If $ω$ is a complex cube root of unity, then the value of the determinant
$Δ=\begin{bmatrix}1 & ω &ω^2\\ω & ω^2 & 1\\ω^2 & 1 & ω\end{bmatrix},$ is
Answer & explanation
Correct answer: option 2
The correct answer is option (2) : 0
Applying $C_1→C_1+C_2+C_3<$ we have
$Δ=\begin{bmatrix}1+ω+ω^2 & ω &ω^2\\ω+ω^2+1 & ω^2 & 1\\ω^2+1+ω & 1 & ω\end{bmatrix}$
$⇒Δ=\begin{bmatrix}0 & ω &ω^2\\0 & ω^2 & 1\\0 & 1 & ω\end{bmatrix}$ $[∵1+ω+ω^2=0]$
$⇒Δ=0$ $[∵C_1 $ consists of all zeroes $]$