Left| {,begin{array}{*{20}{c}}1&1&11&{{omega ^2}}&omega 1&omega &{{omega ^2}

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3 and 4 .Determinants and Matrices

easy

$\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{{\omega ^2}}&\omega \\1&\omega &{{\omega ^2}}\end{array}\,} \right| = $

A

$3\sqrt 3 i$

B

$ - 3\sqrt 3 i$

C

$i\sqrt 3 $

D

$3$

Solution

(a) $\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{{\omega ^2}}&\omega \\1&\omega &{{\omega ^2}}\end{array}\,} \right|\, = 3(\omega – {\omega ^2})$

$ = 3\,\left[ {\frac{{ – 1 + \sqrt 3 i}}{2} – \frac{{ – 1 – \sqrt 3 i}}{2}} \right] = 3\sqrt 3 \,i$.

Standard 12

Mathematics

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