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Question

Prove that ∣ ∣ ∣1ωω2ωω21ω21ω∣ ∣ ∣=0

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Solution

Solving given determinant, we get

[1(ω31)ω(ω2ω2)+ω2(ωω4)] ....... as (ω3=1)

[(11)ω(0)+ω2(ωω)]

0+0+0

0

Hence proved

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