Prove that the determinant | ccc x & & - & -x & 1 & 1 & x | is in… - Vidyadip

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 $\theta.$

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Answer

Let $\Delta=\left[\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right]$
Expanding along first row, $\Delta=x\left|\begin{array}{cc}-x & 1 \\ 1 & x\end{array}\right|-\sin \theta\left|\begin{array}{cc}-\sin \theta & 1 \\ \cos \theta & x\end{array}\right|+\cos \theta\left|\begin{array}{cc}-\sin \theta & -x \\ \cos \theta & 1\end{array}\right|$,
$\Rightarrow \Delta=x\left(-x^2-1\right)-\sin \theta(-x \sin \theta-\cos \theta)+\cos \theta(-\sin \theta+x \cos \theta)$
$\Rightarrow \Delta=-x^3-x+x \sin ^2 \theta+\sin \theta \cos \theta-\sin \theta \cos \theta+x \cos ^2 \theta$
$\Rightarrow \Delta=-x^3-x+x\left(\sin ^2 \theta+\cos ^2 \theta\right)=- x ^3- x + x =- x ^3$
which is independent of $ \theta$

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