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Let $ω \neq 1$ be a cube root of unity and $Δ = |\begin{array}{ccc} 1 - ω - ω^{2} & 2 & 2 \\ 2 ω & ω - ω^{2} - 1 & 2 ω \\ 2 ω^{2} & 2 ω^{2} & ω^{2} - 1 - ω \end{array}|$ then $∆$ equals

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−ω

3ω(1−ω)

1=ω2

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Correct option is C

Using R1→R1+R2+R3 and 1+ω+ω2=0 we obtainΔ=0

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Let ω≠1 be a cube root of unity and Δ=1−ω−ω2222ωω−ω2−12ω2ω22ω2ω2−1−ω then ∆ equals

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Let $ω \neq 1$ be a cube root of unity and $Δ = |\begin{array}{ccc} 1 - ω - ω^{2} & 2 & 2 \\ 2 ω & ω - ω^{2} - 1 & 2 ω \\ 2 ω^{2} & 2 ω^{2} & ω^{2} - 1 - ω \end{array}|$ then $∆$ equals