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જો $ a, b $ અને $c $ એ શૂન્યતર સંખ્યા હોય , તો $\Delta = \left| {\,\begin{array}{*{20}{c}}{{b^2}{c^2}}&{bc}&{b + c}\\{{c^2}{a^2}}&{ca}&{c + a}\\{{a^2}{b^2}}&{ab}&{a + b}\end{array}\,} \right|= .. . .$
$abc$
${a^2}{b^2}{c^2}$
$ab + bc + ca$
એકપણ નહી.
Solution
(d) Multiplying ${R_1}$by $a,\,{R_2}$ by $b$ and ${R_3}$ by $c,$ we have $\Delta = \frac{1}{{abc}}\,\,\left| {\,\begin{array}{*{20}{c}}{a{b^2}{c^2}}&{abc}&{ab + ac}\\{{a^2}b{c^2}}&{abc}&{bc + ab}\\{{a^2}{b^2}c}&{abc}&{ac + bc}\end{array}\,} \right|$
= $\frac{{{a^2}{b^2}{c^2}}}{{abc}}\,\left| {\,\begin{array}{*{20}{c}}{bc}&1&{ab + ac}\\{ac}&1&{bc + ab}\\{ab}&1&{ac + bc}\end{array}\,} \right|\, = \,abc\,\left| {\,\begin{array}{*{20}{c}}{bc}&1&{\Sigma ab}\\{ac}&1&{\Sigma \,ab}\\{ab}&1&{\Sigma \,ab}\end{array}\,} \right|$
{by ${C_3} \to {C_3} + {C_1}$}
= $abc.\Sigma \,ab\,\left| {\,\begin{array}{*{20}{c}}{bc}&1&1\\{ca}&1&1\\{ab}&1&1\end{array}\,} \right| = 0$, [Since ${C_2} \equiv {C_3}$].
Trick : Put $a = 1,\,b = 2,\,c = 3$ and check it.
