2
$\begingroup$

i have to prove: $$\begin{vmatrix} -2a &a+b &a+c \\ b+a&-2b &b+c \\ c+a&c+b &-2c \end{vmatrix} = 4(a+b)(b+c)(c+a)$$

I have tried many calculations between the rows and columns of the determinant to get to the answer i want to solve the exercise however none of them gave me something right. Can someone help?

$\endgroup$
4
  • $\begingroup$ you may seperate it into two matrices $\endgroup$ May 15, 2018 at 16:31
  • $\begingroup$ I know it would be a lot of time to write up, but if we can see your actual work, we may be able to find a mistake $\endgroup$
    – imranfat
    May 15, 2018 at 16:33
  • $\begingroup$ @abhishekchaudhary could you give me an example?Not necessarily with the matrix of the exercise,just something to be able to understand. $\endgroup$ May 15, 2018 at 16:35
  • $\begingroup$ Try $R_1 \to R_1+R_2$ $\endgroup$ May 15, 2018 at 16:38

2 Answers 2

7
$\begingroup$

First of all, note that the determinant is going to be a polynomial in a, b, and c of degree 3. Secondly, note that the polynomial is cyclic.

Substitute $a = -b$, to get, $$\begin{vmatrix} -2a &0 &c+a \\ 0&2a &c-a \\ c+a&c-a &-2c \end{vmatrix} = -2a(-4ac - (c-a)^2 + (c+a)^2) = 0$$

This shows that $a+b$ is factor of the polynomial.

Since $a+b$ is a factor and the polynomial is cyclic, $b+c, c+a$ are also factors. And since the degree of the polynomial is 3, there is at most only a constant factor left. Let the constant factor be $k$.

Thereby, $$\begin{vmatrix} -2a & a+b &c+a \\ a+b& -2b &b+c \\ c+a& b+c &-2c \end{vmatrix} = k(a+b)(b+c)(c+a)$$

Put $a = 0, b = 1, c = 1$ to get, $$\begin{vmatrix} 0 & 1 & 1 \\ 1& -2 & 2 \\ 1& 2 &-2 \end{vmatrix} = k(1)(2)(1) \Rightarrow k = 4$$

Thus, $$\begin{vmatrix} -2a & a+b &c+a \\ a+b& -2b &b+c \\ c+a& b+c &-2c \end{vmatrix} = 4(a+b)(b+c)(c+a)$$

$\endgroup$
1
  • $\begingroup$ Importantly, the polynomial is homogeneous and each term is of degree $3$. $\endgroup$ Jan 31 at 23:11
3
$\begingroup$

Brute force: expand using Cramer's rule: $$ \begin{split} \det A = &-2a\left( (-2b)(-2c) - (b+c)^2 \right)\\ &-(b+a)\left( (-2a)(-2c) - (a+c)^2\right)\\ &+(c+a)\left( (a+b)(b+c)-(-2b)(a+c)\right) \end{split} $$ and simplify out.

Similarly multiply out the RHS. It is painful, but maybe less painful than looking for the clever transformations.

$\endgroup$
1
  • $\begingroup$ This is big and not the examination method $\endgroup$ May 15, 2018 at 16:37

Not the answer you're looking for? Browse other questions tagged .