Let A= 1& &1 - &1& -1&- &1 , where 0 2 . Then 1. Det (A) = 0 2. D… - Vidyadip

Question

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

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

✓

Answer

$\text{A}=\begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix}$
$\therefore|\text{A}|=1(1+\sin^2\theta)-\sin\theta(-\sin\theta+\sin\theta)+1(\sin^2\theta+1)$
$=1+\sin^2\theta+\sin^2\theta+1$
= 2 + 2 sin² $\theta$
= 2 (1 + sin²$\theta$)
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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