If A=cosθ−sinθsinθcosθ, then A2 is

A
$[\begin{matrix} \cos^{2} θ & - \sin^{2} θ \\ \sin^{2} θ & \cos^{2} θ \end{matrix}]$
B
$[\begin{matrix} \cos 2 θ & - \sin 2 θ \\ \sin 2 θ & - \cos 2 θ \end{matrix}]$
C
$[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$
D
None of these

Solution:

$\begin{array}{l} A = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}] \\ A^{2} = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}] \times [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}] \\ = [\begin{matrix} \cos^{2} θ - \sin^{2} θ & - \cos θ \sin θ - \cos θ \sin θ \\ \sin θ \cos θ + \cos θ \sin θ & - \sin^{2} θ + \cos^{2} θ \end{matrix}] \\ = [\begin{matrix} \cos 2 θ & - \sin 2 θ \\ \sin 2 θ & - \cos 2 θ \end{matrix}] \end{array}$

If $A = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]$ , then $A^{2}$ is

Solution:

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