Question

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

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

Answer 1

Answer: (B)

$|A|=\left|\begin{array}{ccc}1& sin\theta & 1\\ -sin\theta & 1& sin\theta \\ -1& -sin\theta & 1\end{array}\right|$
$=2(1+si{n}^2\theta )$
$\theta \in (\frac{3\pi }{4},\frac{5\pi }{4})\Rightarrow \frac{1}{\sqrt{2}}<sin\theta <\frac{1}{\sqrt{2}}$
$\Rightarrow 0\le si{n}^2\theta <\frac{1}{2}$
$\therefore |A|\in [2,3)$