Question

# 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. $3\omega \left( {1 - \omega } \right)$

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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.

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 $\left| {\begin{array}{*{20}{c}} 1&1&1 \\ 1&\omega &{{\omega ^2}} \\ 1&{{\omega ^2}}&{{\omega ^4}} \end{array}} \right|$
$\Rightarrow 1\left[ {\left( \omega \right)\left( {{\omega ^4}} \right) - \left( {{\omega ^2}} \right)\left( {{\omega ^2}} \right)} \right] - 1\left[ {1\left( {{\omega ^4}} \right) - 1\left( {{\omega ^2}} \right)} \right] + 1\left[ {1\left( {{\omega ^2}} \right) - 1\left( \omega \right)} \right] \\ \Rightarrow 1\left[ {{\omega ^5} - {\omega ^4}} \right] - 1\left[ {{\omega ^4} - {\omega ^2}} \right] + 1\left[ {{\omega ^2} - \omega } \right] \\$
$\Rightarrow 1\left[ {{\omega ^3}{\omega ^2} - {\omega ^3}\omega } \right] - 1\left[ {{\omega ^3}\omega - {\omega ^2}} \right] + 1\left[ {{\omega ^2} - \omega } \right]$
We know that ${\omega ^3} = 1$, by using this formula we have
$\Rightarrow 1\left[ {\left( 1 \right){\omega ^2} - \left( 1 \right)\omega } \right] - 1\left[ {\left( 1 \right)\omega - {\omega ^2}} \right] + 1\left[ {{\omega ^2} - \omega } \right] \\ \Rightarrow 1\left[ {{\omega ^2} - \omega } \right] - 1\left[ {\omega - {\omega ^2}} \right] + 1\left[ {{\omega ^2} - \omega } \right] \\ \Rightarrow {\omega ^2} - \omega - \omega + {\omega ^2} + {\omega ^2} - \omega \\$
$\Rightarrow 3{\omega ^2} - \omega \\ \Rightarrow 3\omega \left( {\omega - 1} \right) \\$
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.