Question
Let $$\omega = - \frac{1}{2} + i\frac{{\sqrt 3 }}{2}.$$ Then the value of the determinant \[\left| \begin{array}{l}
1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\
1\,\,\,\,\, - 1 - {\omega ^2}\,\,\,\,\,\,\,\,\,\,{\omega ^2}\\
1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^4}\,\,\,
\end{array} \right|\] is
A.
$$3\omega $$
B.
$$3\omega \left( {\omega - 1} \right)$$
C.
$$3{\omega ^2}$$
D.
$$3\omega \left( {1 - \omega } \right)$$
Answer :
$$3\omega \left( {\omega - 1} \right)$$
Solution :
$$\eqalign{
& {\text{Given that }}\omega = - \frac{1}{2} + i\frac{{\sqrt 3 }}{2},{\omega ^2} = - \frac{1}{2} - i\frac{{\sqrt 3 }}{2} \cr
& {\text{Also 1}} + \omega + {\omega ^2} = 0\,\,{\text{and }}{\omega ^3} = 1 \cr
& {\text{Now given det}}{\text{. is}} \cr} $$
\[\Delta = \left| \begin{array}{l}
1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\
1\,\,\,\,\, - 1 - {\omega ^2}\,\,\,\,\,\,\,\,\,\,{\omega ^2}\\
1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^4}
\end{array} \right| = \left| \begin{array}{l}
1\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\
1\,\,\,\,\,\,\,\,\omega \,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^2}\\
1\,\,\,\,\,\,\,{\omega ^2}\,\,\,\,\,\,\,\,\,\,\,\omega
\end{array} \right|\]
$$\eqalign{
& \left[ {{\text{Using }}\omega = - 1 - {\omega ^2}\,\,{\text{and }}{\omega ^3} = 1} \right] \cr
& {\text{Operating }}\,{C_1} \to {C_1} + {C_2} + {C_3} \cr} $$
\[\Delta = \left| \begin{gathered}
\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,1 \hfill \\
\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\omega \,\,\,\,\,\,\,\,\,\,{\omega ^2} \hfill \\
0\,\,\,\,\,\,\,\,\,{\omega ^2}\,\,\,\,\,\,\,\,\omega \hfill \\
\end{gathered} \right|\,\,\left( {{\text{as 1 + }}\omega {\text{ + }}{\omega ^2} = 0} \right)\]
Expanding along $${C_1},$$ we get
$$3\left( {{\omega ^2} - {\omega ^4}} \right) = 3\left( {{\omega ^2} - \omega } \right) = 3\omega \left( {\omega - 1} \right)$$