A) non negative
B) negative
C) positive
D) non-positive
Correct Answer: B
Solution :
Let \[\Delta =\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|\] \[=a(bc-{{a}^{2}})-b({{b}^{2}}-ac)+c(ab-{{c}^{2}})\] \[=abc-{{a}^{3}}-{{b}^{3}}+abc+abc-{{c}^{3}}\] \[=-({{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc)\] \[=-(a+b+c)\,({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca)\] \[=-\frac{1}{2}(a+b+c)\,\{{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}\}\] < 0: where \[a\ne b\ne c\]You need to login to perform this action.
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