Evaluate: 1 1 1 1 ω 2 ω 1 ω ω 2 (where ω is an imaginary cube root of unity). Share with your friends Share 0 Shruti Tyagi answered this Dear Student, 1111ω2ω1ωω2since ω is cube root of unity then we have1+ω2+ω=0now apply Row transformation on the determinantR1=R1+R2+R3⇒11+ω2+ω1+ω2+ω1ω2ω1ωω2⇒3001ω2ω1ωω2expanding the determinant across R1⇒3×ω2×ω2-ω×ω-0+0⇒3ω4-ω2⇒3×ω3×ω-ω2Now ω3=1⇒3×ω-ω2again 1+ω=-ω2⇒3×ω+1+ω=6ω+3 Regards, 0 View Full Answer