If $R$ is a ring, $E_n(R)$ is the subgroup of the group $GL_n(R)$ generated by matrices obtained from the multiplicative identity matrix by replacing an off-diagonal entry by some $r \in R$. The group of invertible diagonal matrices is $D_n(R)$; it normalizes $E_n(R)$. Say $R$ is a $GE_n$-ring if $GL_n(R) = E_n(R)D_n(R)$, so every invertible $n\times n$ matrix can be row reduced to diagonal form by addition type row operations. This performs a sort of euclidean algorithm on the entries of each column. Say $R$ is a $GE$-ring if it is a $GE_n$-ring for all $n>1$. A ring has matrix completion if for all $n>1$, every left unimodular column of $n$ elements in $R$ occurs as a column in an invertible nxn matrix over $R$. When $R$ is directly finite (e.g. left or right noetherian), this last condition is equivalent to every stably-free $R$-module being free. Computations with integral group rings $ZG$ for finite groups $G$ suggest that $GE$ implies matrix completion, but I see no direct reason for this. The $GE$ condition says the columns of $E_n$ and $GL_n$ matrices are the same, but the matrix completion condition says that set of columns is all left unimodular columns. Does anyone know a counter-example?