I have tried to reason, using multilinear forms, the well-known fact that the determinant of a matrix $[A]$ changes its sign if any two columns of $[A]$ are interchanged. I am not confident if my reason is correct however, and would appreciate a nod or a refute. Thanks.
(Disclaimer: I am confident that if my reasoning is correct, then it is unlikely that I am presenting anything novel here. I haven't however found the below-described argument — using multilinear forms — over math.stackexchange.com in response to similar queries, and therefore thought of seeking a review.)
Here goes the reasoning:
The definition of the determinant ($det$) of a linear operator on a finite-dimensional vector space ($\S 53$ of Finite-Dimensional Vector Spaces, $2^{\text{nd}}$ Ed., by Paul R. Halmos) implies the following.
If $A$ is any linear operator on any $n$-dimensional vector space $\mathcal V$, if $\{x_1, \cdots, x_n\}$ is any basis in $\mathcal V$, and if $w$ is any alternating $n$-linear form on $\mathcal V$, then $$\tag{1}\det \ A = \frac{w(Ax_1, \cdots, Ax_n)}{w(x_1, \cdots, x_n)}.$$
Under the hypothesis of (1), suppose some matrix $[A]$ together with $\{x_1, \cdots, x_n\}$ defines $A$. Suppose $[A_1]$ is the matrix obtained by interchanging two columns, of $[A]$, having any indices $h, k \ (h \neq k, 1 \leq h \leq n, 1 \leq k \leq n)$. Let $A_1$ be the operator defined by $[A_1]$ together with $\{x_1, \cdots, x_n\}$.
It follows from the definition of a matrix ($\S$37 of Finite-Dimensional Vector Spaces, $2^{\text{nd}}$ Ed., by Paul R. Halmos) that $Ax_h = A_1 x_k$, $Ax_k = A_1 x_h$, and that $Ax_i = A_1 x_i$ for $i = 1, \cdots, n$ except when $i$ equals $h$ or $k$. This finding together with the skew-symmetric nature of $w$ ($\S$30 of Finite-Dimensional Vector Spaces, $2^{\text{nd}}$ Ed., by Paul R. Halmos) implies that $$w(A_1 x_1, \cdots, A_1 x_n)= -w(A x_1, \cdots, A x_n).$$
Accordingly, we have $$ \det \ A_1 = \frac{w(A_1x_1, \cdots, A_1x_n)}{w(x_1, \cdots, x_n)} = \frac{-w(Ax_1, \cdots, Ax_n)}{w(x_1, \cdots, x_n)} = -\det \ A.$$
It is clear therefore that $\det\ [A_1] = -\det\ [A]$.
Best Answer
Yay, that is my favourite definition of the determinant haha.
Your reasoning is allright, but you could be more precise. Suppose for example that you swapped columns $1$ and $2$. Then you can show that $A_1x_1=Ax_2$, that $A_1x_2=Ax_1$ and $A_1x_i=Ax_i$ for each $i\neq 1,2$. Similarly picking any two adjacent columns $j,j+1$.