Question

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

• A. $-1$
• B. $1$
• C. $0$
• D. $\omega$

Answer 1

The correct option is C $0$

Let $A=\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right|$

$=\left|\begin{array}{ccc}1+\omega +{\omega }^2& \omega & {\omega }^2\\ 1+\omega +{\omega }^2& {\omega }^2& 1\\ 1+\omega +{\omega }^2& 1& \omega \end{array}\right|$

$C_1\to C_1+C_2+C_3$
$=\left|\begin{array}{ccc}0& \omega & {\omega }^2\\ 0& {\omega }^2& 1\\ 0& 1& \omega \end{array}\right|[\because 1+\omega +{\omega }^2=0]$

$=0$