if $A = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] t h e n A^{2} = I$ is true for

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none of these

$θ = 0$

$θ = \frac{π}{2}$

$θ = \frac{π}{4}$

answer is A.

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Detailed Solution

we are given $A = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] a l s o A^{2} = I$

$A^{2} = A . A = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] A^{2} = [\begin{matrix} \cos^{2} θ - \sin^{2} θ & \cos θ \sin θ + \sin θ \cos θ \\ - \sin θ \cos θ - \cos θ \sin θ & - \sin^{2} θ + \cos^{2} θ \end{matrix}] n o w A^{2} = I \Rightarrow \cos^{2} θ - \sin^{2} θ = 1 \Rightarrow \cos 2 θ = 1 \Rightarrow 2 θ = 0 \Rightarrow θ = 0 a l s o 2 \sin θ \cos θ = 0 \sin 2 θ = 0 \Rightarrow θ = 0 a l s o - 2 \sin θ \cos θ = 0 - 2 \sin 2 θ = 0 a l s o θ = 0$