Correct Answer - Option 1 : 0
Concept:
ω is the cube root of unity.
⇒ω3 = 1 and 1 + ω + ω2 = 0
Calculations:
Given, ω is the cube root of unity.
⇒ω3 = 1 and 1 + ω + ω2 = 0
Consider, \(\begin{vmatrix} 1 & ω & 2ω^2 \\\ 2 & 2ω^2 & 4ω^3 \\\ 3 & 3ω^3 & 6ω^4 \end{vmatrix}\)
= 1(12 ω6 - 12 ω6) - ω (12 ω4 - 12 ω3) + 2ω2 (6 ω3 - 6ω2)
= 0 - ω [12 ω3 ω - 12 (1)] + 2ω2 [6 (1) - 6ω2]
= 12 ω2 + 12 ω - 12ω2 - 12ω4
= 12 ω - 12ω3 ω
= 12 ω - 12 ω
= 0
Hence, the value of \(\begin{vmatrix} 1 & ω & 2ω^2 \\\ 2 & 2ω^2 & 4ω^3 \\\ 3 & 3ω^3 & 6ω^4 \end{vmatrix}\) where ω is the cube root of unity is zero.
