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Question

Let A=1sin θ1-sin θ1sin θ-1- sin θ1, where 0 θ 2π. Then,

(a) Det A=0
(b) Det A 2,
(c) Det A 2, 4
(d) Det A 2, 4

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Solution

(d) Det A 2, 4

1 sin θ 1- sin θ 1 sin θ -1 - sin θ 1= 1 sin θ 2- sin θ 1 0 -1 - sin θ 0 Applying C3C3 + C1= 2 × - sin θ 1 -1 - sin θ Expanding along C3=2 sin2 θ + 1Given: 0θ2π -1sin θ10sin2 θ1A=2sin2 θ + 1A =2 × 1 = 2 θ = 0 =2 × 2 = 4 θ = 2π Det A2, 4

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