If a, b, c are distinct positive real numbers | vedclass

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(A) Let $A = \left|\begin{array}{lll}a & b & c \ b & c & a \ c & a & b\end{array}ight|$.Applying $C_1 ightarrow C_1 + C_2 + C_3$,we get:$A = \le

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$< 0$

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(A) Let $A = \left|\begin{array}{lll}a & b & c \ b & c & a \ c & a & b\end{array}ight|$.Applying $C_1 ightarrow C_1 + C_2 + C_3$,we get:$A = \left|\begin{array}{lll}a+b+c & b & c \ a+b+c & c & a \ a+b+c & a & b\end{array}ight| = (a+b+c) \left|\begin{array}{lll}1 & b & c \ 1 & c & a \ 1 & a & b\end{array}ight|$.Applying $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$A = (a+b+c) \left|\begin{array}{ccc}1 & b & c \ 0 & c-b & a-c \ 0 & a-b & b-c\end{array}ight|$.Expanding along $C_1$:$A = (a+b+c) [(c-b)(b-c) - (a-c)(a-b)]$$A = (a+b+c) [-(b-c)^2 - (a^2 - ab - ac + bc)]$$A = -(a+b+c) [a^2 + b^2 + c^2 - ab - bc - ca]$$A = -\frac{1}{2}(a+b+c) [(a-b)^2 + (b-c)^2 + (c-a)^2]$.Since $a, b, c$ are distinct positive real numbers,$(a+b+c) > 0$ and $[(a-b)^2 + (b-c)^2 + (c-a)^2] > 0$. Therefore,$A < 0$.

If $a, b, c$ are distinct positive real numbers,then the value of the determinant $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is

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