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

Let $A=\left|\begin{array}{ccc}x& sin\theta & cos\theta \\ -sin\theta & -x& 1\\ cos\theta & 1& x\end{array}\right|$
Expanding to corresponding first row, we get
$A=x\left|\begin{array}{cc}-x& 1\\ 1& x\end{array}\right|-sin\theta \left|\begin{array}{cc}-sin\theta & 1\\ cos\theta & x\end{array}\right|+cos\theta \left|\begin{array}{cc}-sin\theta & -x\\ cos\theta & 1\end{array}\right|$
$=x(-x^2-1)-sin\theta (-xsin\theta -cos\theta )+cos\theta (-sin\theta +xcos\theta )=-x^3-x+xsi{n}^2\theta +sin\theta cos\theta -sin\theta cos\theta +xco{s}^2\theta =-x^3-x+x(si{n}^2\theta +co{s}^2\theta )=-x^3-x+x(\because si{n}^2\theta +co{s}^2\theta =1)$
$=-x^3$. Hence, A is independent of $\theta$.