A) \[abc\]
B) \[{{a}^{2}}{{b}^{2}}{{c}^{2}}\]
C) \[ab+bc+ca\]
D) None of these
Correct Answer: D
Solution :
Multiplying \[\left| \,\begin{matrix} 1 & 4 & 9 \\ 3 & 5 & 7 \\ 5 & 7 & 9 \\ \end{matrix}\, \right|\] by \[a,\,{{R}_{2}}\] by \[b\] and \[{{R}_{3}}\] by \[c,\] we have \[\Delta =\frac{1}{abc}\,\,\left| \,\begin{matrix} a{{b}^{2}}{{c}^{2}} & abc & ab+ac \\ {{a}^{2}}b{{c}^{2}} & abc & bc+ab \\ {{a}^{2}}{{b}^{2}}c & abc & ac+bc \\ \end{matrix}\, \right|\] = \[\frac{{{a}^{2}}{{b}^{2}}{{c}^{2}}}{abc}\,\left| \,\begin{matrix} bc & 1 & ab+ac \\ ac & 1 & bc+ab \\ ab & 1 & ac+bc \\ \end{matrix}\, \right|\,=\,abc\,\left| \,\begin{matrix} bc & 1 & \Sigma ab \\ ac & 1 & \Sigma \,ab \\ ab & 1 & \Sigma \,ab \\ \end{matrix}\, \right|\] {by \[{{C}_{3}}\to {{C}_{3}}+{{C}_{1}}\]} = \[abc.\Sigma \,ab\,\left| \,\begin{matrix} bc & 1 & 1 \\ ca & 1 & 1 \\ ab & 1 & 1 \\ \end{matrix}\, \right|=0\], [Since \[{{C}_{2}}\equiv {{C}_{3}}\]]. Trick : Put \[a=1,\,b=2,\,c=3\] and check it.
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