Question

Let $w=-\frac{1}{2}+i\frac{\sqrt{3}}{2}\cdot$ Then the value of the determinant $\left|\begin{array}{ccc}1& 1& 1\\ 1& -1-w^2& w^2\\ 1& w^2& w^4\end{array}\right|$ is

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

Answer 1

The correct option is B $3w(w-1)$

Since w is the cube root of unity, and -1-$w^2$ = w the value of determinant is:

1($w^2$ - $w^4$) + 1($w^2$ - $w^4$) + 1($w^2$ - w )

= $w^2$ - w +$w^2$ -w +$w^2$ - w

=3$w^2$ - 3w = 3w(w - 1)

Hence, option B is correct.