Question

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

Answer 1

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

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

$=-x^3-x+x\times 1=-x^3$ which is independent of $\theta$