A) \[\text{+}\,\text{ve}\]
B) \[-\text{ve}\]
C) zero
D) none of these
Correct Answer: B
Solution :
If \[\Delta =\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|;{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}},\]we get \[\Delta =\left| \begin{matrix} a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b \\ \end{matrix} \right|\] \[=(a+b+c)\left| \begin{matrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \\ \end{matrix} \right|\] \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\]and \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}},\] we get \[=(a+b+c)\left| \begin{matrix} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \\ \end{matrix} \right|\] \[=(a+b+c)\{-{{(c-b)}^{2}}-(a-b)(a-c)\}\] \[=-(a+b+c)\{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca\}\] \[=-\frac{1}{2}(a+b+c)\{2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ab-2bc-2ca\}\] \[=-\frac{1}{2}(a+b+c)\{{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}\}\] Which is always negative.
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