i \sin 3 θ & \cos 3 θ \end{array}] etc. \\ ∴ A^{5} = [\begin{array}{cc} \cos 5 θ & i \sin 5 θ \\ i \sin 5 θ & \cos 5 θ \end{array}] \\ \Rightarrow [\begin{array}{ll} a & b \\ c & d \end{array}] = [\begin{array}{cc} \cos 5 θ & i \sin 5 θ \\ i \sin 5 θ & \cos 5 θ \end{array}] \\ \Rightarrow a = \cos 5 θ, b = i \sin 5 θ, c = i \sin 5 θ and d = \cos 5 θ \\ \Rightarrow a = d = \cos 5 θ and b = c = i \sin 5 θ \\ \Rightarrow a^{2} - d^{2} = 0, a^{2} - c^{2} = \cos^{2} 5 θ + \sin^{2} 5 θ = 1 \\ and a^{2} + b^{2} = \cos^{2} 5 θ - \sin^{2} 5 θ = \cos 10 θ \\ \Rightarrow a^{2} - c^{2} = \cos^{2} 5 θ + \sin^{2} 5 θ = 1 \end{array}$