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]

Answer 1

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]