If 1, ω, ω2 are the roots of unity then is equal to-

Engineering

Mathematics

Introduction to Determinants

Question

If 1, ω, ω² are the roots of unity then $Δ = |\begin{array}{ccc} 1 & ω^{n} & ω^{2 n} \\ ω^{2 n} & 1 & ω^{n} \\ ω^{n} & ω^{2 n} & 1 \end{array}|$ is equal to-

ω²

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$Δ = |\begin{array}{lll} 1 ω^{n} ω^{2 n} \\ ω^{n} ω^{2 n} 1 \\ ω^{2 n} 1 ω^{n} \end{array}|$

= [1(ω³ⁿ – 1) – ωⁿ(ω²ⁿ – ω²ⁿ) + ω²ⁿ(ωⁿ –ω⁴ⁿ)]

= ω³ⁿ – 1 – ω³ⁿ + ω3n + ω³ⁿ – ω⁶ⁿ

= 2ω³ⁿ – 1 – ω⁶ⁿ

n → take any positive integer

n = 1

= 2ω³ – 1 – ω⁶

= 2 – 1 – (ω³)²[ω³ = 1 (given)]

= 1 – (ω³)²

= 1 – 1 = 0

Option A is correct.

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