Question

Let $\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1& -1-w^2& w^2\\ 1& w& w^4\end{array}\right|$, where $w\ne 1$ is a complex number such that $w^3=1$. Then $\Delta$ equals

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

Answer 1

Answer: (B)

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