Matrices And Determinants Ques 8
- 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 $\theta \in\left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right)$, $\operatorname{det}(A)$ lies in the interval
(2019 Main, 12 Jan II)
(a) $\left(\frac{3}{2}, 3\right]$
(b) $\left[\frac{5}{2}, 4\right)$
(c) $\left(0, \frac{3}{2}\right]$
(d) $\left(1, \frac{5}{2}\right]$
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Answer:
Correct Answer: 8.(a)
Solution: (a) Given matrix $A=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$
$ \begin{aligned} & \Rightarrow \operatorname{det}(A)=|A|=\left|\begin{array}{ccc} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{array}\right| \\ & =1\left(1+\sin ^2 \theta\right)-\sin \theta(-\sin \theta+\sin \theta)+1\left(\sin ^2 \theta+1\right) \\ & \Rightarrow|A|=2\left(1+\sin ^2 \theta\right) \end{aligned} $
As we know that, for $\theta \in\left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right)$
$ \sin \theta \in\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) $
$ \begin{aligned} & \Rightarrow \sin ^2 \theta \in\left[0, \frac{1}{2}\right) \Rightarrow 1+\sin ^2 \theta \in\left[0+1, \frac{1}{2}+1\right) \\ & \Rightarrow 1+\sin ^2 \theta \in\left[1, \frac{3}{2}\right) \\ & \Rightarrow 2\left(1+\sin ^2 \theta\right) \in[2,3) \Rightarrow|A| \in[2,3) \subset\left(\frac{3}{2}, 3\right] \end{aligned} $
BETA
