Determinant of matrices without expanding

Question

Show that $$\begin{array}{|ccc|} -2a & a + b & c + a \\ a + b & -2b & b + c \\ c + a & c + b & -2c \end{array} = 4(a+b)(b+c)(c+a)\text{.}$$

I added the all rows but couldn't get it.

Answer 1

We can deduce from the structure of the matrix that (a) the determinant will be a symmetric polynomial in $a,b,c$ with every monomial with non-zero coefficient having degree $3$, and (b) that the coefficient of $a^3$, $b^3$, and $c^3$ will be $0$. So $$ \det= k\,(a^2 b+ a^2 c+ b^2 a+ b^2 c+ c^2 a+ c^2 b)+m\,abc $$ for some constants $k,m$. To find $k,m$, we substitute in values of $(a,b,c)$ into the matrix.

• When $a=0$, $b=1$, $c=1$ we get $8=\begin{vmatrix} 0 & 1 & 1 \\ 1 & -2 & 2 \\ 1 & 2 & -2 \\ \end{vmatrix}=\det=2k$, so $k=4$.

• When $a=1$, $b=-1$, $c=1$ we get $0=\begin{vmatrix} -2 & 0 & 2 \\ 0 & 2 & 0 \\ 2 & 0 & -2 \\ \end{vmatrix}=\det=4 \times 2-m$, so $m=8$.

Thus \begin{align*} \det &= 4(a^2 b+ a^2 c+ b^2 a+ b^2 c+ c^2 a+ c^2 b)+8abc \\ &= 4(a+b)(b+c)(c+a) \end{align*} as desired.

Answer 2

Let $x=b+c,y=c+a,z=a+b$. We claim that $$ \left|\begin{pmatrix} x-y-z & z & y\\ z & y-z-x & x\\ y & x & z-x-y \end{pmatrix}\right|=4xyz. $$ When $x=0$, add column 1 to columns 2 and 3 to obtain $$ \left|\begin{pmatrix} -y-z & -y & -z\\ z & y & z\\ y & y & z \end{pmatrix}\right|=0. $$ Thus by symmetry, $xyz$ divides the determinant. Setting $x=y=z=1$ yields $$ \left|\begin{pmatrix} -1 & 1 & 1\\ 1 & -1 & 1\\ 1 & 1 & -1 \end{pmatrix}\right|= \left|\begin{pmatrix} -1 & 0 & 0\\ 1 & 0 & 2\\ 1 & 2 & 0 \end{pmatrix}\right|=4. $$ Therefore the determinant is $4xyz$, since it is of degree 3.