Question

If the $\Delta =\left|\begin{array}{ccc}1& 3\cos \theta & 1\\ \sin \theta & 1& 3\cos \theta \\ 1& \sin \theta & 1\end{array}\right|$ , then the value of $\left({\Delta }_{max}\right)/2$ is

Answer 1

$\Delta =\left|\begin{array}{ccc}1& 3\cos \theta & 1\\ \sin \theta & 1& 3\cos \theta \\ 1& \sin \theta & 1\end{array}\right|$

$=1(1-3\sin \theta \cos \theta )-3\cos \theta (\sin \theta -3\cos \theta )+4({\sin }^2\theta -1)$

$=1-3\sin \theta \cos \theta -3\sin \theta \cos \theta +9{\cos }^2\theta -{\cos }^2\theta$

$=1-3\sin 2\theta +8{\cos }^2\theta$

${\Delta }_{max}$ is at $\theta =0$ which ${\Delta }_{max}=9$

$\therefore \frac{{\Delta }_{max}}{2}=\frac{9}{2}=4.5$