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Question
Let A = `[(1, sin theta, 1),(-sin theta,1,sin 1),(-1, -sin theta, 1)]` where 0 ≤ θ≤ 2π, then ______.
Options
Det (A) = 0
Det (A) ∈ (2, ∞)
Det (A) ∈ (2, 4)
Det (A)∈ [2, 4]
Solution
Let A = `[(1, sin theta, 1),(-sin theta,1,sin 1),(-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 + sin2θ] - sinθ[-sinθ + sinθ] + 1[sin2θ + 1]
= `1 + sin^2theta + sin^2theta + 1`
= `2 + 2sin^2theta = 2(1 + sin^2theta)`
When θ = 0, π, 2π then sinθ = 0
⇒ |A| = 2
When `theta = pi/2, (3pi)/2` then sin2θ = 1,
|A| = 2(1 + 1) = 2 × 2 = 4
∴ det A ∈ [2, 4]
Hence, option det (A) ∈ [2, 4] is correct.
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