Question

Let $\Delta= \begin{vmatrix}1&1&1\\ 1&-1-w^{2}&w^{2}\\ 1&w&w^{4}\end{vmatrix}$, where $w \neq 1$ is a complex number such that $w^3 = 1$. Then $\Delta$ equals

• A. $3 w + w^2$
• B. $3w^2$
• C. $3(w = w^2)$
• D. $-3 w^2$
• E. $3 w^2 + 1 $

Answer 1

Answer: (B)

We have,
$\Delta=\begin{vmatrix}1 & 1 & 1 \\ 1 & -1-w^{2} & w^{2} \\ 1 & w & w^{4}\end{vmatrix}$
$=\begin{vmatrix}1 & 1 & 1 \\ 1 & w & w^{2} \\ 1 & w & w\end{vmatrix}$
$\left[\because 1+w+w^{2}=0, w^{3}=1\right]$
$=1\left(w^{2}-w^{3}\right)-1\left(w-w^{2}\right)+1(w-w)$
$=w^{2}-1-w+w^{2}$
$=2 \,w^{2}-(1+w)$
$=2\, w^{2}-\left(-w^{2}\right)$
$=3 \,w^{2}$