If $ω$ is an imaginary cube root of unity, then the value of $\begin{vmatrix}1+ω&ω^2&-ω\\1+ω^2&ω&-ω^2\\ω^2+ω&ω&-ω^2\end{vmatrix}$ is equal to
Answer & explanation
Correct answer: option 4
Using $1+ω+ω^2 = 0$, we get
$\begin{vmatrix}1+ω&ω^2&-ω\\1+ω^2&ω&-ω^2\\ω^2+ω&ω&-ω^2\end{vmatrix}$
$=\begin{vmatrix}1+ω+ω^2&ω^2&-ω\\1+ω^2+ω&ω&-ω^2\\ω^2+2ω&ω&-ω^2\end{vmatrix}$ [Applying $C_1→C_1 + C_2$]
$=\begin{vmatrix}0&ω^2&-ω\\0&ω&-ω^2\\ω-1&ω&-ω^2\end{vmatrix}$
$=(ω-1)\begin{vmatrix}ω^2&-ω\\ω&-ω^2\end{vmatrix}$
$=(ω-1)=(-ω^4+ω^2)=(ω-1)=(-ω+ω^2)$
$=-ω^2+ω^3+ω-ω^2=-ω^2+(1+ω)-ω^2=-3ω^2$