I drew a parallelepiped that is spanned by three vectors, and we know the volume is given by the absolute value of the determinant of the matrix – with the three vectors arranged in rows (or columns, since $detA = detA^T$).
How can I draw the tetrahedron from this parallelepiped to convince myself it is indeed 1/6th of the volume, i.e., 1/6 * |detA|?
Thanks,
Best Answer
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.