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.
\[ \Rightarrow {R_1} = {R_1} + {R_2} + {R_3}\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {1 + {\omega ^n} + {\omega ^{2n}}}&{1 + {\omega ^n} + {\omega ^{2n}}}&{1 + {\omega ^n} + {\omega ^{2n}}} \\
  {{\omega ^n}}&{{\omega ^{2n}}}&1 \\
  {{\omega ^{2n}}}&1&{{\omega ^n}}
\end{array}} \right|\]
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.