If 1, \omega ,{\omega ^2} are the cube roots of unity, then \Delta = \left| {\begin{matrix} 1 & {{\omega ^n}} & {{\omega ^{2n}}} \ {{\omega ^n}} & {{\omega ^{2n}}} & 1 \ {{\omega ^{2n}}} & 1 & {{\omega ^n}} \ \end{matrix} } ight| is equal to

Question

If 1, \omega ,{\omega ^2} are the cube roots of unity, then \Delta = \left| {\begin{matrix} 1 & {{\omega ^n}} & {{\omega ^{2n}}} \ {{\omega ^n}} & {{\omega ^{2n}}} & 1 \ {{\omega ^{2n}}} & 1 & {{\omega ^n}} \ \end{matrix} } ight| is equal to

Accepted Answer

Answer: B. 0 \Delta = \left| {\begin{matrix} 1 & {{\omega ^n}} & {{\omega ^{2n}}} \ {{\omega ^n}} & {{\omega ^{2n}}} & 1 \ {{\omega ^{2n}}} & 1 & {{\omega ^n}} \ \end{matrix} } ight| = 1\left( {{\omega ^{3n}} - 1} ight) - {\omega ^n}\left( {{\omega ^{2n}} - {\omega ^{2n}}} ight) + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^{2n}}\left( {{\omega ^n} - {\omega ^{4n}}} ight) = {\omega ^{3n}} - 1 - 0 + {\omega ^{3n}} - {\omega ^{6n}} = 1 - 1 + 1 - 1 = 0 \left[ {} ight. as \,\,\,\,\, {\omega ^{3n}} = 1 \left. {} ight]

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

Answer

Explanation

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