Question

If 1, $w$, $w^2$ are the cube roots of unity, then $\left|\begin{array}{ccc}1& w^n& w^{2n}\\ w^n& w^{2n}& 1\\ w^{2n}& 1& w^n\end{array}\right|$ is equal to

• A. $w^2$
• B. 0
• C. 1
• D. w

Answer 1

Answer: (B)

0

Expanding the determinant, we get
$1(w^{3n}-1)+w^n(w^{2n}-w^{2n})+w^{2n}(w^n-w^{4n})$
$=1\left(0\right)+w^n\left(0\right)+w^{2n}(w^n-w^n)$
$w^{4n}={\left(w^4\right)}^n$
$={(w^3.w)}^n$
$=i{w}^n$
$=1\left(0\right)+w^n\left(0\right)+w^{2n}\left(0\right)$
$=0$