Let \[\Delta =\left| \,\begin{matrix} 1 & \omega & 2{{\omega }^{2}} \\ 2 & 2{{\omega }^{2}} & 4{{\omega }^{3}} \\ 3 & 3{{\omega }^{3}} & 6{{\omega }^{4}} \\ \end{matrix}\, \right|\] where \[\omega \] is the cube root of unity, then

Let \[\Delta =\left| \,\begin{matrix} 1 & \omega & 2{{\omega }^

1.	Let \[\Delta =\left\| \,\begin{matrix} 1 & \omega & 2{{\omega }^{2}} \\ 2 & 2{{\omega }^{2}} & 4{{\omega }^{3}} \\ 3 & 3{{\omega }^{3}} & 6{{\omega }^{4}} \\ \end{matrix}\, \right\|\] where \[\omega \] is the cube root of unity, then
A.	\[\Delta =0\]
B.	\[\Delta =1\]
C.	\[\Delta =2\]
D.	\[\Delta =3\]
Answer» B. \[\Delta =1\]

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Let \[\Delta =\left| \,\begin{matrix} 1 & \omega & 2{{\omega }^{2}} \\ 2 & 2{{\omega }^{2}} & 4{{\omega }^{3}} \\ 3 & 3{{\omega }^{3}} & 6{{\omega }^{4}} \\ \end{matrix}\, \right|\] where \[\omega \] is the cube root of unity, then

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