# Let A = $\begin{bmatrix} 1 & sin\theta & 1 \\ -sin\theta & 1 & sin\theta \\ -1 & -sin\theta & 1 \end{bmatrix}$, where $0 \leq \theta \leq 2\pi$. Then: $\begin{array} ((A) \, Det(A) = 0 \quad& (B) \, Det(A) \in (2, \infty) \\[0.5em] (C) \, Det(A) \in (2, 4) \quad &(D) Det(A) \in [2,4] \end{array}$

Toolbox:
• 0$\leq sin\theta\leq 1$,when $0\leq \theta \leq 2\pi$
• Determinant of a matrix of order three can be determined by expressing it in terms of second order determinant.This is known as expansion of a determinant along a row (or a column).
Let A=$\begin{bmatrix} 1 & sin\theta & 1 \\ -sin\theta & 1 & sin\theta \\ -1 & -sin\theta & 1 \end{bmatrix}$,

Expanding along $R_1$

$|A|=1(1+sin^2\theta)-sin\theta(-sin\theta+sin\theta)+1(sin^2\theta+1)$

$\;\;\;=1+sin^2\theta-0+1+sin^2\theta$

$\;\;\;=2+2sin^2\theta$

|A|=2($1+sin^2\theta$).

It is given $0\leq \theta \leq 2\pi$ (But we know the interval of $sin\theta$ is between 0 and 1)

$\Rightarrow 0\leq sin\theta\leq 1$

$\Rightarrow 0\leq sin^2\theta\leq 1$

$1\leq 1+ sin^2\theta\leq 2$

Multiplying by 2,

$2\leq 2( 1+ sin^2\theta)\leq 4$

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

Hence the correct answer is D.