I have the following determinant :

Row1: (-2a) (a+b) (a+c)

Row2: (a+b) (-2b) (b+c)

Row3: (a+c) (b+c) (-2c)

I need to prove that the determinant equals 4(a+b)(b+c)(c+a).

I have tried using all the standard determinant properties, but I wasn't able to arrive at the final expression.

The result can be verified by expanding the determinant, but that is obviously too long. Another method is to realize that the determinant must be a homogeneous expression in a,b and c of third degree. Moreover, on replacing (a with -b) the determinant evaluates to zero, indicating that (a+b) must be a factor. Using this approach I concluded that the final answer must be of the form:


The value of k can be determined by substituting some easy numbers for a,b and c such as 0,1 and 2.

But I am still not satisfied as I want to solve the determinant using its basic properties (such as row operations and factoring out terms). Please explain how that can be done. It should be possible right ?

  • $\begingroup$ This is a rather nice question.you have sufficient reputation why don't you consider placing a bounty on this question. $\endgroup$
    – Navin
    Dec 29 '16 at 11:09

Applying the Rule of Sarrus we obtain $$ \det (A)=4(a^2b + a^2c + ab^2 + 2abc + ac^2 + b^2c + bc^2)=4(a+b)(b+c)(c+a). $$ I don't think that this computation is "obviously too long". It is rather nice.

  • $\begingroup$ That was nice. But is there still another way? $\endgroup$
    – Newton
    Oct 12 '16 at 16:03
  • $\begingroup$ Yes, many other ways. One way is Laplace's formula. $\endgroup$ Oct 12 '16 at 18:08
  • $\begingroup$ How is that different from just expanding it? $\endgroup$
    – Newton
    Oct 13 '16 at 2:07
  • $\begingroup$ It is better than just expanding it, because it uses the adjugate matrix, which has good properties. $\endgroup$ Dec 29 '16 at 11:49

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