Concept: ω is a cube root of unity. Property of cube root of unity: ω3=1, and 1+ω+ω2=0 Calculations: Given, ω is a cube root of unity. ⇒ω3=1,ad1+ω+ω2=0 Now, consider the determinant |
1+ω
ω2
−ω
1+ω2
ω
−ω2
ω2+ω
ω
−ω2
| Taking ω and −ω common from C2 and C3 respectively, =−ω2|
1+ω
ω
1
1+ω2
1
ω
ω2+ω
1
ω
| =−ω2[(1+ω)(ω−ω)−ω(ω+1−1−ω2)+(1+ω2−ω2−ω)] =−ω2[−ω2+1+1−ω] =ω−2ω2+1 =1+ω+ω2−3ω2 =0−3ω2 =−3ω2 Hence, If ω is a cube root of unity, then find the value of the determinant |