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Let $A=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$ where $0 \leq \theta \leq 2 p$. Then
$\begin{array}{ll}\text { (a) } \operatorname{Det}(A)=0 & \text { (b) } \operatorname{Det}(A) \in(2, \infty)\end{array}$
(c) $\operatorname{Det}(\mathrm{A}) \in(2,4)$
(d) $\operatorname{Det}(\mathrm{A}) \in[2,4]$
MathematicsDeterminants
Solution:
1305 Upvotes Verified Answer
(d) $A=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$
$\operatorname{Det}(A)=\left|\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right|=2\left(1+\sin ^2 \theta\right)$
For $\theta=0, \pi, 2 \pi \Rightarrow \operatorname{Det}(A)=2$
For $\theta=\frac{\pi}{2}, \frac{3 \pi}{2}$
$\Rightarrow \operatorname{Det}(A)=2 \cdot(1+1)=4$

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