Determinant of a matrix and linear independence (explanation needed) It is written on Wikipedia that:
$n$ vectors in $\mathbb R^n$ are linearly independent if and only if the determinant of the matrix formed by taking the vectors as its columns is non-zero
Can someone explain this to me? You do not have to give a complete proof, just in simple terms explain what the determinant of that matrix has to do with linear independence? And why it has to be non-zero? And are vectors allowed to be rows instead of columns in that matrix?

The determinant is an n-linear (multilinear) alternating form.
(Let us assume that our determinants are with respect to an arbitrary base of the space considered, it's just a technical aspect, for rigor, but you should consider the determinant of a family with respect to a certain base.)
Alternating characteristic
What is relevant here is the alternating characteristic, let's take $(x_1, \ldots, x_n)$ a family of vectors and $f : \mathbb{K}^n \to \mathbb{K}$ a $n$-linear alternating form from the $\mathbb{K}$-vector space of dimension $n$ to $\mathbb{K}$.
If there is $(i, j) \in \{ 1, 2, \ldots, n \}^2$ such that $i \neq j$ and $x_i = x_j$, then $f(x_1, \ldots, x_n) = 0$.
Use the $n$-linear characteristic, and you get that:
If $(x_1, \ldots, x_n)$ is not linearly independent, then $f(x_1, \ldots, x_n) = 0$.
The case of the determinant
Well, this applies to $\det$ also, so that, if $\det (x_1, \ldots, x_n) \neq 0$, then $(x_1, \ldots, x_n)$ is linearly independent.
Finally, we define the determinant of a matrix as the determinant of columns (or lines, because $\det$ is invariant with respect to the transposed of a matrix).
