# If a,b,c are positive and unequal, show that the following determinant is negative. $\Delta = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}$

Toolbox:
• Elementary transformation in a determinant can be made by
• (a) Interchanging the rows or columns.
• (b) The addition to the elements of any rows or columns.
• If $A=\begin{bmatrix}a_{11} & a_{12} & a_{13}\\a_{21}& a_{22} & a_{23}\\a_{31} & a_{32} &a_{33}\end{bmatrix}$
• $|A|=a_{11}(a_{22}\times a_{33}-a_{23}\times a_{32})-a_{12}(a_{21}\times a_{33}-a_{23}\times a_{31})+a_{13}(a_{21}\times a_{32}-a_{22}\times a_{31})$
Step 1:
$\Delta=\begin{bmatrix}a & b& c\\b & c & a\\c & a & b\end{bmatrix}$
Apply $C_1\rightarrow C_1+C_2+C_3$
Hence $\Delta=\begin{bmatrix}a+b+c & b& c\\a+b+c & c & a\\a+b+c & a & b\end{bmatrix}$
Take (a+b+c) as the common factor from $C_1$
$\Delta=(a+b+c)\begin{bmatrix}1 & b& c\\1 & c & a\\1 & a & b\end{bmatrix}$
Apply $R_2\rightarrow R_2-R_1$ and $R_3\rightarrow R_3-R_1$
$\Delta=(a+b+c)\begin{bmatrix}1 & b& c\\0 & c-b & a-c\\0 & a-b & b-c\end{bmatrix}$
Step 2:
Now let us expand along $C_1$
$\Delta=(a+b+c)[(c-b)(b-c)-(a-c)(a-b)]$
$\;\;\;=(a+b+c)(-a^2-b^2-c^2+ab+bc+ca)$
Now multiply and divide by -2
$\Delta=\frac{-1}{2}(a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ca)$
But we know $(a+b)^2=a^2+2ab+b^2$
$\Delta$ can be written as
$\Delta=\frac{-1}{2}(a+b+c)(a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ca)$
$\;\;\;=\frac{-1}{2}(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]$
This is negative,since a+b+c > 0 and $(a-b)^2+(b-c)^2+(c-a)^2$ > 0
Hence proved.