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Question
Let A = `[(1, sin theta, 1),(-sin theta,1,sin theta),(-1, -sin theta, 1)]` where 0 ≤ θ ≤ 2π, then ______.
Options
Det (A) = 0
Det (A) ∈ (2, ∞)
Det (A) ∈ (2, 4)
Det (A) ∈ [2, 4]
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Solution
Let A = `[(1, sin theta, 1),(-sin theta,1,sin theta),(-1, -sin theta, 1)]` where 0 ≤ θ ≤ 2π, then Det (A) ∈ [2, 4].
Explanation:
Let, A = `[(1,sintheta,1),(-sintheta,1,sintheta),(-1,-sintheta,1)]`
∴ |A| = `[(1,sintheta,1),(-sintheta,1,sintheta),(-1,-sintheta,1)]`
= 1[1 + sin2 θ] − sin θ[−sin θ + sin θ] + 1[sin2 θ + 1]
= 1 + sin2 θ + sin2 θ + 1
= 2 + 2 sin2 θ
= 2(1 + sin2 θ)
When θ = 0, π, 2π then sin θ = 0
⇒ |A| = 2
When θ = `pi/2, (3pi)/2` then sin2 θ = 1
|A| = 2(1 + 1)
= 2 × 2
= 4
∴ Det A ∈ [2, 4]
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