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Let \[\omega = \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}\], then the value of the determinant \[\left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\
  1&{{\omega ^2}}&{{\omega ^4}}
\end{array}} \right|\] is
A. \[3\omega \]
B. \[3\omega \left( {\omega - 1} \right)\]
C. \[3{\omega ^2}\]
D. \[3\omega \left( {1 - \omega } \right)\]

Answer
VerifiedVerified
162.3k+ views
Hint:
In this problem, \[\omega \] is the cube root of unity. And \[1 + \omega + {\omega ^2} = 0\] i.e., \[ - 1 - {\omega ^2} = \omega \] is the property of cube roots of unity. So, use this concept to reach the solution of the problem.


Complete step-by-step answer:
Given determinant is \[\left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\
  1&{{\omega ^2}}&{{\omega ^4}}
\end{array}} \right|\]
By using the property \[ - 1 - {\omega ^2} = \omega \], the determinant can be converted into \[\left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&\omega &{{\omega ^2}} \\
  1&{{\omega ^2}}&{{\omega ^4}}
\end{array}} \right|\]
By solving the determinant, we have
\[
   \Rightarrow 1\left[ {\left( \omega \right)\left( {{\omega ^4}} \right) - \left( {{\omega ^2}} \right)\left( {{\omega ^2}} \right)} \right] - 1\left[ {1\left( {{\omega ^4}} \right) - 1\left( {{\omega ^2}} \right)} \right] + 1\left[ {1\left( {{\omega ^2}} \right) - 1\left( \omega \right)} \right] \\
   \Rightarrow 1\left[ {{\omega ^5} - {\omega ^4}} \right] - 1\left[ {{\omega ^4} - {\omega ^2}} \right] + 1\left[ {{\omega ^2} - \omega } \right] \\
\]
This can be rewrite as
\[ \Rightarrow 1\left[ {{\omega ^3}{\omega ^2} - {\omega ^3}\omega } \right] - 1\left[ {{\omega ^3}\omega - {\omega ^2}} \right] + 1\left[ {{\omega ^2} - \omega } \right]\]
We know that \[{\omega ^3} = 1\], by using this formula we have
\[
   \Rightarrow 1\left[ {\left( 1 \right){\omega ^2} - \left( 1 \right)\omega } \right] - 1\left[ {\left( 1 \right)\omega - {\omega ^2}} \right] + 1\left[ {{\omega ^2} - \omega } \right] \\
   \Rightarrow 1\left[ {{\omega ^2} - \omega } \right] - 1\left[ {\omega - {\omega ^2}} \right] + 1\left[ {{\omega ^2} - \omega } \right] \\
   \Rightarrow {\omega ^2} - \omega - \omega + {\omega ^2} + {\omega ^2} - \omega \\
\]
Cancelling the common terms, we have
\[
   \Rightarrow 3{\omega ^2} - \omega \\
   \Rightarrow 3\omega \left( {\omega - 1} \right) \\
\]
Therefore, the determinant of \[\left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\
  1&{{\omega ^2}}&{{\omega ^4}}
\end{array}} \right|\] is \[3\omega \left( {\omega - 1} \right)\]
Thus, the correct option is B. \[3\omega \left( {\omega - 1} \right)\]
Note: In the given question ‘\[\omega \]’is the imaginary root. Always use the properties of cube roots of unity whenever you find ‘\[\omega \]’ in the question if possible so that you can solve the problem more easily.