Prove that the determinant x& & - &-x&1 &1&x is independent of θ. - Vidyadip

Question

Prove that the determinant $\begin{vmatrix}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{vmatrix}$ is independent of θ.

✓

Answer

$\triangle=\begin{vmatrix}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{vmatrix}$
$= x(-x^2-1)-\sin\theta(-x\sin\theta-\cos\theta)+\cos\theta(-\sin\theta+x\cos\theta)$
$=-x^3-x+x\sin^2\theta+\sin\theta\cos\theta-\sin\theta\cos\theta+x\cos^2\theta$
$=-x^3-x+x(\sin^2\theta+\cos^2\theta)$
$=-x^3-x+x$
$=-x^3$ (Independent of $\theta$)
Hence, $\triangle$ is independent of $\theta.$

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