Question

If $\omega \ne 1$ is a complex cube root of unity, then value of
$\Delta =\left|\begin{array}{ccc}{a}_1+b_1\omega & a_1{\omega }^2+b_1& c_1+b_1\overline{\omega }\\ a_2+b_2\omega & a_2{\omega }^2+b_2& c_2+b_2\overline{\omega }\\ a_3+b_3\omega & a_3{\omega }^2+b_3& c_3+b_3\overline{\omega }\end{array}\right|$
is

• A. 0
• B. -1
• C. 2
• D. none of these

Answer 1

Answer: (A)

0

$\Delta =\left|\begin{array}{ccc}{a}_1+b_1\omega & a_1{\omega }^2+b_1& c_1+b_1\overline{\omega }\\ a_2+b_2\omega & a_2{\omega }^2+b_2& c_2+b_2\overline{\omega }\\ a_3+b_3\omega & a_3{\omega }^2+b_3& c_3+b_3\overline{\omega }\end{array}\right|={\Delta }_1{\Delta }_2$
Where ${\Delta }_1=\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|$ and ${\Delta }_2=\left|\begin{array}{ccc}1& \omega & 0\\ {\omega }^2& 1& 0\\ 0& \overline{w}& 1\end{array}\right|$
But ${\Delta }_2=\left|\begin{array}{ccc}1& \omega & 0\\ {\omega }^2& 1& 0\\ 0& \overline{w}& 1\end{array}\right|=1(1-\overline{w})-\omega \left({\omega }^2\right)=0$
Hence $\Delta =0$