Let A= 1& &1 - &1& -1&- &1 , where 0 2 . Then - Vidyadip

MCQ

$\text{Let A}=\begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix},$ where $0\leq\theta\leq2\pi.$ Then

A
Det $(A) = 0$
B
Det $(A) \in (2, \infty)$
C
Det $(A) \in (2, 4)$
✓
Det $(A) \in [2, 4]$

✓

Answer

Correct option: D.

Det $(A) \in [2, 4]$

$\text{A}=\begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix}$ $\therefore|\text{A}|=1(1+\sin^2\theta)-\sin\theta(-\sin\theta+\sin\theta)+1(\sin^2\theta+1)$ $=1+\sin^2\theta+\sin^2\theta+1$
$=2+2 \sin ^2 \theta$
$=2\left(1+\sin ^2 \theta\right)$
Now, $0 \leq \theta \leq 2 \pi$
$\Rightarrow 0 \leq \sin \theta \leq 1$
$\Rightarrow 0 \leq 1+\sin ^2 \theta \leq 2$
$\Rightarrow 2 \leq 2\left(1+\sin ^2 \theta\right) \leq 4$
$\therefore \operatorname{Det}(A) \in[2,4]$
The correct answer is d.

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