Question

If $\omega$ is a complex cube root of unity, then a root of the equation $\left|\begin{array}{ccc}x+1& \omega & {\omega }^2\\ \omega & x+{\omega }^2& 1\\ {\omega }^2& 1& x+\omega \end{array}\right|=0$, is

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

Answer 1

The correct option is D $x=0$

$\left|\begin{array}{ccc}x+1& \omega & {\omega }^2\\ \omega & x+{\omega }^2& 1\\ {\omega }^2& 1& x+\omega \end{array}\right|=0$
$C_1\to C_1+C_2+C_3$
$\left|\begin{array}{ccc}x+1+\omega +{\omega }^2& \omega & {\omega }^2\\ x+1+\omega +{\omega }^2& x+{\omega }^2& 1\\ x+1+\omega +{\omega }^2& 1& x+\omega \end{array}\right|=0$
$x\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ 1& x+{\omega }^2& 1\\ 1& 1& x+\omega \end{array}\right|=0$ $(1+\omega +{\omega }^2=0)$
$R_1\to R_1-R_2,R_2\to R_2-R_3$
$x\left|\begin{array}{ccc}0& \omega -x-{\omega }^2& {\omega }^2-1\\ 0& x+{\omega }^2-1& 1-x-\omega \\ 1& 1& x+\omega \end{array}\right|=0$
$x[x^2+x\omega -{\omega }^2+\omega ]=0$
$\Rightarrow x=0$ is a root of the equation