Reading at Mathworld, I came across the subject of tetrahedrons. Particularly calculating the volume with four known vertices. There's a formula which uses the triple product to calculate the volume of a parallelepiped. I'm very aware of why the triple product represents the volume of a parallelepiped.
However, I don't understand how to derive the relation that the volume of a tetrahedrons is one-sixth of a parallelepiped. I can accept it, sure. And graphically, I'm sure that it is true.
Is there any way to prove it using linear algebra?
Wikipedia provides me with one proof. However, I don't understand how they went from pyramid volume being $1/3\,A_0h$ to tetrahedrons being $1/6\,A_0h$. It must mean that the volume of a tetrahedron is half that of a pyramid, must it not?