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