How to evaluate this determinant by just using row and column operations ? I'm stuck.Help please!
\begin{vmatrix} -2a & a+b & a+c \\ b+a & -2b & b+c \\ c+a & c+b & -2c \end{vmatrix}
BTW is there any general method to simplify symmetric determinants like this?
Add the first column to the second and the third column. Follow up by adding the first row to the second and the third row. You then have the matrix $$\pmatrix{-2a & b-a & c-a \cr b-a & 0 & 2(b+c) \cr c-a & 2(b+c) & 0}$$ It follows that $b+c$ divides the determinant. (Expand by the first row.) Since the determinant must be symmetric in $a,b,c$, it also has $a+b$ and $c+a$ as factors. Since it is a homogenous third degree polynomial in $a,b,c$, it must be on the form $$ k(a+b)(b+c)(c+a) $$ for some constant $k$. Consider the case $a=b=c={1\over 2}$, and compute the determinant to obtain $k=4$.