Question 9
Let A = 1 sin 1 sin 1 sin 1 sin 1 θ − θ θ − − θ , where 0 ≤ θ ≤ 2π. Then (A) Det (A) = 0 (B) Det (A) ∈ (2, ∞) (C) Det (A) ∈ (2, 4) (D) Det (A) ∈ [2, 4]
SOLUTION
→→→→
Let A = 1 sin 1 sin 1 sin 1 sin 1 θ − θ θ − − θ , where 0 ≤ θ ≤ 2π. Then (A) Det (A) = 0 (B) Det (A) ∈ (2, ∞) (C) Det (A) ∈ (2, 4) (D) Det (A) ∈ [2, 4]