Question

If $\omega \ne 1$ is a complex cube root of unity, and
$x+iy=\left|\begin{array}{ccc}1& i& -\omega \\ -i& 1& {\omega }^2\\ \omega & -{\omega }^2& 1\end{array}\right|$
then

• A. $x=-1,y=0$
• B. $x=1,y=-1$
• C. $x=1,y=1$
• D. none of these

Answer 1

Answer: (A)

$x=-1,y=0$

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

$=1(1+{\omega }^4)-i(-i-{\omega }^3)-\omega (-i-\omega )$

$=1+\omega -i(-i+\omega )+i\omega +{\omega }^2$

$=1+\omega -1-i\omega +i\omega +{\omega }^2=\omega +{\omega }^2=-1$

$\Rightarrow x=-1,y=0$