Algebraically-nice general solution for last step of Gaussian elimination to Smith Normal Form? My question takes a little bit of preamble: it concerns a well-known and solved problem, but I am looking for a solution with a particularly nice property.
$\newcommand{\matrix}[4]{\left( \begin{array}{cc} #1 & #2 \\ #3 & #4 \end{array} \right)}
 \DeclareMathOperator{\lcm}{lcm}$In using Guassian elimination to put a matrix into Smith normal form over $\mathbb{Z}$ (or, more generally, some PID), the last step is to make sure that successive diagonal entries divide each other.  Solving this reduces to the following problem (with all matrices over $\mathbb{Z}$):


*

*given a diagonal matrix $M = \left( \begin{array}{cc} a & 0 \\ 0 & b \end{array} \right)$, find invertible matrices $L$, $R$ such that 
$$ M = L \left( \begin{array}{cc} \gcd(a,b) & 0 \\ 0 & \lcm(a,b) \end{array} \right) R $$


This is of course well-known and not hard to do.  For instance, using the Euclidean algorithm to find $(x,y)$ such that $ax + by = d = \gcd(a,b)$, one can define
$$ L = \left( \begin{array}{cc} a/d & -y \\ b/d & x \end{array} \right),
 \quad R = \left( \begin{array}{cc} 1-yb/d & yb/d \\ -1 & 1 \end{array} \right)$$
or alternatively
$$ L = \left( \begin{array}{cc} a/d & -1 \\ 1-xa/d & x \end{array} \right),
\quad R = \left( \begin{array}{cc} 1-yb/d & b/d \\ -y & 1 \end{array} \right).$$
However, in the special case where $a$ divides $b$, we know that $M$ is already as desired and so we could simply take $L = R = I$.  My question is: can we find a general algebraic solution like the ones above (i.e. algebraic definitions of $L$, $R$ in terms of the integers $(x,y,a/d,b/d)$), but with the additional property that when $a$ divides $b$ (and so $x=1$, $y=0$, $a/d=1$), the solution yields $L = R = I$?  Roughly: can we find an algebraic solution which only does something non-trivial if it needs to?
 A: Yes.  We just need to come up with a sequence of integer row and column operations that reduces to Smith normal form, and which is trivial in the special case where $a$ divides $b$.  Here is one such sequence:


*

*$\begin{bmatrix}a & 0 \\ 0 & b\end{bmatrix} = E_1 \begin{bmatrix}a & 0 \\ a & b\end{bmatrix}$ for some elementary matrix $E_1$.

*$\begin{bmatrix}a & 0 \\ a & b\end{bmatrix} = \begin{bmatrix}r & s \\ d & b\end{bmatrix} M_2$ for some matrix $M_2$.(Note that $r=d=a$, $s=0$, and $M_2$ is the identity matrix in the special case.)

*$\begin{bmatrix}r & s \\ d & b\end{bmatrix} = E_3\begin{bmatrix}d & t \\ d & b\end{bmatrix}$ for some elementary matrix $E_3$. 
(Note that this row operation is trivial in the special case, with $t=0$.)

*$\begin{bmatrix}d & t \\ d & b\end{bmatrix} = E_4\begin{bmatrix}d & t \\ 0 & m\end{bmatrix}$ for some elementary matrix $E_4$.
(In the special case, $m=b$, and this row operation is the inverse of the row operation in the first step.)

*$\begin{bmatrix}d & t \\ 0 & m\end{bmatrix} = \begin{bmatrix}d & 0 \\ 0 & m\end{bmatrix}E_5$ for some elementary matrix $E_5$.
(Note that this column operation is trivial in the special case, since $t=0$.)
Computing gives
$$
E_1 = \begin{bmatrix}1 & 0 \\ -1 & 1\end{bmatrix},\qquad 
M_2 = \begin{bmatrix}(a+by)/d & (b-bx)/d \\ -y & x\end{bmatrix},\qquad
E_3 = \begin{bmatrix}1 & ax/d-1 \\ 0 & 1\end{bmatrix},\qquad 
$$
$$
E_4 = \begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix},\qquad 
E_5 = \begin{bmatrix}1 & b(d-a)/d^2 \\ 0 & 1\end{bmatrix}.
$$
Multiplying $L=E_1E_3E_4$ and $R=E_5M_2$ gives us
$$
L \;=\; \begin{bmatrix}ax/d & ax/d-1 \\ 1-ax/d & 2-ax/d\end{bmatrix},
\qquad
R \;=\; \begin{bmatrix}a(d+by)/d^2 & b(d-ax)/d^2 \\ -y & x\end{bmatrix}.
$$
In the case where $x=1$, $y=0$, and $d=a$, both of these matrices are the identity matrix.
