if $$A=1sin$$θ$$1$$−$$sin$$θ$$1sin$$θ−$$1$$−$$sin$$θ$$1$$ then for all θ∈$$3$$$$π$$$$4$$,$$5$$$$π$$$$4$$ minimum value of detA is

We have |A|$$=2+2sin2$$θ, $$Now3$$$$π$$$$4$$<θ<$$5$$$$π$$$$4$$ ⇒$$-12$$

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if $A = [\begin{matrix} 1 & \sin θ & 1 \\ - \sin θ & 1 & \sin θ \\ - 1 & - \sin θ & 1 \end{matrix}]$ then for all $θ \in (\frac{3 π}{4}, \frac{5 π}{4})$ minimum value of det $(A)$ is

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$\begin{array}{ll} W e h a v e | A | = 2 + 2 \sin^{2} θ, N o w & \frac{3 π}{4} < θ < \frac{5 π}{4} \\ \Rightarrow - \frac{1}{\sqrt{2}} < \sin θ < \frac{1}{\sqrt{2}} \\ \Rightarrow 0 \leq \sin^{2} θ < \frac{1}{2} \\ \Rightarrow 2 \leq 2 + 2 \sin^{2} θ < 3 \\ \Rightarrow 2 \leq | A | < 3 \end{array}$
$∴ T h e m i n i m u m v a l u e o f |A| i s 2 .$

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Introduction to Determinants

Remember concepts with our Masterclasses.

if A=1sinθ1−sinθ1sinθ−1−sinθ1 then for all θ∈3π4,5π4 minimum value of detA is

We have |A|=2+2sin2θ, Now3π4<θ<5π4 ⇒-12<sinθ<12 ⇒ 0≤sin2θ<12 ⇒ 2≤2+2sin2θ<3 ⇒ 2≤|A|<3∴The minimum value of A is 2.

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if $A = [\begin{matrix} 1 & \sin θ & 1 \\ - \sin θ & 1 & \sin θ \\ - 1 & - \sin θ & 1 \end{matrix}]$ then for all $θ \in (\frac{3 π}{4}, \frac{5 π}{4})$ minimum value of det $(A)$ is