MCQ
Let $\omega $ be a complex number such that  $2\omega + 1 = z$ where $z = \sqrt { - 3} $ . If $\left| {\begin{array}{*{20}{c}}1&1&1\\1&{ - {\omega ^2} - 1}&{{\omega ^2}}\\1&{{\omega ^2}}&{{\omega ^7}}\end{array}} \right| = 3k$ then $k$ is equal to :
  • A
    $1$
  • $-z$
  • C
    $z$
  • D
    $-1$

Answer

Correct option: B.
$-z$
Given $2\omega  + 1 = z;$$z = \sqrt {3i} $
$ \Rightarrow \omega  = \frac{{\sqrt {3i}  - 1}}{2}$
$ \Rightarrow \omega $ is complex cube root of unity 
Applying ${R_1} \to {R_1} + {R_2} + {R_3}$
$ = \left| \begin{array}{l}
3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\
1\,\,\,\,\, - {\omega ^2} - 1\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^2}\\
1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\omega 
\end{array} \right|\,$
$ = 3\left( { - 1 - \omega - \omega } \right) $
$= - 3\left( {1 + 2\omega } \right)\, = - 3z$
$ \Rightarrow k = - z$

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