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If $a \ne b \ne c,$ the value of $x$ which satisfies the equation $\left| {\,\begin{array}{*{20}{c}}0&{x - a}&{x - b}\\{x + a}&0&{x - c}\\{x + b}&{x + c}&0\end{array}\,} \right| = 0$, is
$x = 0$
$x = a$
$x = b$
$x = c$
Solution
(a) Obviously, on putting $x = 0$, we observe that the determinant becomes
${\Delta _{x = 0}} = \left| {\,\begin{array}{*{20}{c}}0&{ – a}&{ – b}\\a&0&{ – c}\\b&c&0\end{array}\,} \right| = a(bc) – b(ac) = 0$
$\therefore $ $x = 0$ is a root of the given equation.
Aliter : Expanding $\Delta $, we get
$\Delta \equiv – (x – a)\,[ – (x + b)(x – c)] + (x – b)\,[(x + a)\,(x + c)] = 0$
$ \Rightarrow $ $2{x^3} – (2\,\Sigma ab)x = 0$
$ \Rightarrow $ Either $x = 0$ or ${x^2} = \sum {ab} $ (i.e., $x = \pm \sum {ab} )$
Again $x = 0$ satisfies the given equation.
