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At what value of $x,$ will $\left| {\,\begin{array}{*{20}{c}}{x + {\omega ^2}}&\omega &1\\\omega &{{\omega ^2}}&{1 + x}\\1&{x + \omega }&{{\omega ^2}}\end{array}\,} \right| = 0$
$x = 0$
$x = 1$
$x = - 1$
None of these
Solution
(a) $\left| {\,\begin{array}{*{20}{c}}{x + {\omega ^2}}&\omega &1\\\omega &{{\omega ^2}}&{1 + x}\\1&{x + \omega }&{{\omega ^2}}\end{array}\,} \right| = 0$
Check at $x = 0,$ we get $\left| {\,\begin{array}{*{20}{c}}{{\omega ^2}}&\omega &1\\\omega &{{\omega ^2}}&1\\1&\omega &{{\omega ^2}}\end{array}\,} \right|$
= ${\omega ^2}({\omega ^4} – \omega ) – \omega ({\omega ^3} – 1) + 1({\omega ^2} – {\omega ^2})$
= ${\omega ^2}(\omega – \omega ) – \omega \,(1 – 1) + 0 = 0$ Or
$\Delta = \left| {\,\begin{array}{*{20}{c}}{1 + \omega + {\omega ^2} + x}&\omega &1\\{1 + \omega + {\omega ^2} + x}&{{\omega ^2}}&{1 + x}\\{1 + \omega + {\omega ^2} + x}&{x + \omega }&{{\omega ^2}}\end{array}\,} \right|$
by ${C_1} \to {C_1} + {C_2} + {C_3}$
$ = \,\left| {\,\begin{array}{*{20}{c}}x&\omega &1\\x&{{\omega ^2}}&{1 + x}\\x&{x + \omega }&{{\omega ^2}}\end{array}\,} \right|$, ($\because \,1 + \omega + {\omega ^2} = 0$)
$= 0$, if $x = 0$.
