A) Non-negative
B) Non-positive
C) Negative
D) Positive
Correct Answer: C
Solution :
[c] \[\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|=\left| \begin{matrix} a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b \\ \end{matrix} \right|\] \[(\because \,{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}})\] \[=(a+b+c)\left| \begin{matrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \\ \end{matrix} \right|\] [on taking (a+b+c) common from \[{{C}_{1}}\]] \[=(a+b+c)[1(bc-{{a}^{2}})-b(b-a)+c(a-c)]\] \[=(a+b+c)[bc-{{a}^{2}}-{{b}^{2}}+ab+ac-{{c}^{2}}]\] \[=(a+b+c)[-({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca)]\] \[=-\frac{1}{2}(a+b+c)[{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}]\] Hence, the determinant is negative value
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