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Let $ω = \frac{1}{2} (- 1 + \sqrt{3} i)$ and $Δ = |\begin{array}{ccc} 1 & 1 & 1 \\ 1 & - 1 - ω^{2} & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{array}|$ then $Δ$ equals

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3ω

3ω(ω−1)

3ω2

3ω(1−ω)

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detailed solution

Correct option is B

Using ω3=1 and 1+ω+ω2=0, and applying C1→C1+C2+C3, we getΔ=3110ωω20ω2ω=3ω2−ω=3ω(ω−1)

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Let ω=12(−1+3i) and Δ=1111−1−ω2ω21ω2ω4 then Δ equals

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Let $ω = \frac{1}{2} (- 1 + \sqrt{3} i)$ and $Δ = |\begin{array}{ccc} 1 & 1 & 1 \\ 1 & - 1 - ω^{2} & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{array}|$ then $Δ$ equals