Question

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

• A. $3\omega$
• B. $3\omega (\omega -1)$
• C. $3{\omega }^2$
• D. $3\omega (1-\omega )$

Answer 1

Answer: (B)

$3\omega (\omega -1)$

As $\omega$ is a cube root of unity, $1+\omega +{\omega }^2=0$ and ${\omega }^3=1$ Thus, we can write $\Delta$ as
$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right|=\left|\begin{array}{ccc}3& 1& 1\\ 1+\omega +{\omega }^2& \omega & {\omega }^2\\ 1+{\omega }^2+\omega & {\omega }^2& \omega \end{array}\right|$ --------Using $C_1\to C_1+C_2+C_3$

$=\left|\begin{array}{ccc}3& 1& 1\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right|=3({\omega }^2-{\omega }^4)=3\omega (\omega -1)$