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An undergraduate came to me with an identity for 4x4 determinants that is actually correct:


where R cyclically permutes the last three rows of the matrix A. I wont define h here but, except for the signs of the terms, it is the usual incorrect extension of Sarrus Rule that is familiar to anyone who has has taught linear algebra (His identity is not the Laplace expansion, as it has three 4x4 matrices, rather than four 3x3 matrices. )

Is something like this known? Mathscinet lists a paper Monaco and Monaco that might be relevant, but my library couldnt get it. I haven't even found the original reference for Sarrus' rule itself.

Eric Schmutz

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So the trick is that every permutation $\pi\in S_4$ can be uniquely written as $\sigma\xi$, where $\sigma\in Z_3$ (here, $Z_3$ is seen as a subgroup of $S_4$ acting on the four-element set by cyclically permuting its last three elements) and $\xi\in D_4$ (where $D_4$ is the dihedral group generated by the shift $\left(x\mapsto x+1\mod 4\right)$ and the reflection $\left(x\mapsto -x\mod 4\right)$). The $\xi$ corresponds to the application of the Sarrus rule, and the $\sigma$ corresponds to $1$, $R$ rsp. $R^2$. If I find some generalization of this, I'll post it as an answer. – darij grinberg Mar 21 '10 at 22:46
I should have denoted $Z_3$ by $C_3$; it's the cyclic group on $3$ elements. Note that $D_4$ is the dihedral group with $8$ (not $4$) elements; I know that there are some conflicting notations here. – darij grinberg Mar 21 '10 at 22:47
There is an amazing book on determinants and their history by Thomas Muir. That's the first place I'd look. – Mariano Suárez-Alvarez Mar 21 '10 at 22:57
And if it's not in Muir, try Dodgson, An Elementary Treatise On Determinants. Dodgson, of course, is better known as Lewis Carroll. – Gerry Myerson Mar 21 '10 at 23:11
My quotes are actually from 1923 (the fourth volume of Muir). And I must say I have seen better notations in literature from that time. van der Waerden's Modern Algebra is just 7 years younger! – darij grinberg Mar 21 '10 at 23:29
up vote 1 down vote accepted

I have found references to the Sarrus determinant rule and to an extension of it. In The Quarterly journal of pure and applied mathematics, Volume 38 which is available on Google books in the article "A fourth list of writings on determinants". On page 239 There is a reference to what I believe is Sarrus original result. Then on page 350 there is a reference to an extension of Sarrus' original result. I believe that part one of Muir's book on determinants and their history is available on Google books

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What is the extension? (I think you mean page 250.) – Jonas Meyer Mar 22 '10 at 3:02

The logic behind this does extend to general $n\times n$ determinants, though probably not as nicely as you wish. Note that I am taking the liberty to replace "last" by "first" in "where R cyclically permutes the last three rows of the matrix A". It doesn't matter, because Sarrus' rule is invariant under cyclic shift, and a simple cyclic shift turns the last three rows to the first three rows.

Consider the alternating group $A_{n-1}$ embedded into the symmetric group $S_n$: every element of $A_{n-1}$ is a permutation of the set $\left\lbrace 1,2,...,n-1\right\rbrace$, and thus can be seen as a permutation of the set $\left\lbrace 1,2,...,n\right\rbrace$ which leaves $n$ fixed.

Also consider the dihedral group $D_n$ defined as the subgroup of $S_n$ generated by the cyclic shift $\left(x\mapsto x+1\mod n\right)$ and the reflection $\left(x\mapsto n+1-x\right)$.

Then, every element $\pi\in S_n$ can be uniquely written as $\pi=\sigma\xi$ with $\sigma\in A_{n-1}$ and $\xi\in D_n$. In fact, $\xi$ is uniquely determined by the conditions $\left(\pi\xi^{-1}\right)\left(n\right)=n$ and $\mathrm{sign}\left(\pi\xi^{-1}\right)=1$, and then $\sigma$ results.

Now, write the determinant of an $n\times n$ matrix in the form $\sum_{\pi\in S_n}\mathrm{sign}\pi\cdot\prod ...=\sum_{\sigma\in A_{n-1}}\sum_{\xi\in D_n}\mathrm{sign}\xi\cdot\prod ...$. Each inner sum $\sum_{\xi\in D_n}\mathrm{sign}\xi\cdot\prod ...$ is the naive "Sarrus determinant" of some permutation of the matrix; which permutation it actually is is decided by the $\sigma$.

For $n=3$, we have $A_{n-1}=A_2=1$, so the outer sum $\sum_{\sigma\in A_{n-1}}$ has only one term, and the "Sarrus determinant" is the real determinant.

For $n=4$, we have $A_{n-1}=A_3=C_3$ (the cyclic group with $3$ elements), so the outer sum $\sum_{\sigma\in A_{n-1}}$ has three terms, and it follows that the determinant of a $4\times 4$ matrix can be written as a sum of three "Sarrus determinants". A closer look at the sum shows which ones.

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I've posted a diagram version of 4x4 Sarrus here:

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