linear algebra - Sarrus determinant rule: references, extensions

The logic behind this does extend to general $ntimes n$ 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 $leftlbrace 1,2,...,n-1rightrbrace$, and thus can be seen as a permutation of the set $leftlbrace 1,2,...,nrightrbrace$ which leaves $n$ fixed.

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

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

Now, write the determinant of an $ntimes n$ matrix in the form $sum_{piin S_n}mathrm{sign}picdotprod ...=sum_{sigmain A_{n-1}}sum_{xiin D_n}mathrm{sign}xicdotprod ...$. Each inner sum $sum_{xiin D_n}mathrm{sign}xicdotprod ...$ 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 $sum_{sigmain A_{n-1}}$ 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 $sum_{sigmain A_{n-1}}$ has three terms, and it follows that the determinant of a $4times 4$ matrix can be written as a sum of three "Sarrus determinants". A closer look at the sum shows which ones.

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