Evaluate: 1 1 1 1 ω 2 ω 1 ω ω 2 (where ω is an imaginary cube root of unity).

Dear Student,


1111ω2ω1ωω2since ω is cube root of unity then we have1+ω2+ω=0now apply Row transformation on the determinantR1=R1+R2+R311+ω2+ω1+ω2+ω1ω2ω1ωω23001ω2ω1ωω2expanding the determinant across R13×ω2×ω2-ω×ω-0+03ω4-ω23×ω3×ω-ω2Now ω3=13×ω-ω2again 1+ω=-ω23×ω+1+ω=6ω+3
Regards,

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