A) \[{{x}_{1}}{{x}_{2}}{{x}_{3}}+{{y}_{1}}{{y}_{2}}{{y}_{3}}\]
B) \[{{x}_{1}}{{x}_{2}}{{x}_{3}}{{y}_{1}}{{y}_{2}}{{y}_{3}}\]
C) \[{{x}_{2}}{{x}_{3}}{{y}_{2}}{{y}_{3}}+{{x}_{3}}{{x}_{1}}{{y}_{3}}{{y}_{1}}+{{x}_{1}}{{x}_{2}}{{y}_{1}}{{y}_{2}}\]
D) 0
Correct Answer: D
Solution :
[d] We can write \[\Delta ={{\Delta }_{1}}+{{y}_{1}}{{\Delta }_{2}},\] where \[{{\Delta }_{1}}=\left| \begin{matrix} 1 & 1+{{x}_{1}}{{y}_{2}} & 1+{{x}_{1}}{{y}_{3}} \\ 1 & 1+{{x}_{2}}{{y}_{2}} & 1+{{x}_{2}}{{y}_{3}} \\ 1 & 1+{{x}_{3}}{{y}_{2}} & 1+{{x}_{3}}{{y}_{3}} \\ \end{matrix} \right|\] and \[{{\Delta }_{2}}=\left| \begin{matrix} {{x}_{1}} & 1+{{x}_{1}}{{y}_{2}} & 1+{{x}_{1}}{{y}_{3}} \\ {{x}_{2}} & 1+{{x}_{2}}{{y}_{2}} & 1+{{x}_{2}}{{y}_{3}} \\ {{x}_{3}} & 1+{{x}_{3}}{{y}_{2}} & 1+{{x}_{3}}{{y}_{3}} \\ \end{matrix} \right|\] In \[{{\Delta }_{1}},\] use \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\] so that, \[{{\Delta }_{1}}=\left| \begin{matrix} 1 & {{x}_{1}}{{y}_{2}} & {{x}_{1}}{{y}_{3}} \\ 1 & {{x}_{2}}{{y}_{2}} & {{x}_{2}}{{y}_{3}} \\ 1 & {{x}_{3}}{{y}_{2}} & {{x}_{3}}{{y}_{3}} \\ \end{matrix} \right|=0\] [\[\because \,\,\,{{C}_{2}}\] and \[{{C}_{3}}\] are proportional] In \[{{\Delta }_{2}},\] us \[{{C}_{2}}\to {{C}_{2}}-{{y}_{2}}{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{y}_{3}}{{C}_{1}}\] to get \[{{\Delta }_{2}}=\left| \begin{matrix} {{x}_{1}} & 1 & 1 \\ {{x}_{2}} & 1 & 1 \\ {{x}_{3}} & 1 & 1 \\ \end{matrix} \right|=0\] [\[\because \,\,\,{{C}_{2}}\] and \[{{C}_{3}}\] are identical] \[\therefore \,\,\,\Delta =0\]
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