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Suppose that $A = (a_{i, j})_{i, j \in [n]}$ is any square matrix. This question is about collecting the known formulas for $\det A$ (which don't depend on $A$ having a special form or dimension) in one place. I suppose I should start with a few:

  • The complete expansion says that $\det A = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{k=1}^n a_{k, \sigma(k)}$.
  • The Laplace expansion at the $j$-th row is $\det A = \sum_{i=1}^n (-1)^{i+j} a_{i, j} \det(A_{i, j})$, where $A_{i, j}$ is formed from $A$ by deleting the $j$-th column and the $i$-th row (in any order).
  • One may form the characteristic polynomial; the determinant will then be the product of its roots (which are, of course, by definition precisely the eigenvalues of $A$ with algebraic multiplicity), or alternatively the constant coefficient times $(-1)^n$.
Cloudscape
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    There are formulas for $A\in M_n(K)$ for small $n$ known, for example the Rule of Sarrus, for $n=3$. For higher $n$ we also know explicit formulas, but they are just "huge" polynomials in $n^2$ variables. Furthermore you probably know determinant formulae with the adjugate matrix - see here for example. I don't think you can say "all known formulas". – Dietrich Burde Jan 25 '23 at 22:22
  • Indeed: $A \operatorname{adj}(A) = \det A \operatorname{I}$. I wasn't at all claiming that my list is complete; to the contrary, it was supposed to be a starting point and the purpose of my question is to give a big list of most known formulas. – Cloudscape Jan 25 '23 at 22:29
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    Here is one more. – Anne Bauval Jan 25 '23 at 22:39
  • Your "complete expansion" is also known as the "Leibniz formula" – Ben Grossmann Jan 25 '23 at 22:45
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    @BenGrossmann I know, but apparently it was contemporaneously discovered in Japan (by Seki Takakazu). – Cloudscape Jan 25 '23 at 22:47
  • Cramer's rule is helpful also. It is recursive but you can apply it to $n$-th size matrices. – John Alexiou Jan 26 '23 at 00:31
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    I still think you should change your title. One cannot ask for all known formulas, and also it is not defined exactly what you accept as a "general" formula. – Dietrich Burde Jan 26 '23 at 09:10
  • Why can't one ask for all known formulas? If each formula is known by a sufficient number of mathematicians, chances are that one of them will turn up here. (And even if they didn't we could ask.) As for the generality, I suppose I could clarify this point. – Cloudscape Jan 26 '23 at 10:40
  • I am not sure how you are going to find the characteristic polynomial except by computing a more complex determinant, so I don't think your 3rd formula really is one. – ancient mathematician Jan 26 '23 at 15:50
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    @ancientmathematician For symmetric real matrices, you may successively compute the largest eigenvalue using the classical ("von Mises") iterative method and then quotient out the corresponding eigenvector. I'm not saying that this is an efficient method though. Alternatively, you could perform the QR algorithm. I'm insufficiently acquainted with today's NUMLA community in order to tell you whether people actually do these things. – Cloudscape Jan 26 '23 at 21:37
  • Thanks @Cloudscape I had intended to delete the comment after overnight thought. My reason was that one can use the usual Smith Normal Form algorithm to compute the elementary divisors of $A-xI$: and if by "formula" you allow an algorithm then we'll get the determinant this way. – ancient mathematician Jan 27 '23 at 07:46
  • I was entirely unaware of that, very interesting. It is not immediately obvious to me though why the latter determinant should be a lot more difficult to compute by the Smith normal form. – Cloudscape Jan 27 '23 at 13:51

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