TITLE: The determinant function is the only one satisfying the conditions
QUESTION [11 upvotes]: How can I prove that the determinant function satisfying the following properties is unique:
$\det(I)=1$ where  $I$ is identity matrix, the function $\det(A)$ is linear in the rows of the matrix and if two adjacent rows of a matrix $A$ are equal, then $\det A=0$.
This is how Artin has stated the properties.I find Artin's first chapter rough going and would appreciate some help on this one.

REPLY [6 votes]: An alternative way is the  Gaussian elimination: for a given $n\times n$ matrix $A$ with rows $r_1,..,r_n$, the following steps are allowed to use, in order to arrive to the identity matrix or one with a zero row (by the linearity, if $A$ has a zero row, the 'Artinian determinant' has to be zero).

Add a scalar multiple of a row $r_j$ to another row $r_i$, i.e.: $i\ne j$ and
$$r_i':= r_i+\lambda r_j$$
Multiply a row by a nonzero scalar, i.e.: $\lambda\ne 0$ and
$$r_i':=\lambda\cdot r_i$$
Exchange 2 rows (can also be obtained by 1. and 2.)

Let's assume, we have two 'Artinian determinants': $D$ and $D'$. Using the above mentioned fact that every matrix can be transformed to the indentity or with a zero row, we will have $D=D'$, because 1. keeps both $D$ and $D'$ (why?), 2. multiplies both $D$ and $D'$ by $\lambda$, and 3. by $-1$.