If  ω is the cube root of unity, then what is the value of

\(\rm\begin{vmatrix} 1 & \omega &\omega^2 \\ \omega & \omega^2 &1 \\ \omega^2& 1 & \omega \end{vmatrix}\)

Concept:

If the 1, ω and ω²  are the cube roots of unity, then 1 + ω + ω² = 0

If any row and column have all the elements as zero in the square matrix, then the determinant is zero.

Calculation:

D = \(\rm\begin{vmatrix} 1 & ω &ω^2 \\ ω & ω^2 &1 \\ ω^2& 1 & ω \end{vmatrix}\)

R₃ = R₃ + R₁ + R₂

D = \(\rm\begin{vmatrix} 1 & ω &1+ω+ω^2 \\ ω & ω^2 &1+ω+ω^2 \\ ω^2& 1 & 1+ω+ω^2 \end{vmatrix}\)

D = \(\rm\begin{vmatrix} 1 & ω &0 \\ ω & ω^2 &0 \\ ω^2& 1 & 0 \end{vmatrix}\) = 0