If $α$ is a non-real cube root of -2, then the value of $\begin{vmatrix}1&2α&1\\α^2&1&3α^2\\2&2α&1\end{vmatrix}$, is
Answer & explanation
Correct answer: option 3
We have, $α = (−2)^{1/3} w$ or, $α = (− 2)^{1/3} w^2$
$∴α^3=-2$
Now, $\begin{vmatrix}1&2α&1\\α^2&1&3α^2\\2&2α&1\end{vmatrix}$
$=\begin{vmatrix}1&2α&1\\α^2&1&3α^2\\1&0&0\end{vmatrix}$ [Applying $R_3 → R_3 - R_1$]
$=6α^3-1$ [On expanding along $R_3$]
$=-13$ $[∵ α^3 = -2]$