So a quick consultation of wikipedia suggests that the determinant was used for linear systems long before it was understood those systems could written in terms of matrices.
Nowadays, of course, we know we can write such systems in terms of matrices, or more broadly as linear operators. It turns out that a simple extension of a linear operator under the exterior algebra helps replicate the determinant:
The exterior algebra uses a wedge product, denoted with $\wedge$, and wedge products of several vectors allow us to treat planes, volumes, and such as algebraic elements.
The natural extension of a linear operator across the wedge product is to wedge every vector in that product: so given a wedge product $a \wedge b \wedge c \wedge \ldots$, the natural extension of a linear operator $\underline T$ is
$$\underline T(a \wedge b \wedge c \wedge \ldots) \equiv \underline T(a) \wedge \underline T(b) \wedge \underline T(c) \wedge \ldots$$
That is, have the operator act on each vector individually, then compute the wedge.
When you do this with the highest-graded wedge product, a wedge product of $n$ vectors in an $n$-dimensional space, the action of the linear operator reduces to a scalar multiplication. Let the $n$-vector be denoted $i$, and we get
$$\underline T(i) = (\det \underline T)i$$
This expresses how the "volume" of the space changes orientation and is dilated or shrunk under the transformation. The antisymmetry of the wedges captures exactly the same alternating plus and minus signs typically used to compute the determinant.
Edit: Why does the change in the volume matter? Well, you can grasp how this relates to invertibility: if the linear operator maps volume to volume, then you can see how there may be a bijection between vectors, but if the operator maps volume to zero volume, then the image is at most something smaller (in dimensionality) than volume, and as with any projection, that means multiple vectors are mapped to the same output vector---such a map cannot be invertible.