3 and 4 .Determinants and Matrices
easy

If $\omega $is a cube root of unity, then $\left| {\,\begin{array}{*{20}{c}}{x + 1}&\omega &{{\omega ^2}}\\\omega &{x + {\omega ^2}}&1\\{{\omega ^2}}&1&{x + \omega }\end{array}\,} \right| = $

A

${x^3} + 1$

B

${x^3} + \omega $

C

${x^3} + {\omega ^2}$

D

${x^3}$

Solution

(d) $\Delta = \left| {\,\begin{array}{*{20}{c}}{x + 1}&\omega &{{\omega ^2}}\\\omega &{x + {\omega ^2}}&1\\{{\omega ^2}}&1&{x + \omega }\end{array}\,} \right|$

=$\left| {\,\begin{array}{*{20}{c}}{x + 1 + \omega + {\omega ^2}}&\omega &{{\omega ^2}}\\{x + 1 + \omega + {\omega ^2}}&{x + {\omega ^2}}&1\\{x + 1 + \omega + {\omega ^2}}&1&{x + \omega }\end{array}\,} \right|$,$({C_1} \to {C_1} + {C_2} + {C_3})$

= $x\,\left| {\,\begin{array}{*{20}{c}}1&\omega &{{\omega ^2}}\\1&{x + {\omega ^2}}&1\\1&1&{x + \omega }\end{array}\,} \right|$    $  (1 + {\omega} + {\omega^2} = 0) $

= $x\,[1\{ (x + {\omega ^2})\,(x + \omega ) – 1\} + \omega \{ 1 – (x + \omega )\} $

$ + {\omega ^2}\{ 1 – (x + {\omega ^2})\} ]$

= $x({x^2} + \omega x + {\omega ^2}x + {\omega ^3} – 1 + \omega – \omega x – {\omega ^2}$

$ + {\omega ^2} – {\omega ^2}x – {\omega ^4})$

= ${x^3}$ , .$(\because \,\,{{\omega }^{3}}=1)$

Standard 12
Mathematics

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