# Determinant as an alternating multilinear map.

Hello
I'm studying the properties of the determinant,
on various books they say the determinant of a matrix is an alternating multilinear map.
For clarity, I will use the following notation:

${Let \ A \in Mat_{\ n \ \times \ n}(K)\ be\ a\ n \times n\ square\ matrix,\ \\ \\ A = (A^{1}|A^{2}|...|A^{n}), \\where \ A^{i}\in K^n \ is \ the \ i\!-\!eth \ column \ of \ A, \ \forall \ i \in \{1,2,...,n \} }$

They give this definition for alternating multilinear map:
${Let\ f: K^{n} \times ...\times K^{n}\longrightarrow K, \ we\ say\ f\ is\ \textbf {multilinear}\ if: \\ \small \mathsf i)\ f(A^1|...|\lambda A^j|...|A^n) = \lambda\ f(A^1|...|A^j|...|A^n) \\ \hspace{0.3mm} \mathsf i\mathsf i)\small f(A^1|...|A^i|X\! +\! Y|A^{i+2}|...|A^n)\! =\!f(A^1|...|A^i|X|A^{i+2}|...|A^n)+ f(A^1|...|A^{i}|Y|A^{i+2}|...|A^n) \\ Also\ we\ say\ f\ is\ \textbf { alternating}\ if\ whenever \ A^i=A^{i+1}\, \\we\ get\ \small f (A^1|...|A^i|A^{i+1}|...|A^n)=0\hspace{5mm} \forall i \in \{1,...,n\}. \\[4mm]}$

Then while studying the properties of the determinant I came across this proof:
${Let\ A=(A^1|...|A^{i}|A^{j}|...|A^n),\\ let\ B=(A^1|...|A^{j}|A^{i}|...|A^n)\\ then\ detB = -detA\\ \textbf{Proof}:\\ Let\ C=(A^1|...|A^{i}+A^{j}|A^{i}+A^{j}|...|A^n), we\ know\ det(C)=0.\\ Furthermore\ det(C)=det(A)+det(B)\ hence\ det(B)=-det(A).\square }$

Now since the determinant is an alternating multilinear map, I don't understand why this happens: ${Let\ A=(A^1|...|X|Y|...|A^n)\ such\ that\ det(A) \ne0, \\let \ B= (A^1|...|\underline O|Y|...|A^n),\\ let\ C = (A^1|...|X|\underline O|...|A^n),\\ det(A) \ne\ det (B)+det(C)=0}$

Can anybody help me understand how my last example is different from the approach of the proof?

Using the approach of the proof, in your ''counter-example'' we must have: $$B= (A^1|...|\underline O|Y|...|A^n)$$ and $$C=(A^1|...|Y|\underline O|...|A^n)$$ so that $$\det A= \det(A^1|...|Y|Y|...|A^n)=0$$
From: $$B= (A^1|...|X|Y|...|A^n) \quad \mbox{and}\quad C= (A^1|...|Y|X|...|A^n)$$ and $$A= (A^1|...|X+Y|Y+X|...|A^n)$$ we have: $\det A=0$ because the determinant is an alternating form, and, from multilinearity: $$\det A=0= \det (A^1|...|X+Y|Y+X|...|A^n)=\det(A^1|...|X|Y+X|...|A^n)+\det(A^1|...|Y|Y+X|...|A^n)=$$ $$=\det(A^1|...|X|Y|...|A^n)+\det(A^1|...|X|X|...|A^n)+\det(A^1|...|Y|Y|...|A^n)+\det(A^1|...|Y|X|...|A^n)=$$ $$=\det(A^1|...|X|Y|...|A^n)+0+0+\det(A^1|...|Y|X|...|A^n)=$$ $$=\det B + \det C$$ so: $\det B=-\det C$.