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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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    $\begingroup$ You might find my post here to be helpful. The video that I refer to in my comment can be found here. $\endgroup$ Oct 27, 2020 at 15:20
  • $\begingroup$ See also the explanation of the determinant given here. $\endgroup$ Oct 27, 2020 at 15:22
  • $\begingroup$ 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. $\endgroup$ Oct 27, 2020 at 15:26
  • $\begingroup$ 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. $\endgroup$
    – pll04
    Oct 27, 2020 at 19:14
  • $\begingroup$ Can this equivalence be shown without using the cross product (as if it is used, the proof would seem circular)? $\endgroup$
    – pll04
    Oct 27, 2020 at 19:15

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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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    $\begingroup$ For some it's-not-coincidence intuition, see e.g. this video, which math.se has discussed before. $\endgroup$
    – J.G.
    Oct 27, 2020 at 15:41

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