Question

Choose the correct option:

Let `|(0, sin theta, 1),(-sintheta, 1, sin theta),(1, -sin theta, 1 - a)|` where 0 ≤ θ ≤ 2n, then

Options

• Det (A) = 0

• Det (A) ∈ (2, `oo`)

• Det (A) ∈ (2, 4)

• Det (A) ∈ (2, 4)

Answer 1

Det (A) ∈ (2, 4)

Explanation:

Given A = `|(0, sin theta, 1),(- sin theta, 1, sin theta),(1, - sin theta, 1 - a)|`

∴ |A| `1(1 + sin^2theta) - sin theta (- sin theta + sin theta) + 1(sin^2theta + 1)`

= `1 + sin^2theta + sin^2theta + 1`

= `2 + 2sin^2theta`

= `2(1 + sin^2theta)`

Now, 0 ≤ θ ≤ 2π

⇒ 0 ≤ sin θ ≤ 1

⇒ 0 ≤ sin²θ ≤ 1

⇒ 1 ≤ 1 + sin²θ ≤ 2

⇒ 2 ≤ 2 (1 + sin²θ) ≤ 4

∴ Det (A) [2, 4)]