A) \[{{a}_{1}}\Delta \]
B) \[{{a}_{1}}{{a}_{3}}\Delta \]
C) \[({{a}_{1}}+{{b}_{1}})\Delta \]
D) None of these
Correct Answer: A
Solution :
\[{{B}_{2}}=\left| \,\begin{matrix} {{a}_{1}} & {{c}_{1}} \\ {{a}_{3}} & {{c}_{3}} \\ \end{matrix} \right|={{a}_{1}}{{c}_{3}}-{{c}_{1}}{{a}_{3}}\] \[{{C}_{2}}=-\left| \,\begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{3}} & {{b}_{3}} \\ \end{matrix}\, \right|=-({{a}_{1}}{{b}_{3}}-{{a}_{3}}{{b}_{1}})\] \[{{B}_{3}}=-\left| \,\begin{matrix} {{a}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{c}_{2}} \\ \end{matrix}\, \right|=-({{a}_{1}}{{c}_{2}}-{{a}_{2}}{{c}_{1}})\] \[{{C}_{3}}=\left| \,\begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix}\, \right|={{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}\] \[\left| \,\begin{matrix} {{B}_{2}} & {{C}_{2}} \\ {{B}_{3}} & {{C}_{3}} \\ \end{matrix}\, \right|=\left| \begin{matrix} {{a}_{1}}{{c}_{3}}-{{a}_{3}}{{c}_{1}} & -({{a}_{1}}{{b}_{3}}-{{a}_{3}}{{b}_{1}}) \\ -({{a}_{1}}{{c}_{2}}-{{a}_{2}}{{c}_{1}}) & {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}} \\ \end{matrix}\, \right|\] \[=\left| \,\begin{matrix} {{a}_{1}}{{c}_{3}} & -{{a}_{1}}{{b}_{3}} \\ -{{a}_{1}}{{c}_{2}} & {{a}_{1}}{{b}_{2}} \\ \end{matrix}\, \right|+\left| \,\begin{matrix} {{a}_{1}}{{c}_{3}} & {{a}_{3}}{{b}_{1}} \\ -{{a}_{1}}{{c}_{2}} & -{{a}_{2}}{{b}_{1}} \\ \end{matrix}\, \right|\]\[+\left| \,\begin{matrix} -{{a}_{3}}{{c}_{1}} & -{{a}_{1}}{{b}_{3}} \\ \,\,\,{{a}_{2}}{{c}_{1}} & {{a}_{1}}{{b}_{2}} \\ \end{matrix}\, \right|+\left| \,\begin{matrix} -{{a}_{3}}{{c}_{1}} & {{a}_{3}}{{b}_{1}} \\ {{a}_{2}}{{c}_{1}} & -{{a}_{2}}{{b}_{1}} \\ \end{matrix}\, \right|\] \[=a_{1}^{2}({{b}_{2}}{{c}_{3}}-{{b}_{3}}{{c}_{2}})+{{a}_{1}}{{b}_{1}}(-{{c}_{3}}{{a}_{2}}+{{a}_{3}}{{c}_{2}})\] \[+{{a}_{1}}{{c}_{1}}(-{{a}_{3}}{{b}_{2}}+{{a}_{2}}{{b}_{3}})+{{c}_{1}}{{b}_{1}}({{a}_{3}}{{a}_{2}}-{{a}_{2}}{{a}_{3}})\]\[={{a}_{1}}\Delta \].
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