If aᵣ = cos 2rπ/9 + i sin 2rπ/9, r = 1, 2, 3, ....., i = √(- 1), then
the determinant | a₁, a₂, a₃; a₄, a₅, a₆; a₇, a₈, a₉ | is equal to :

Solution

Maths sir · 4 steps

Every entry is a power of one complex number, so the rows come out proportional and the determinant is zero.

1. Rewrite each entry. By De Moivre's theorem, aᵣ = (cos 2π/9 + i sin 2π/9)ʳ = ωʳ, where ω = e^(2π i/9). Also ω⁹ = 1, so a₉ = 1.
2. Compare the rows. Row 2 is ω⁴, ω⁵, ω⁶, which is ω³ times row 1 (ω, ω², ω³). Row 3 is ω⁶ times row 1.

   |ω, ω², ω³; ω⁴, ω⁵, ω⁶; ω⁷, ω⁸, ω⁹| = ω³ |ω, ω², ω³; ω, ω², ω³; ω⁷, ω⁸, ω⁹| = 0

3. Check the options. (c): a₁a₉ - a₃a₇ = ω¹⁰ - ω¹⁰ = 0. (a): ω⁸ - ω¹² = ω⁸ - ω³ ≠ 0. (b) is a₉ = 1, and (d) is a₅ = ω⁵ ≠ 0.
4. So. The determinant is 0, and only option (c) also equals 0.

The idea to keep. When entries are powers of one number, check whether one row is a fixed multiple of another. Two proportional rows make the determinant zero at once.

Answer from the NTA final answer key. Solution by PrepWiser.