Let omega = - frac{1}{2} + ifrac{{√ 3 }}{2} . Then value of determinant left| {,begin{array}{*{20}

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3 and 4 .Determinants and Matrices

easy

Let $\omega = - \frac{1}{2} + i\frac{{\sqrt 3 }}{2}$. Then the value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{ - 1 - {\omega ^2}}&{{\omega ^2}}\\1&{{\omega ^2}}&{{\omega ^4}}\end{array}\,} \right|$ is

A

$3\omega $

B

$3\omega (\omega - 1)$

C

$3{\omega ^2}$

D

$3\omega (1 - \omega )$

(IIT-2002)

Solution

(b) $\Delta = \left| {\,\begin{array}{*{20}{c}}3&1&1\\0&{ – 1 – {\omega ^2}}&{{\omega ^2}}\\0&{{\omega ^2}}&\omega \end{array}\,} \right|$

$({C_1} \to {C_1} + {C_2} + {C_3})$

$(\because\,\,1 + \omega + {\omega ^2} = 0)$

$ = 3\,[\omega .\omega – {\omega ^4}] = 3({\omega ^2} – \omega )$ $ = 3\omega (\omega – 1)$.

Standard 12

Mathematics

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