Question

If $\omega$ is a cube root of unity, then $\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right|$ is equal to

• A. 1
• B. 0
• C. $\omega$
• D. ${\omega }^2$

Answer 1

Answer: (B)

0

$\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right|$
applying $R_1\to R_1+R_2+R_3$ gives
$\left|\begin{array}{ccc}0& 0& 0\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right|=0$ ($\because 1+\omega +{\omega }^2=0$ )
$\therefore \left|\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right|=0$
Hence, option B.