If A = 1sinθ1-sinθ1sinθ-1-sinθ1; then for all θ ∈ 3π4, 5π4, det(A) lies in the interval

R3→R3+R1|A|=1sinθ1-sinθ1sinθ002 =2(1+sin2θ) =2+2sin2θsince θ∈3π4, 5π4, 32<|A|≤3

If A = 1sinθ1-sinθ1sinθ-1-sinθ1; then for all θ ∈ 3π4, 5π4, det(A) lies in the interval

R3→R3+R1|A|=1sinθ1-sinθ1sinθ002 =2(1+sin2θ) =2+2sin2θsince θ∈3π4, 5π4, 32<|A|≤3

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If $A = [\begin{matrix} 1 & sinθ & 1 \\ - sinθ & 1 & sinθ \\ - 1 & - sinθ & 1 \end{matrix}]$ ; then for all $θ \in (\frac{3 π}{4}, \frac{5 π}{4})$ , det(A) lies in the interval

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$(1, \frac{5}{2})$

$(\frac{5}{2}, 4)$

$(0, \frac{3}{2})$

$(\frac{3}{2}, 3)$

answer is D.

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Detailed Solution

$\begin{array}{rcl} R_{3} & \to & R_{3} + R_{1} \\ | A | & = & |\begin{matrix} 1 & sinθ & 1 \\ - sinθ & 1 & sinθ \\ 0 & 0 & 2 \end{matrix}| \\ = & 2 (1 + \sin^{2} θ) \\ = & 2 + 2 \sin^{2} θ \\ since θ & \in & (\frac{3 π}{4}, \frac{5 π}{4}), \\ \frac{3}{2} & < & | A | \leq 3 \end{array}$