Question

The determinant $\left|\begin{array}{ccc}x& \sin \theta & \cos \theta \\ -\sin \theta & -x& 1\\ \cos \theta & 1& x\end{array}\right|$ is

• A. independent of $\theta$ only
• B. independent of $x$ only
• C. independent of both $\theta$ and $x$
• D. None of the above

Answer 1

Answer: (A)

Let $\Delta =\left|\begin{array}{ccc}x& \sin \theta & \cos \theta \\ -\sin \theta & -x& 1\\ \cos \theta & 1& x\end{array}\right|$
By expanding along first row, we get
$\Delta =x\left|\begin{array}{cc}-x& 1\\ 1& x\end{array}\right|-\sin \theta \left|\begin{array}{cc}-\sin \theta & 1\\ \cos \theta & x\end{array}\right|+\cos \theta \left|\begin{array}{cc}-\sin \theta & -x\\ \cos \theta & 1\end{array}\right|$
$=x\left(-x^2-1\right)-\sin \theta (-x\sin \theta -\cos \theta )+\cos \theta (-\sin \theta +x\cos \theta )$
$=-x^3-x+x{\sin }^2\theta +\sin \theta \cos \theta -\sin \theta \cos \theta +x{\cos }^2\theta$
$=-x^3-x+x\left({\sin }^2\theta +{\cos }^2\theta \right)=-x^3-x+x$
$\left(\because {\sin }^2\theta +\cos {\theta }^2=1\right)$
$=-x^3$
Hence, $\Delta$ is independent of $\theta$.