Question

$\Delta =\left|\begin{array}{ccc}1& sinQ& 1\\ -sinQ& 1& sinQ\\ -1& -sinQ& 1\end{array}\right|$

prove that:$2⩽\Delta ⩽4$

Answer 1

given

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

Expanding the determinant by $1st$ row

$=1\left|\begin{array}{cc}1& \sin \theta \\ -\sin \theta & 1\end{array}\right|-\sin \theta \left|\begin{array}{cc}-\sin \theta & \sin \theta \\ 1& 1\end{array}\right|+1\left|\begin{array}{cc}-\sin \theta & 1\\ -1& -\sin \theta \end{array}\right|$

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

$=2+2{\sin }^2\theta +{\sin }^2\theta -{\sin }^2\theta$

$=2+2{\sin }^2\theta =\Delta$

We know range of ${\sin }^2\theta$ is given as

$0\le {\sin }^2\theta \le 1\left(multiplying2\right)$

$0\le {\sin }^2\theta \le 2\left(adding2\right)$

$2\le 2+2{\sin }^2\theta \le 4$

$2\le \Delta \le 4$ proved