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3 and 4 .Determinants and Matrices
medium
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
A
$0$
B
$1$
C
$\omega $
D
${\omega ^2}$
(AIEEE-2003)
Solution
(a) $\Delta = 1\,({\omega ^{3n}} – 1) + {\omega ^n}({\omega ^{2n}} – {\omega ^{2n}}) + {\omega ^{2n}}({\omega ^n} – {\omega ^{4n}})$
$\Delta = \,[{({\omega ^3})^n} – 1] + 0 + {\omega ^{2n}}[{\omega ^n} – {({\omega ^3})^n}.{\omega ^n}]$
$\Delta = 1 – 1 + 0 + {\omega ^{2n}}[{\omega ^n} – {\omega ^n}] = 0$.
Standard 12
Mathematics
