Determinant/Recursively/Multilinear/Fact

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

Then the

determinant

${\mathrm{Mat}}_n(K)=(K^n{)}^n⟶K,M⟼detM,$

is

multilinear.

This means that for every $k\in \{1,\dots ,n\}$, and for every choice of $n-1$ vectors $v_1,\dots ,v_{k-1},v_{k+1},\dots ,v_n\in K^n$, and for any $u,w\in K^n$, the identity

$det\left(\begin{array}{c}{v}_1\\ ⋮\\ v_{k-1}\\ u+w\\ v_{k+1}\\ ⋮\\ v_n\end{array}\right)=det\left(\begin{array}{c}{v}_1\\ ⋮\\ v_{k-1}\\ u\\ v_{k+1}\\ ⋮\\ v_n\end{array}\right)+det\left(\begin{array}{c}{v}_1\\ ⋮\\ v_{k-1}\\ w\\ v_{k+1}\\ ⋮\\ v_n\end{array}\right)$

holds, and for $s\in K$, the identity

$det\left(\begin{array}{c}{v}_1\\ ⋮\\ v_{k-1}\\ su\\ v_{k+1}\\ ⋮\\ v_n\end{array}\right)=sdet\left(\begin{array}{c}{v}_1\\ ⋮\\ v_{k-1}\\ u\\ v_{k+1}\\ ⋮\\ v_n\end{array}\right)$

holds.

Proof, Write another proof