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[SOLVED] Determinant property

dwsmith

Well-known member
Feb 1, 2012
1,673
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-known member
MHB Math Scholar
Jan 30, 2012
2,503
Well, determinants are linear w.r.t. addition in any single row or column. (Wink)
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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:

Determinants - Uniqueness and Properties