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

MCQ

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

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

✓

Answer

Correct option: D.

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

$\begin{vmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix}$
$=\begin{vmatrix}1&\sin\theta&2\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix}$ [Applying $C_3 → C_3 + C_1$]
$=2\times\begin{vmatrix}-\sin\theta&1\\-1&-\sin\theta\end{vmatrix} [$Expanding along $C_3]$
$=2(\sin^2\theta+1)$
Given, $0\leq\theta\leq2\pi$
$-1\leq\sin\theta\leq1$
$0\leq\sin^2\theta\leq1$
$|\text{A}|=2(\sin^2\theta+1)$
$|\text{A}|=2\times1=2[\theta=0]$
$|\text{A}|=2\times2=4[\theta=2\pi]$
$\text{Det (A)}\in[2,4]$

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