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
- ADet $(A) = 0$
- BDet $(A) \in (2, \infty)$
- CDet $(A) \in (2, -4)$
- ✓Det $(A) \in [2, 4]$
✓
Answer
Correct option: D.
Det $(A) \in [2, 4]$
$\text{A}=\begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix}$
$\therefore|\text{A}|=1(1+\sin^2\theta)-\sin\theta(-\sin\theta+\sin\theta)+1(\sin^2\theta+1)$
$=1+\sin^2\theta+\sin^2\theta+1$
$= 2 + 2 \sin^2 \theta$
$= 2 (1 + \sin^2\theta$)
Now, $0\leq\theta\leq2\pi$
$\Rightarrow 0\leq\sin\theta\leq1$
$\Rightarrow 0\leq1+\sin^2\theta\leq2$
$\Rightarrow2\leq2(1+\sin^2\theta)\leq4$
$\therefore$ Det $(A) \in [2, 4]$
The correct answer is $d.$
$\therefore|\text{A}|=1(1+\sin^2\theta)-\sin\theta(-\sin\theta+\sin\theta)+1(\sin^2\theta+1)$
$=1+\sin^2\theta+\sin^2\theta+1$
$= 2 + 2 \sin^2 \theta$
$= 2 (1 + \sin^2\theta$)
Now, $0\leq\theta\leq2\pi$
$\Rightarrow 0\leq\sin\theta\leq1$
$\Rightarrow 0\leq1+\sin^2\theta\leq2$
$\Rightarrow2\leq2(1+\sin^2\theta)\leq4$
$\therefore$ Det $(A) \in [2, 4]$
The correct answer is $d.$
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