Choose the correct answer.
Let $A=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$, where $0 \leq \theta \leq 2 \pi$, then
A. $\operatorname{Det}(A)=0$
B. $\operatorname{Det}(\mathrm{A}) \in\left(2,^{\infty}\right)$
C. $\operatorname{Det}(A) \in(2,4)$
D. Det $(\mathrm{A}) \in[2,4]$
Answer: D
$A=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$
$\therefore|A|=1\left(1+\sin ^{2} \theta\right)-\sin \theta(-\sin \theta+\sin \theta)+1\left(\sin ^{2} \theta+1\right)$
$=1+\sin ^{2} \theta+\sin ^{2} \theta+1$
$=2+2 \sin ^{2} \theta$
$=2\left(1+\sin ^{2} \theta\right)$
Now, $0 \leq \theta \leq 2 \pi$
$\Rightarrow-1 \leq \sin \theta \leq 1$
$\Rightarrow 0 \leq \sin ^{2} \theta \leq 1$
$\Rightarrow 1 \leq 1+\sin ^{2} \theta \leq 2$
$\Rightarrow 2 \leq 2\left(1+\sin ^{2} \theta\right) \leq 4$
$\therefore \operatorname{Det}(A) \in[2,4]$
The correct answer is D.