A) \[a+b+c\]
B) \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}\]
C) 0
D) 1
Correct Answer: C
Solution :
\[\left| \begin{matrix} 1 & a & {{a}^{2}}-bc \\ 1 & b & {{b}^{2}}-ac \\ 1 & c & {{c}^{2}}-ab \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 1 & a & {{a}^{2}}-bc \\ 0 & b-a & {{b}^{2}}-ac-{{a}^{2}}+bc \\ 0 & c-a & {{c}^{2}}-ab-{{a}^{2}}+bc \\ \end{matrix} \right|\] \[[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}]\] \[=1[(b-a)({{c}^{2}}-ab-{{a}^{2}}+bc)-(c-a)\] \[({{b}^{2}}-ac-{{a}^{2}}+bc)]\] \[=b{{c}^{2}}-a{{b}^{2}}-{{a}^{2}}b+{{b}^{2}}c-a{{c}^{2}}+{{a}^{2}}b\] \[+{{a}^{3}}-abc-{{b}^{2}}c+a{{c}^{2}}+{{a}^{2}}c-b{{c}^{2}}+a{{b}^{2}}\] \[-{{a}^{2}}c-{{a}^{3}}+abc\] \[=0\]You need to login to perform this action.
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