# Determinant/Recursively/Multilinear/Fact

Let ${\displaystyle {}K}$ be a field, and ${\displaystyle {}n\in \mathbb {N} _{+}}$.

Then the
${\displaystyle \operatorname {Mat} _{n}(K)=(K^{n})^{n}\longrightarrow K,M\longmapsto \det M,}$

is

multilinear.

This means that for every ${\displaystyle {}k\in \{1,\ldots ,n\}}$, and for every choice of ${\displaystyle {}n-1}$ vectors ${\displaystyle {}v_{1},\ldots ,v_{k-1},v_{k+1},\ldots ,v_{n}\in K^{n}}$, and for any ${\displaystyle {}u,w\in K^{n}}$, the identity

${\displaystyle {}\det {\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\u+w\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}=\det {\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\u\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}+\det {\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\w\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}\,}$

holds, and for ${\displaystyle {}s\in K}$, the identity

${\displaystyle {}\det {\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\su\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}=s\det {\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\u\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}\,}$

holds.