A) \[\frac{1}{abc}(ab+bc+ca)\]
B) \[ab+bc+ca\]
C) 0
D) \[a+b+c\]
Correct Answer: C
Solution :
\[\left| \begin{matrix} {{b}^{2}}{{c}^{2}} & bc & b+c \\ {{c}^{2}}{{a}^{2}} & ca & c+a \\ {{a}^{2}}{{b}^{2}} & ab & a+b \\ \end{matrix} \right|\] Multiplying \[{{R}_{1}},\,{{R}_{2}},\,{{R}_{3}}\] by a, b, c respectively and divide the whole by abc \[\frac{1}{abc}\left| \begin{matrix} a{{b}^{2}}{{c}^{2}} & abc & a\,(b+c) \\ b{{c}^{2}}{{a}^{2}} & bca & b\,(c+a) \\ {{a}^{2}}{{b}^{2}}c & abc & c\,(a+b) \\ \end{matrix} \right|\] Take common abc from \[{{C}_{1}}\] and \[{{C}_{2}}\] \[\frac{(abc)\,(abc)}{abc}\left| \begin{matrix} bc & 1 & ab+ac \\ ca & 1 & bc+ab \\ ab & 1 & ca+ab \\ \end{matrix} \right|\] Now \[{{C}_{1}}\to {{C}_{2}}+{{C}_{3}}\] \[abc\left| \begin{matrix} ab+bc+ca & 1 & ab+ac \\ ca+bc+ab & 1 & bc+ab \\ ab+bc+ca & 1 & ca+ab \\ \end{matrix} \right|\] Taking common \[ab+be+ca\] from \[{{C}_{1}}\], \[(abc)\,(ab+bc+ca)\,\left| \begin{matrix} 1 & 1 & ab+ac \\ 1 & 1 & bc+ab \\ 1 & 1 & ca+ab \\ \end{matrix} \right|\] \[(abc)\,(ab+bc+ca)\,.\,\,(0)\] As \[{{C}_{1}}={{C}_{2}}\], so value of A is zero \[\therefore \] \[\left| \begin{matrix} {{b}^{2}}{{c}^{2}} & ba & b+c \\ {{c}^{2}}{{a}^{2}} & ca & c+a \\ {{a}^{2}}{{b}^{2}} & ab & a+b \\ \end{matrix} \right|=0\]
You need to login to perform this action.
You will be redirected in
3 sec
