I have seen the formula for a determinant derived using cross product and vice versa. Is there a reasoning behind the actual computation?

I know both relate to the volume of a parallelepiped, and have seen a simple, geometric derivation of the formula for the determinant of a 2×2 matrix (finding the area of a parallelogram). Is there an intuitive derivation for the 3×3 case, that isn't circular (as with the cross product)?

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    You might find [my post here](https://math.stackexchange.com/a/3875351/81360) to be helpful. The video that I refer to in my comment can be found [here](https://www.youtube.com/watch?v=BaM7OCEm3G0). – Ben Grossmann Oct 27 '20 at 15:20
  • See also the explanation of the determinant [given here](https://math.stackexchange.com/a/669/81360). – Ben Grossmann Oct 27 '20 at 15:22
  • If there is something that you're after that is not addressed by these posts, please try to explain what it is that you're looking for that is missing. – Ben Grossmann Oct 27 '20 at 15:26
  • Thank you! I like your summary of the 3b1b video! I think I didn't phrase my question clearly - I am looking for an explanation of how the formula for the determinant of a 3×3 matrix (where you sum the products of the elements of a row/column and their corresponding cofactors) relates to the volume of the parallelepiped formed by the three vectors in the matrix (V=Bh). Similarly to how you can derive the formula for the determinant of a 2×2 matrix by finding the area of the corresponding parallelogram. – pll04 Oct 27 '20 at 19:14
  • Can this equivalence be shown without using the cross product (as if it is used, the proof would seem circular)? – pll04 Oct 27 '20 at 19:15
  • I think your question was clear enough; I pointed to the link in my second comment because it I thought it addresses your question more directly. Did you try to read through that answer? – Ben Grossmann Oct 27 '20 at 19:18
  • To be a bit more explicit about the connection, the point is that properties 1,2,3,4 from the answer are enough to deduce the formula for the determinant in terms of the matrix's cofactors – Ben Grossmann Oct 27 '20 at 19:19
  • So with that said: are you satisfied with the answer's justification that the determinant has properties 1-4? If not, what do you want clarified? Would you want to see a derivation of the cofactor formula using these properties? – Ben Grossmann Oct 27 '20 at 19:22
  • Thank you - I read it. I was wondering if there is another derivation of the cofactor formula, other than using the fact that it satisfies the properties of a determinant. Perhaps a proof that takes into account the more intuitive view of determinants as volume scaling factors. – pll04 Oct 27 '20 at 20:25
  • As with the linked answer, most derivations of the formula determinant that start with the geometric (volume scaling) view being by using the geometric view to deduce the algebraic properties of the determinant, then using those properties to get the formula. I don't know of another way to approach the proof that makes things any more intuitive – Ben Grossmann Oct 27 '20 at 20:39

1 Answers1


In one way you can just think of the fact that the cross product can be computed as a 3x3 determinant as a happy conincidence, or a handy way of remembering the formula. The formulas simply coincide.

Determinants for $n \times n$ matrices are defined inductively, independtly of the cross product. The cross product is defined differently in different textbooks, certainly depending on the level of the book. Some books define it by its formula and derive its geometric properties, while others define it by its geometric properties.

As you mention, there is a neat connection between the cross product and a $3 \times 3$ determinant through the area of a parallellepiped. It turns out that the area of a parallellepiped spanned by the vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ is $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$, and that it can be computed by taking the determinant of the matrix whose columns are $\mathbf{a}, \mathbf{b}, \mathbf{c}$. However, this does not really have anything to do with parallellepipeds. Once again, the formulas simply coincide. In general we have

$$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \det[\mathbf{a} \, \mathbf{b} \, \mathbf{c}].$$

This can, rather boringly, be proven by simply varifying that both sides of the equations are the same.

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  • For some it's-not-coincidence intuition, see e.g. [this video](https://www.youtube.com/watch?v=eu6i7WJeinw), which math.se has discussed before. – J.G. Oct 27 '20 at 15:41