# Choose the correct answer.

Question:

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 $(\mathrm{A}) \in[2,4]$

Solution:

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.

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