invertibility of matrix over free associative algebra For a commutative ring $R$, a matrix $A \in M_n(R)$ is invertible iff $\det (A)$ is a unit in $R$. Is there a similar criterion to determine invertibility (having two-sided inverse) of a matrix over a noncommutative ring ? I am particularly interested in determining invertibility of matrices in $M_n(F \langle x, y \rangle )$, the free associative algebra in $2$ indeterminates over a field $F$.
Passing onto the polynomial ring using the relation $xy - yx = 0$, we may conclude that the `determinant' should be in $F$ which serves as a necessary condition to filter out certain matrices as being non-invertible. Based on this, I have a question as stated below.
Let $f \in F \langle x, y \rangle$ and $a \in F - \{ 0 \}$. Clearly matrices of the form 
$$\begin{bmatrix}
1 & f\\
0 & 1
\end{bmatrix},
\begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix},
\begin{bmatrix}
a & 0\\
0 & 1
\end{bmatrix}
$$ are invertible in $M_2(F \langle x, y \rangle)$. By composing such matrices, one gets other examples of invertible matrices in $M_2(F \langle x, y \rangle)$.
Question :  Can any invertible matrix in $M_2(F \langle x, y \rangle)$ be decomposed as a product of matrices of the above $3$ types ?
 A: You are asking whether the following holds.


Claim. Let $F$ be a field. Then the free associative algebra $F\langle x,y\rangle$ is a $GE_2$-ring in the sense of P. M. Cohn [2].


The answer is yes. Indeed, free associative algebras over a field are more generally $GE$-rings [Theorem 3.4, 2]. This fact has been used by Bass [1] to prove that a finitely generated projective module over a free algebra with coefficients in a PID is free.
By contrast, free commutative algebras over a field, i.e., polynomial rings over a field aren't $GE$-rings if the number of indeterminates exceeds $1$ [Proposition 7.3, 2]. (I have made this remark in another MO post).
We recall below the definition of a $GE$-ring in the sense of P. M. Cohn and we give subsequently a proof of the claim. 
Let $R$ be an associative ring with identity and let $E_n(R)$ be the subgroup of $GL_n(R)$ generated by matrices obtained from the identity matrix by replacing an off-diagonal entry by some $r∈R$. Let $D_n(R)$ be the group of invertible diagonal matrices; note that it normalizes $E_n(R)$. 
A ring $R$ is termed a $GE_n$-ring if $GL_n(R)=E_n(R)D_n(R)$. 
The ring $R$ is a $GE$-ring in the sense of P. M. Cohn if it is a $GE_n$-ring for every $n > 1$. 
Let us turn to the proof of the claim. Let $S$ be the free monoid on $\{x, y\}$ endowed with the shortlex order, i.e., words are ordered according to word length and words of the same length are ordered lexicographically. Then $S$ is a bi-ordered monoid and it is not difficult to show that $F\langle x,y \rangle = F[S]$ is an integral domain with unit group $F^{\times}$. Let $\varphi: S \rightarrow \omega$ be an order-preserving bijection. The following lemma is instrumental.


Lemma. Let $R = F\langle x,y \rangle$ and let $a_1, a_2 \in R$. Assume there are $b_1, b_2 \in F\langle x,y \rangle$, not both zero, such that $a_1b_1 + a_2 b_2 = 0$. Then we can find $E \in E_2(R)$ and $a \in R$ such that $(a_1, a_2)E = (a, 0)$.
Proof. For $r \in R$, write $r = \sum_{w \in S} r_w w$, where each $r_w \in F$ and $\text{supp}(r) \Doteq \{w \in S \,\vert\, r_w \neq 0\}$ is a finite subset of $S$. Since $R$ is an integral domain, we can assume, without loss of generality, that none of $a_1, a_2, b_1$ and $b_2$ is zero. Setting $n(r) \Doteq \varphi(\max(\text{supp}(r))) \in \omega$ for $r \neq 0$, we argue by induction on $n(a_1) + n(a_2)$. Let $w_i = \max(\text{supp}(a_i)) \in S$, for $i = 1,2$.
    Since $a_1b_1 + a_2 b_2 = 0$, we can find $v_1, v_2 \in S$ such that $w_1 v_1 = w_2 v_2$. Swapping $a_1$ and $a_2$ if needed, we can assume that $w_1 \le w_2$, so that $w_1$ is an initial subword of $w_2$. Therefore we can find $s \in S$ and $\lambda \in F$ such that either $a_2' = 0$ or
    $n(a_2') < n(a_2)$ where $a_2' = a_2 - \lambda a_1 s$. We set $a_1' = a_1$, 
    $b_1' = b_1 + \lambda sb_2$ and $b_2' = b_2$. Then we have $a_1' b_1' + a_2'b_2' = 0$ and $(a_1', a_2')$ is obtained from $(a_1, a_2)$ by right-multiplication of a matrix in $E_2(R)$. If $a_2' = 0$, we are done. Otherwise the induction hypothesis applies to $(a_1', a_2')$, which yields the result.


We are now in position to prove the claim.


Proof of the claim. Let $A = (a_{ij}) \in GL_2(R)$ with $R = F\langle x,y \rangle$. As $A$ has a right inverse, we can find $b_1, b_2 \in R$, not both zero, such that $a_{11} b_1 + a_{12}b_2 = 0$. By the above lemma, we can assume that $a_{12} = 0$. It follows that $a_{11} \in F^{\times}$, so that $A$ can be reduced to an invertible diagonal matrix through an elementary row operation. Therefore $A \in E_2(R)D_2(R)$. 



[1] H. Bass, "Projective modules over free groups are free", 1964.
[2] P. M. Cohn, "On the structure of the $GL_2$ of a ring", 1966.  
