**Context:** some probably know that there are Capelli identities which state
det(A)det(B) = det(AB+correction) for some matrices with non-commuting elements, they go back to 19-th century, but also investigated up to present due relations with repesentation theory, quantum integrable systems, etc. In forthcoming paper I will discuss a more general question/conjecture on
det(A+correction)det(B+correction) = det(AB+correction). But what just I understood that
there is quite a simple to ask question, which seems to be open and let me ask it here.

**Notations** Consider sets of commuting variables $a_i$ and another set $b_i$, but $[a_i, b_j] \ne 0$, more precisely let require the following commutation relations:

Denote $E_{ij}=a_i b_j$ and require that $[E_{ij}, E_{kl} ] = \delta_{jk}E_{il}-\delta_{li}E_{kj}$, i.e. the same relations as satisfied by the "matrix units" $e_{ij}$, i.e. matrices having 1 at position (i,j) and zeros everywhere else.

**Question 1** Is it true that $det^{column}(E+diag(n-1,n-2, ..., 1,0)) = 0 $ ? (n>1).

Where $E$ is a matrix with elements $E_{ij}$ at position (ij), $det^{column}$ means we use usual definition of the determinant, but the elements of first column comes first in the products, second column - second, etc.

**Motivations:** If $a,b$ would be commuting variables then $E$ is matrix of rank 1
and its determinat equal to zero. The correction $+diag(n-1,n-2, ..., 1,0)$ is usual
correction for the Capelli type identities.

**Checks** For 2x2 matrix I have checked it. If commutation relations between
$[a_i, b_j]=-\delta_{ij}$ then it is also known to be true as a part of the modern Capelli story. I.e. if $b_j = \partial_{a_j}$ we get standard construction of the $S^*(C^n)$
representation of $gl_n$ by differential operators.

**Subquestion:** For 3x3 case it is not easy to check by hands, may be it is possible to check by some software ?