A) \[{{a}^{2}}\]
B) \[{{b}^{2}}\]
C) \[{{c}^{2}}\]
D) \[{{x}^{2}}\]
Correct Answer: D
Solution :
\[\left| \,\begin{matrix} {{a}^{2}}+{{x}^{2}} & ab & ca \\ ab & {{b}^{2}}+{{x}^{2}} & bc \\ ca & bc & {{c}^{2}}+{{x}^{2}} \\ \end{matrix}\, \right|\] Multiply \[{{C}_{1}},\,{{C}_{2}},{{C}_{3}}\] by \[a,\,\,b,\,c\] respectively and hence divide by abc \ \[\Delta =\frac{1}{abc}\,\left| \,\begin{matrix} a({{a}^{2}}+{{x}^{2}}) & a{{b}^{2}} & {{c}^{2}}a \\ {{a}^{2}}b & b({{b}^{2}}+{{x}^{2}}) & b{{c}^{2}} \\ c{{a}^{2}} & {{b}^{2}}c & c({{c}^{2}}+{{x}^{2}}) \\ \end{matrix}\, \right|\] Now take out a, b and c common from \[{{R}_{1}},\,{{R}_{2}}\] and \[{{R}_{3}}\], \ \[\Delta =\left| \,\begin{matrix} {{a}^{2}}+{{x}^{2}} & {{b}^{2}} & {{c}^{2}} \\ {{a}^{2}} & {{b}^{2}}+{{x}^{2}} & {{c}^{2}} \\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}}+{{x}^{2}} \\ \end{matrix}\, \right|\] Now applying \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}\] \[\Rightarrow \] \[\Delta =({{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{x}^{2}})\,\left| \,\begin{matrix} 1 & {{b}^{2}} & {{c}^{2}} \\ 1 & {{b}^{2}}+{{x}^{2}} & {{c}^{2}} \\ 1 & {{b}^{2}} & {{c}^{2}}+{{x}^{2}} \\ \end{matrix}\, \right|\] Þ \[\Delta ={{x}^{4}}({{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{x}^{2}})\] Hence, it is divisible by \[{{x}^{2}}\].
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