Question

If $\omega$ is a complex cube root of unity, then $\begin{vmatrix} 1 & \omega& \omega^2 \\\[0.3em] \omega& \omega^2 & 1 \\\[0.3em] \omega^2 & 1& \omega \end{vmatrix}$ is equal to

• A. -1
• B. 1
• C. 0
• D. $\omega$

Answer 1

Answer: (C)

0

Answer 2

$Let\,A= \begin{vmatrix} 1 & \omega& \omega^2 \\\[0.3em] \omega& \omega^2 & 1 \\\[0.3em] \omega^2 & 1& \omega \end{vmatrix}$
$Let\,A= \begin{vmatrix} 1 +\omega +\omega^2& \omega& \omega^2 \\\[0.3em] 1+ \omega+\omega^2 & \omega^2 & 1 \\\[0.3em] 1+\omega+\omega^2 & 1& \omega \end{vmatrix}$
$C_1 \to C_1+C_2+C_3$
$Let\,A= \begin{vmatrix} 0& \omega& \omega^2 \\\[0.3em] 0 & \omega^2 & 1 \\\[0.3em] 0 & 1& \omega \end{vmatrix} [=1+\omega+\omega+\omega^2=0]$ =0