UPSC NDA Math Model Paper 1

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 |

1+ω	ω2	−ω
1+ω2	ω	−ω2
ω2+ω	ω	−ω2

| is −3ω2