Question

This question is most probably not research level, but I thought that the MO folks might like it... Feel free to close.

Here is the motivation: If you have ever teached a maths course for engineers which covered determinants and which included a written exam then you often ask the students to calculate a 4 by 4 determinant to check if they got the basic rules (e.g. using Laplace's formula to reduce to 3 by 3 if the structure is favorable or use Gauß elimination). If you did this you most probably have seen a student solving this problem by applying the "Sarrus rule for 4 by 4 matrices". Usually, the students memorize for 3 by 3 a pattern like

(of course, the lines mean that you should multiply the numbers along the lines; green lines get a $+$, red lines get a $-$, finally add everything up). My colleagues told me, that in every exam there is at least one smart guy who happily generalizes this rule to 4 by 4 matrices with a scheme like this:

which I am going to refer to as False Sarrus Rule. Indeed, one could turn this into a working rule by assigning the right signs and repeating the procedure two times more in a different way. I wrote a small blog post here (and there is even paper on this (German description, Russian description)). Basically, I wrote this blog post to give the people who search the net for a generalized Rule of Sarrus some visual reminder that there is no easy "Sarrus Type Rule" available. Believe it or not: The post is found frequently via search terms like "sarrus rule", sarrus 4*4", "sarrus matrice 4 4" or the like. Discussing this with a colleague today, we asked ourselves the following question:

How does the set of 4 by 4 matrices for which this "False Sarrus Rule" gives the correct determinant looks like?

Basic thoughts: Obviously, a matrix $A=(a_{ij})$ is in this set, if and only if the following equation is fulfilled $$\sum_{\text{eight special permutations}\ \pi_j} \pm a_{1\pi_j(1)}\cdots a_{4\pi_j(4)} = \det(A).$$ Four out the the eight summands on the left have the right sign, the other four have the wrong sign, and hence, one could simplify a bit. However, the bottom line is: There is just this one equation which has to be fulfilled for all the sixteen entries of a 4 by 4 matrix (and this equation is a homogeneous polynomial of degree four) and hence, the set of matrices for which the False Sarrus Rule gives the right result is a 15-dimensional variety, but I have no clue how it looks like. Probably some algebraic geometers could step in and provide some insight?

Final remark: I do not plan to include this discussion in a math course for engineers (although it may help to scare some people away from the thought that "there could be an easy Rule of Sarrus).

Answer 1

Here is one result suggested by Gerhard Paseman's comments. The False Sarrus Rule is correct on all matrices of rank $1$ and $4\times 4$ and $5\times 5$ matrices of rank $2$. It does not hold in general on matrices of rank $3$ for $n\times n$ matrices with $n\gt 3$. It also fails for some matrices of rank $2$ and dimension $6$ or greater.

To see that it fails on matrices of rank $3$, consider block diagonal matrices with a nonsingular $2\times 2$ block $A = {a~~b \choose c~~d}$ and a $J_{n-2}$ block. This is singular for $n \gt 3$ but the False Sarrus Rule produces $\det(A) \ne 0.$

$$\begin{pmatrix} a & b & 0 & \cdots & 0 \\ c & d & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 1 &\cdots &1\end{pmatrix}$$

If $M$ has rank $\le 2$, let the columns be linear combinations of $\vec{v}$ and $\vec w$, $a_i \vec{v} + b_i\vec{w}.$ Then the monomials of the False Sarrus Rule are of the form $\prod_{i\in I} a_{\pi(i)} v_i \prod_{i \in I^c} b_{\pi(i)} w_i.$ If $|I| \le 2$ or $|I^c| \le 2$ then the coefficient of the monomial is $0$ by collecting terms ($n-n$ for ranks $0$ and $1$, and $1-1$ for rank $2$).

This fails for $|I|,|I^c| \ge 3$. For example, the False Sarrus Rule evaluates to $1$ on $P (J_3 \oplus J_{n-3}) P $, where $P$ is the permutation matrix for the $(3 ~4)$ transposition since only the main diagonal contributes. For $n=6$, this is

$$\begin{pmatrix}1 & 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 \end{pmatrix}.$$