JEE Main 2003Annual, Paper 1MathematicsMediumMatrices and Determinants245× since 2002If 1,1,1, ω,ω2\omega ,{\omega ^2}ω,ω2 are the cube roots of unity, then Δ=∣1ωnω2nωnω2n1ω2n1ωn∣\Delta = \left| {\matrix{ 1 & {{\omega ^n}} & {{\omega ^{2n}}} \cr {{\omega ^n}} & {{\omega ^{2n}}} & 1 \cr {{\omega ^{2n}}} & 1 & {{\omega ^n}} \cr } } \right|Δ=1ωnω2nωnω2n1ω2n1ωn is equal toAOption A. ω2{\omega ^2}ω2BOption B. 000COption C. 111DOption D. ω\omegaωCheckSkip · show solution
