Question

Prove that the determinant $\left|\begin{array}{ccc}x& \sin \theta & \cos \theta \\ -\sin \theta & -x& 1\\ \cos \theta & 1& x\end{array}\right|$ is independent of $\theta$.

Answer 1

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

Expanding along the first row, we get

$\Delta =x(-x^2-1)-\sin \theta (-x\sin \theta -\cos \theta )+\cos \theta (-\sin \theta +x\cos \theta )$

$=-x^3-x+x{\sin }^2\theta +\sin \theta \cos \theta -\sin \theta \cos \theta +x{\cos }^2\theta$

$=-x^3-x+x({\sin }^2\theta +{\cos }^2\theta )$

$\Rightarrow \Delta =-x^3$ ($\because {\sin }^2\theta +{\cos }^2\theta =1$)

Hence, the given determinant is independent of $\theta$