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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?

  • If you add any column to another column, a row operation will suggest itself.... – Greg Martin May 21 '15 at 06:13
  • I added the 1 st and 2nd column...and then added the second row to the first....but after that ? (I got only one zero in the left top after these operations) @GregMartin –  May 21 '15 at 06:19
  • Basically the same question: http://math.stackexchange.com/questions/1299090/determinant-of-matrices-without-expanding – Martin Sleziak May 26 '15 at 08:43

1 Answers1

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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 $k=4$.

Per Erik Manne
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