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).