Linear Algebra: row reducing in order to find the determinant When looking for the determinant of some square matrix, one can row reduce until the matrix is in upper triangular form.
My question is: we are allowed to add or subtract multiples of one row to another row, but when we want to multiply/divide a singular row by a constant, we need to multiply the entire determinant by this constant. Why is this?
Thanks!
 A: The determinant is linear in each of its $n$ arguments, e.g.:
$$
\mbox{det} (c a_1, \ldots, a_n) = c \,\mbox{det} (a_1, \ldots, a_n)
$$
This is called multilinear.
It is a form, because it is a map fron $V^n$ to its field $K$.
Then it switches sign, if you switch two arguments, this is a consequence of it being alternating, i.e. vanishing if two arguments are the same.
This gives: alternating multilinear form
This can be verified with its definition:
$$
\mbox{det} A =
\sum_{\pi\in S_n} \mbox{sgn}(\pi) \,a_{1\,\pi(1)}\cdots a_{n\,\pi(n)}
$$
where $S_n$ is the symmetric group consisting of the permutations of the set 
$\{1,\ldots,n\}$.
Or you think of the determinant as signed volume of the paralel epiped spawned by the argument vectors. Scaling a single vector will scale the volume.
And 
$$
\mbox{det}(a_1 + c a_j, \cdots, a_n) =
\mbox{det}(a_1, \cdots, a_n) + c \, \mbox{det}(a_j, \cdots, a_n)
$$
where the second term vanishes because $a_j$ shows up twice, which is the alternating property. This cancels the multiplier $c$.
In the volume picture this would result in a deformation which is not volume changing, which is less obvious for higher dimensions to me.
