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Question: Let \[\omega = \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}\], then the value of the determinant \[\left| {...

Let ω=12+i32\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. 3ω3\omega
B. 3ω(ω−1)3\omega \left( {\omega - 1} \right)
C. 3ω23{\omega ^2}
D. 3ω(1−ω)3\omega \left( {1 - \omega } \right)

Explanation

Solution

Hint:
In this problem, ω\omega is the cube root of unity. And 1+ω+ω2=01 + \omega + {\omega ^2} = 0 i.e., −1−ω2=ω- 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−ω2=ω- 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

⇒1[(ω)(ω4)−(ω2)(ω2)]−1[1(ω4)−1(ω2)]+1[1(ω2)−1(ω)] ⇒1[ω5−ω4]−1[ω4−ω2]+1[ω2−ω]  \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
⇒1[ω3ω2−ω3ω]−1[ω3ω−ω2]+1[ω2−ω]\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 ω3=1{\omega ^3} = 1, by using this formula we have

⇒1[(1)ω2−(1)ω]−1[(1)ω−ω2]+1[ω2−ω] ⇒1[ω2−ω]−1[ω−ω2]+1[ω2−ω] ⇒ω2−ω−ω+ω2+ω2−ω  \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

⇒3ω2−ω ⇒3ω(ω−1)  \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 3ω(ω−1)3\omega \left( {\omega - 1} \right)
Thus, the correct option is B. 3ω(ω−1)3\omega \left( {\omega - 1} \right)
Note: In the given question ‘ω\omega ’is the imaginary root. Always use the properties of cube roots of unity whenever you find ‘ω\omega ’ in the question if possible so that you can solve the problem more easily.