determinantgeometrylinear algebra
The absolute value of a $2 \times 2$ matrix determinant is the area of a corresponding parallelogram with the $2$ row vectors as sides.
The absolute value of a $3 \times 3$ matrix determinant is the volume of a corresponding parallelepiped with the $3$ row vectors as sides.
Can it be generalized to $n-D$? The absolute value of an $n \times n$ matrix determinant is the volume of a corresponding $n-$parallelotope?
A determinant is always a member of the field (or ring) that the matrix entries comes from -- for any given size of the matrix the determinant is a particular polynomial in the entries. Thus, if the matrix entries are all real, then so is the determinant.
But with, say, complex entries in the matrix it is easy to find a matrix with a non-real determinant:
$$\left|\begin{matrix}i & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right|=i $$
By the way, it doesn't really work well to define the determinant as the product of the eigenvalues, because some matrices have fewer different eigenvalues than the size of the matrix, and then you need to include some eigenvalues in the product several times, according to their (algebraic) multiplicity. But the way to define the multiplicity of eigenvalues is through the characteristic polynomial which in itself is defined using determinants!
This image was taken from Dune-Project.org
Notice that each piece in fact has the same volume as the tetrahedron described in anorton's comment above (by connecting tips of the three vectors) by the same reason that the area of a triangle relies solely on base times height, a three dimensional object relies on the area of the base times the shortest distance from the final vertex to the base. The green tetrahedron for example originally has vertices 4,5,7,0 but you could imagine dragging the vertex from 0 to the vertex at 1 without changing the volume at any time along it's journey.
More on the idea of how you can slide things as I suggest with this applet here.
Best Answer
Yes it can. In fact, as Jamie Banks noted, a determinant is an intuitive way of thinking about volumes. To summarise the argument, if we consider the vectors as a matrix, switching two rows, multiplying one by a constant or adding a linear combination will have the same effect on the volume as on the determinate. We can use these operations to transform any n-parallelotope to cube and note that the determinate matches the signed volume here, so it will match it everywhere as well.