If $A=\left[\begin{array}{ccc}1& \sin \theta & 1\\ -\sin \theta & 1& \sin \theta \\ -1& -\sin \theta & 1\end{array}\right]$; then for all $\theta \in \left(\frac{3\pi }{4},\frac{5\pi }{4}\right),\det \left(A\right)$ lies in the interval

Here, $\det \left(A\right)=\left|\begin{array}{ccc}1& \sin \theta & 1\\ -\sin \theta & 1& \sin \theta \\ -1& -\sin \theta & 1\end{array}\right|$

$\begin{array}{l}1\left(1+{\sin }^2\theta \right)-\sin \theta \left(0\right)+1\left({\sin }^2\theta +1\right)=2\left(1+{\sin }^2\theta \right)\\ \because \theta \in \left(\frac{3\pi }{4},\frac{5\pi }{4}\right)\Rightarrow \sin \theta \in \left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\\ \Rightarrow {\sin }^2\theta \in \left[0,\frac{1}{2}\right)\therefore \det \left(A\right)\in [2,3)\end{array}$