Given $A\in Mat\_{n\times n}(R)$ where $R$ is a non-commutative associative ring are there exist any (non-zero) matrices $X, Y\in Mat\_{n\times n}(R)$ such that $XAY=diag(a\_1, \ldots , a\_n)$ for some $a\_i$?
The interesting answer for me is if $A=(x\_{i,j})$ and 
$R=\mathbb Z [x\_{i,j}]$
(free associative non-commutative algebra on $x\_{i,j}$ over $\mathbb Z$).


For example if $R$ is commutative then we put $X=Id$, $Y=(det(A\_{i,j}))$ and get $XAY=det A\cdot Id$. What about non-commutative polynomials?


Upd1: I want to have a non-commutative polynomial equality. Also I want $X$ and $Y$ to be in general invertible.


Upd2: Ok, I've understood that Update1 wasn't correct. I'm interested in having such matrices over $R=Mat\_{m\times m}(A)$ where $A$ is a commutative ring with $1$.



on existence of matrices X, Y s.t. XAY is diagonal over non-commutative ring
I've understood that if $R=Mat\_n(K)$ then for every $A\in Mat\_m(R)$ exist $B\in Mat\_m(R)$, s.t. $AB=\lambda Id$. $A=(a\_{ij,kl})$ is a $mn\times mn$-matrix over $K$ and $B=A^{V}$ is a $m\times m$-matrix over $R$.


