Determinant Property: Seen it Before? True?

[Dustinsfl]

Has anyone seen this before? Is this true?
$$
\begin{vmatrix}
a & b+c & 1\\
b & a+c & 1\\
c & a+b & 1
\end{vmatrix} =
\begin{vmatrix}
a & b & 1\\
b & a & 1\\
c & a & 1
\end{vmatrix} +
\begin{vmatrix}
a & c & 1\\
b & c & 1\\
c & b & 1
\end{vmatrix}
$$
In this example this works but I don't know if this just a coincidence.

 

[Evgeny.Makarov]

Well, determinants are linear w.r.t. addition in any single row or column. (Wink)

 

[Deveno]

Determinants are multilinear, alternating functions of row or column vectors. If one adds the stipulation that:

$\det(I_n) = 1$

then these properties completely determine the determinant function.

Clearly one such function (the determinant function) with these properties exists. For a proof that the determinant function is the ONLY function with these properties, see:

http://www.millersville.edu/~bikenaga/linear-algebra/det-unique/det-unique.html