Question

$If \ A \ = \begin {bmatrix} 1 & sin\theta & 1 \\ -sin\theta & 1 & sin\theta \\ -1 & -sin\theta & 1 \end {bmatrix};$ then for all $\theta \ \in \bigg( \frac{3\pi}{4} , \frac{5\pi}{4}\bigg)$, $det (A)$ lies in the interval :

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

Answer 1

Answer: (B)

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