If $1,\;\omega ,{\omega ^2}$ are the cube roots of unity, then $\Delta = \left| {\begin{array}{*{20}{c}} 1&{{\omega ^n}}&{{\omega ^{2n}}}\\ {{\omega ^n}}&{{\omega ^{2n}}}&1\\ {{\omega ^{2n}}}&1&{{\omega ^n}} \end{array}} \right|$ is equal to

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Application +4 | -1

1,\;\omega ,{\omega ^2}

are the cube roots of unity, then

\Delta = \left| {\begin{array}{cccccccccccccccccccc} 1&{{\omega ^n}}&{{\omega ^{2n}}}\\ {{\omega ^n}}&{{\omega ^{2n}}}&1\\ {{\omega ^{2n}}}&1&{{\omega ^n}} \end{array}} \right|

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because the determinant simplifies to 0 using the properties of cube roots of unity, where

\omega^{3n} = 1