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Let   A = $$\left[\begin{array}{ccc} {1} & {\sin \theta} & {1} \\ {-\sin \theta} & {1} & {\sin \theta} \\ {-1} & {-\sin \theta} & {1} \end{array}\right]$$ where $$0 \leq \theta \leq 2 \pi$$. Then
• Det(A) $$\in$$ (2, $$\infty$$)
• Det(A) = 0
• Det(A) $$\in$$ (2, 4)
• Det(A) $$\in$$ [2, 4]
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## QuestionMathsClass 12

Let   A = $$\left[\begin{array}{ccc} {1} & {\sin \theta} & {1} \\ {-\sin \theta} & {1} & {\sin \theta} \\ {-1} & {-\sin \theta} & {1} \end{array}\right]$$ where $$0 \leq \theta \leq 2 \pi$$. Then
• Det(A) $$\in$$ (2, $$\infty$$)
• Det(A) = 0
• Det(A) $$\in$$ (2, 4)
• Det(A) $$\in$$ [2, 4]

$$A=\left[\begin{array}{ccc} {1} & {\sin \theta} & {1} \\ {-\sin \theta} & {1} & {\sin \theta} \\ {-1} & {-\sin \theta} & {1} \end{array}\right]$$
|A| = 1 (1 $$\times$$ 1 – sin $$\theta$$ $$\times$$ (-sin $$\theta$$)) – sin $$\theta$$ (-sin $$\theta$$+$$sin\theta$$)  + 1 [(- sin $$\theta)~\times (-sin \theta )-(-1)\times1]$$
|A| = $$1+sin^2\theta+sin^2\theta +1$$
|A| = $$2 + 2~sin^2\theta$$
|A| = $$2(1 + sin2\theta)$$
Now, 0 $$\leq$$ $$\theta$$ $$\leq$$ 2$$\pi$$
$$\Rightarrow$$ sin 0 $$\leq$$ sin $$\theta$$ $$\leq$$ sin 2$$\pi$$
$$\Rightarrow$$ 0 $$\leq$$ sin2$$\theta$$ $$\leq$$ 1
$$\Rightarrow$$ 1 + 0 $$\leq$$ 1 + sin2$$\theta$$ $$\leq$$ 1 + 1
$$\Rightarrow$$ 2 $$\leq$$ 2(1 + sin2$$\theta$$) $$\leq$$ 4
$$\therefore$$ Det (A) $$\in$$ [2, 4]
Therefore the choice is: D