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# Prove that the determinant is independent of θ.

Question:

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

Solution:

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

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

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

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

$=-x^{3}-x+x$

$=-x^{3}($ Independent of $\theta)$

Hence, $\Delta$ is independent of $\theta$.