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I just encountered a very curious relation in an algebra. A bit simplified, I am working in a (particular) non-commutative algebra, with some relations.

One particular relation is the following: For (some particular choices of) elements $a,b,c,x,y,z$ in the algebra, the relation is of the form $bz+cx+ay = az+bx+cy$. This can be stated as a vanishing 3x3-determinant,

$$ \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ x & y & z \end{vmatrix}=0. $$

We can think of $a,...,z$ as words, and multiplication is simply concatenation. Has this type of relation been observed/used in some algebras before? Does it fit into some larger picture?

In my concrete case, this relation does not hold in general, but only for certain particular combinations of $a,b,c,x,y,z$, so examples where there are some restrictions on when the relation applies are also interesting to me.

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    $\begingroup$ What is your definition of the determinant in a non-commutative algebra? $\endgroup$
    – abx
    Commented Apr 9, 2020 at 12:05
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    $\begingroup$ The formula stated below the determinant. That is, "Sarrus' rule". I am mainly interested in the 3x3 case (as that's what I encountered). $\endgroup$
    – Per Alexandersson
    Commented Apr 9, 2020 at 12:52
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    $\begingroup$ I don't understand the question at all. The free non-commutative algebra on $\{a,b,c,x,y,z\}$ has as a basis all ordered monomials in the letters. So $bz$ is linearly independent of $cz, ay, az, bx, cy$, ruling out the relation you claim holds. What algebra are we working in? Is your question really: `if this relation holds, what can deduce about related determinants?' $\endgroup$
    – Mark Wildon
    Commented Apr 9, 2020 at 13:42
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    $\begingroup$ But your determinant has plenty of non-equivalent ways to be read, unless you specify precisely rules. Formally: you want to choose a lift from the determinant formula from the polynomial algebra in $n^2$ variables to the non-commutative one. Probably you implicitly have such a lift in your mind. $\endgroup$
    – YCor
    Commented Apr 9, 2020 at 21:41
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    $\begingroup$ @PerAlexandersson : I have the feeling it would be easier if you would explain the concrete example you are working with in more details. $\endgroup$
    – Libli
    Commented Apr 9, 2020 at 22:39

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