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If the matrix $\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]$ is singular, then what is one of the values of $\theta$?
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The correct answer is:
$\frac{\pi}{4}$
A matrix is singular if value of its determinant is zero.
Given that matrix $\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]$ is singular,
$\left|\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ll}\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right|=0 \Rightarrow \cos ^{2} \theta-\sin ^{2} \theta=0=\cos \frac{\pi}{2}$
$\Rightarrow \cos 2 \theta=\cos \frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{4}$
Given that matrix $\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]$ is singular,
$\left|\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ll}\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right|=0 \Rightarrow \cos ^{2} \theta-\sin ^{2} \theta=0=\cos \frac{\pi}{2}$
$\Rightarrow \cos 2 \theta=\cos \frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{4}$
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