Proposition 4

For $n$ vectors $x_1, \ldots, x_n$ of $A^n$, let $X(x_1, \ldots, x_n)$ denote the square matrix of order $n$ whose $i$-th column is $x_i$ for $1 \leq i \leq n$. Then the mapping

of $(A^n)^n$ into $A$ is alternating and $n$-linear.

In particular, the determinant of a matrix two of whose columns are equal is zero. If a permutation is performed on the columns of a matrix, the determinant of the new matrix is equal to that of the old multiplied by $\varepsilon_{\sigma}$. If to one column of a matrix is added a scalar multiple of a column of a different index, the determinant of the new matrix is equal to that of the old.

More generally, let $M$ be a free $A$-module of finite dimension $n$ and let $(e_i)_{i \in I}$ be a basis of $M$; for every automorphism $u$ of $M$, if $X$ is the matrix of $u$ with respect to the basis $(e_i)$, then

as follows immediately from the definitions.

When $I = \{1, n\}$, the determinant of $X$ is also denoted by

or simply $\det(\xi_{ij})$ if this causes no confusion, or also

When $X = 1$, the matrix $X$ is called unimodular.