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

Answer 1

Answer: (D)

$0$

Applying $C_1\to C_1+C_2+C_2$, we get
$\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right|$

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

$=\left|\begin{array}{ccc}0& \omega & {\omega }^2\\ 0& {\omega }^2& 1\\ 0& 1& \omega \end{array}\right|$ ...... [$\because \omega$ is a cube root of unity, then ${\omega }^2+\omega +1=0$]

$=0$