Question

Prove that 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

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

$\Rightarrow △=x\left|\begin{array}{cc}-x& 1\\ 1& x\end{array}\right|-\sin \theta \left|\begin{array}{cc}-\sin \theta & 1\\ \cos \theta & x\end{array}\right|+\cos \theta \left|\begin{array}{cc}-\sin \theta & -x\\ \cos \theta & 1\end{array}\right|$

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

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

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

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

$\Rightarrow △=-x^3-x+x\left(1\right)$

$(\because {\sin }^2\theta +{\cos }^2\theta =1)$

$\Rightarrow △=-x^3$

Hence, ​​​​​​​$△=-x^3$

Thus, the determinant is independent of $\theta$

Hence Proved.

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