Question

If $\left[\begin{array}{ccc}1& \sin \theta & 1\\ -\sin \theta & 1& \sin \theta \\ -1& -\sin \theta & 1\end{array}\right]$ ; then for all

$\theta \in (\frac{3\pi }{4},\frac{5\pi }{4}),det\left(A\right)$ lies in the interval :

• A. $(0,\frac{3}{2}]$
• B. $(1,\frac{5}{2}]$
• C. $(\frac{3}{2},3]$
• D. $[\frac{5}{2},4)$

Answer 1

The correct option is D $[\frac{5}{2},4)$

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

$R_1\to R_1+R_3$
$\Rightarrow |A|=\left|\begin{array}{ccc}0& 0& 2\\ -\sin \theta & 1& \sin \theta \\ -1& -\sin \theta & 1\end{array}\right|$

$\Rightarrow |A|=2({\sin }^2\theta +1)$

$\theta \in (\frac{3\pi }{4},\frac{5\pi }{4})\Rightarrow \sin \theta \in (\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}})$
$\therefore 0\le {\sin }^2\theta <\frac{1}{2}$
$\Rightarrow 2\le 2(1+{\sin }^2\theta )<3$
$|A|\in [2,3)$