If $A = [\begin{matrix} \sin θ & - \cos θ \\ \cos θ & \sin θ \end{matrix}]$ and $B = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]$ , then $(\sin θ) A + (\cos θ) B$

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$[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$[\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

$[\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix}]$

$[\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}]$

answer is B.

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Detailed Solution

$\begin{array}{l} (\sin θ) [\begin{matrix} \sin θ & - \cos θ \\ \cos θ & \sin θ \end{matrix}] + (\cos θ) [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] \\ = [\begin{matrix} \sin^{2} θ + \cos^{2} θ & - \sin θ \cos θ + \cos θ \sin θ \\ \sin θ \cos θ - \cos θ \sin θ & \sin^{2} θ + \cos^{2} θ \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \end{array}$