Question

PROVE THAT DETERMINANT IS INDEPENDENT OF THETA​

Answer 1

Solution:-

Let

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

Now simplify we get

$\Delta = \rm \: x\Bigg|\begin{array}{ccc} \sf -x&\sf 1\\\sf 1&\sf x\end{array}\Bigg| - \sin \theta\Bigg|\begin{array}{ccc} \sf - \sin \theta&\sf 1\\\sf \cos \theta&\sf x\end{array}\Bigg| + \cos \theta \Bigg|\begin{array}{ccc} \sf - \sin \theta &\sf - x\\\sf \cos\theta&\sf 1\end{array}\Bigg|$

Now

$\rm = \rm \: x \{( - x)(x) - (1)(1) \} - \sin \theta \{( - \sin \theta)(x) - (1) (\cos \theta) \} + \cos\theta \{( - \sin\theta)(1) - ( - x) \cos \theta \}$

$\rm \: = x( - x - 1) - \sin\theta( - x \ \sin\theta - \cos\theta) + \cos\theta( - \sin \theta + x \cos \theta)$

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

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

$\rm \: = - {x}^{2} - x + x( \sin ^{2} \theta + \cos ^{2} \theta)$

we know that

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

We get

$= \rm - {x}^{2} - x + x$

$= \rm- {x}^{2}$

So

$\rm \therefore\rm \: \theta \: is \: independent \: because \: thier \: is \: no \: term \: of \: \theta \: in \: answer$

:- HENCE PROVED