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,

Answer 1

In this particular question use the concept of cube root of unity, and use some of the properties of cube root of unity which is given as, ${\omega ^3} = 1,1 + \omega + {\omega ^2} = 0$, so, first expand the determinant then apply these properties to simplify so use these concepts to reach the solution of the question.

Complete step by step answer:
$\omega = - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}$
Now we have to find out the value of \left| {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\\ 1&{{\omega ^2}}&{{\omega ^4}} \end{array}} \right|
Now as we know that the properties of cube root of unity are ${\omega ^3} = 1,1 + \omega + {\omega ^2} = 0$, so first simplify the determinant we have,
\Rightarrow \left| {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\\ 1&{{\omega ^2}}&{{\omega ^4}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\\ 1&{{\omega ^2}}&{\omega .{\omega ^3}} \end{array}} \right|
\Rightarrow \left| {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\\ 1&{{\omega ^2}}&\omega \end{array}} \right|
Now expand the determinant we have,
\Rightarrow \left| {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\\ 1&{{\omega ^2}}&\omega \end{array}} \right| = 1\left| {\begin{array}{*{20}{c}} { - 1 - {\omega ^2}}&{{\omega ^2}} \\\ {{\omega ^2}}&\omega \end{array}} \right| - 1\left| {\begin{array}{*{20}{c}} 1&{{\omega ^2}} \\\ 1&\omega \end{array}} \right| + 1\left| {\begin{array}{*{20}{c}} 1&{ - 1 - {\omega ^2}} \\\ 1&{{\omega ^2}} \end{array}} \right|
Now expand the mini determinant we have,
\Rightarrow \left| {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\\ 1&{{\omega ^2}}&\omega \end{array}} \right| = 1\left( {\omega \left( { - 1 - {\omega ^2}} \right) - \left( {{\omega ^2}} \right)\left( {{\omega ^2}} \right)} \right) - 1\left( {\omega - {\omega ^2}} \right) + 1\left( {{\omega ^2} - \left( { - 1 - {\omega ^2}} \right)} \right)
Now simplify it we have,
\Rightarrow \left| {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\\ 1&{{\omega ^2}}&\omega \end{array}} \right| = \left( { - \omega - {\omega ^3}} \right) - \left( {{\omega ^4}} \right) - \omega + {\omega ^2} + {\omega ^2} + 1 + {\omega ^2}
\Rightarrow \left| {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\\ 1&{{\omega ^2}}&\omega \end{array}} \right| = - \omega - {\omega ^3} - \left( {\omega .{\omega ^3}} \right) - \omega + {\omega ^2} + {\omega ^2} + 1 + {\omega ^2}
\Rightarrow \left| {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\\ 1&{{\omega ^2}}&\omega \end{array}} \right| = - \omega - 1 - \left( \omega \right) - \omega + {\omega ^2} + {\omega ^2} + 1 + {\omega ^2}, $\left[ {\because {\omega ^3} = 1} \right]$
\Rightarrow \left| {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\\ 1&{{\omega ^2}}&\omega \end{array}} \right| = - 3\omega + 3{\omega ^2} = 3\left( {{\omega ^2} - \omega } \right)
Now as we know that $1 + \omega + {\omega ^2} = 0$
$\Rightarrow {\omega ^2} = - \left( {1 + \omega } \right)$
\Rightarrow \left| {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\\ 1&{{\omega ^2}}&\omega \end{array}} \right| = 3\left( { - 1 - \omega - \omega } \right)
\Rightarrow \left| {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\\ 1&{{\omega ^2}}&\omega \end{array}} \right| = 3\left( { - 1 - 2\omega } \right) = - 3 - 6\omega
Now substitute the value of $\omega = - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}$ we have,
\Rightarrow \left| {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\\ 1&{{\omega ^2}}&\omega \end{array}} \right| = - 3 - 6\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)
\Rightarrow \left| {\begin{array}{*{20}{c}} 1&1&1 \\\ 1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\\ 1&{{\omega ^2}}&\omega \end{array}} \right| = - 3 - \left( { - 3 + i3\sqrt 3 } \right) = - i3\sqrt 3
So this is the required value of the determinant.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic properties of cube root of unity which is stated above, and also recall how to expand the determinant, so first simplify the determinant using the properties then expand the determinant as above then again apply the properties and simplify and at last substitute the value of $\omega$ we will get the required answer.