Question

If $1,\omega {\text{ }}and{\text{ }}{\omega ^2}$ are the cube roots of unity, then \left| {\begin{array}{*{20}{c}} 1&{{\omega ^n}}&{{\omega ^{2n}}} \\\ {{\omega ^n}}&{{\omega ^{2n}}}&1 \\\ {{\omega ^{2n}}}&1&{{\omega ^n}} \end{array}} \right| is equal to
A. ${\omega ^2}$
B. 0
C. 1
D. $\omega$

Answer 1

Hint : We have to only use the identity of complex function which states that if $1,\omega {\text{ }}and{\text{ }}{\omega ^2}$ are the cube roots of the unity then $1 + \omega + {\omega ^2} = 0$. So, we had to reduce the given determinant such that we can take the term $1 + {\omega ^n} + {\omega ^{2n}}$ common form any row or column.

Complete step by step solution :
So, now let us solve the above given determinant.
So, adding row 2 and row 3 of the determinant to row 1.
$\Rightarrow {R_1} = {R_1} + {R_2} + {R_3}$

{1 + {\omega ^n} + {\omega ^{2n}}}&{1 + {\omega ^n} + {\omega ^{2n}}}&{1 + {\omega ^n} + {\omega ^{2n}}} \\\ {{\omega ^n}}&{{\omega ^{2n}}}&1 \\\ {{\omega ^{2n}}}&1&{{\omega ^n}} \end{array}} \right|$$ Now take $$1 + {\omega ^n} + {\omega ^{2n}}$$ common from the first row of the determinant because it is present in each column of the first row. $$ \Rightarrow \left( {1 + {\omega ^n} + {\omega ^{2n}}} \right)\left| {\begin{array}{*{20}{c}} 1&1&1 \\\ {{\omega ^n}}&{{\omega ^{2n}}}&1 \\\ {{\omega ^{2n}}}&1&{{\omega ^n}} \end{array}} \right|$$ (1) Now let us find the value of $$1 + {\omega ^n} + {\omega ^{2n}}$$. So, as we know that according to the identity of the cube root of unity $${\omega ^3} = 1$$. So, $${\left( 1 \right)^n} = {\left( {{\omega ^3}} \right)^n} = {\omega ^{3n}}$$ So, now putting the value of 1 in $$1 + {\omega ^n} + {\omega ^{2n}}$$. $$ \Rightarrow {\omega ^{3n}} + {\omega ^n} + {\omega ^{2n}} = {\omega ^n}\left( {{\omega ^{2n}} + 1 + {\omega ^n}} \right)$$ So, $$1 + {\omega ^n} + {\omega ^{2n}} = {\omega ^n}\left( {1 + {\omega ^n} + {\omega ^{2n}}} \right)$$ $$ \Rightarrow \left( {1 + {\omega ^n} + {\omega ^{2n}}} \right)\left( {1 - {\omega ^n}} \right) = 0$$ Now as we know that $${\omega ^n} = 1$$ when n is a multiple of 3. So, $$1 + {\omega ^n} + {\omega ^{2n}} = 0$$ Now as we know that $$1 + {\omega ^n} + {\omega ^{2n}}$$ = 0. So, equation 1 becomes $$ \Rightarrow \left( 0 \right)\left| {\begin{array}{*{20}{c}} 1&1&1 \\\ {{\omega ^n}}&{{\omega ^{2n}}}&1 \\\ {{\omega ^{2n}}}&1&{{\omega ^n}} \end{array}} \right| = 0$$ So, the value of a given determinant becomes equal to zero. Hence, the correct option will be B. **Note** : Whenever we come up with this type of problem then we had to use only one identity to solve the determinant and that is $$1 + \omega + {\omega ^2} = 0$$ and if the powers of $$\omega $$ are in terms of n then the cube root of unity equation will become $$1 + {\omega ^n} + {\omega ^{2n}}$$ = 0. So, here we first, add all the rows to first row and then we had to take the term $$1 + {\omega ^n} + {\omega ^{2n}}$$ common from first row then the equation becomes $$1 + {\omega ^n} + {\omega ^{2n}}$$* (some determinant) and as we know that $$1 + {\omega ^n} + {\omega ^{2n}}$$ = 0. So, the value of determinant will be zero. This will be the easiest and efficient way to find the solution of the problem.