Mathematics - Determinant Question with Solution | TestHub

Given, $x^2=\left|\begin{array}{ccc}\sin \theta & \cos \theta & 0\\ -\cos \theta & \sin \theta & 1\\ \sin \theta & \cos \theta & 2\end{array}\right|$

On expanding $\left|\begin{array}{ccc}\sin \theta & \cos \theta & 0\\ -\cos \theta & \sin \theta & 1\\ \sin \theta & \cos \theta & 2\end{array}\right|$ along $C_3$, we get

$x^2=0-1\left|\begin{array}{cc}\sin \theta & \cos \theta \\ \sin \theta & \cos \theta \end{array}\right|+2\left|\begin{array}{cc}\sin \theta & \cos \theta \\ -\cos \theta & \sin \theta \end{array}\right|=-1\left(\sin \theta \cos \theta -\sin \theta \cos \theta \right)+2\left({\sin }^2\theta +{\cos }^2\theta \right)$

$=-1\times 0+2\times 1$

$\Rightarrow x^2=2\Rightarrow x=\pm \sqrt{2}$

If $x=\sqrt{2}$, then

$4x^2+x\sin \frac{3\pi }{2}+5=4\times {\left(\sqrt{2}\right)}^2-\sqrt{2}+5$

$=8-\sqrt{2}+5=\left(13-\sqrt{2}\right)$

If $x=-\sqrt{2}$, then

$4x^2+x\sin \frac{3\pi }{2}+5=4\times {\left(-\sqrt{2}\right)}^2+\sqrt{2}+5$

$=8+\sqrt{2}+5=\left(13+\sqrt{2}\right)$