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Let A = [(1, sin theta, 1),(-sin theta,1,sin theta),(-1, -sin theta, 1)] where 0 ≤ θ ≤ 2π, then ______. - Mathematics

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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]

MCQ
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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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Chapter 4: Determinants - Exercise 4.7 [Page 143]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 4 Determinants
Exercise 4.7 | Q 19 | Page 143

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