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3 and 4 .Determinants and Matrices
easy
If $\omega $ is the cube root of unity, then $\left| {\begin{array}{*{20}{c}}1&\omega &{{\omega ^2}}\\\omega &{{\omega ^2}}&1\\{{\omega ^2}}&1&\omega \end{array}} \right|$=
A
$1$
B
$0$
C
$\omega $
D
${\omega ^2}$
Solution
(b) $\left| {\,\begin{array}{*{20}{c}}1&\omega &{{\omega ^2}}\\\omega &{{\omega ^2}}&1\\{{\omega ^2}}&1&\omega \end{array}\,} \right| = \left| {\,\begin{array}{*{20}{c}}{1 + \omega + {\omega ^2}}&\omega &{{\omega ^2}}\\{1 + \omega + {\omega ^2}}&{{\omega ^2}}&1\\{1 + \omega + {\omega ^2}}&1&\omega \end{array}\,} \right|$
$ = \,\left| {\,\begin{array}{*{20}{c}}0&\omega &{{\omega ^2}}\\0&{{\omega ^2}}&1\\0&1&\omega \end{array}\,} \right| = 0$.
Standard 12
Mathematics
