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Question

If ω be imaginary root of unity then evaluate ∣ ∣ ∣2+3ω100+ω2032ω2111+ω100+3ω2032ω2ω2ω23+ω100+ω203∣ ∣ ∣

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Solution

ω100=(ω3)33.ω=ω,ω203=(ω100)2.ω3
=ω2.1=ω2
Apply C1+C2+C3 then new C1 becomes a column of zeros as 3(1+ω+ω2)=0

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