Idea behind the factorization of the matrix $\operatorname{diag}(a,a^{-1})$ in algebraic K-Theory If $a \in S$ is some invertible element in a ring $S$, then a computation shows
$$\pmatrix{a & 0 \\ 0 & a^{-1}} = \pmatrix{1 & a \\ 0 & 1} \pmatrix{1 & 0 \\ -a^{-1} & 1} \pmatrix{1 & a \\ 0 & 1} \pmatrix{0 & -1 \\ 1 & 0}.$$
If $R \to S$ is a surjective homomorphism, we see that this invertible matrix over $S$ may be lifted to some invertible matrix over $R$. This observation is important in algebraic K-Theory; for example it is used in the exactness of the relative $K_0$-sequence (see Rosenberg's book, 1.5.4 -  1.5.5).
Questions. What is the idea behind this factorization? Of course it makes no problem to verify this identity, but how can you come up with such a nontrivial factorization? Does it have a geometric interpretation? Who was the first one to find and use this identity?
PS: Isn't it sad that only few textbooks and papers offer explanations of the important insights, rather than only proof verifications?
 A: I have no answer about the history or geometric interpretation, but this is how I would come up with a similar identity in a more or less algorithmic fashion. The goal is to express
$$\left(\begin{array}{cc}a&0\\0&a^{-1}\end{array}\right)$$
as a product of elementary matrices. To do this it suffices to "reduce it to $I$ elementary row and column operations". I can do that as follows:


*

*Add $a$ times column 2 to column 1, getting 
$$\left(\begin{array}{cc}a&0\\1&a^{-1}\end{array}\right).$$

*Add $1-a^{-1}$ times column 1 to column 2, getting
$$\left(\begin{array}{cc}a&a-1\\1&1\end{array}\right).$$

*Subtract column 2 from column 1, getting
$$\left(\begin{array}{cc}1&a-1\\0&1\end{array}\right).$$

*Subtract $a-1$ times column 1 from column 2, getting $I$.
Working backwords, I find the factorisation
$$\left(\begin{array}{cc}a&0\\0&a^{-1}\end{array}\right) = \left(\begin{array}{cc}1&a-1\\0&1\end{array}\right)\left(\begin{array}{cc}1&0\\1&1\end{array}\right)\left(\begin{array}{cc}1&a^{-1}-1\\0&1\end{array}\right)\left(\begin{array}{cc}1&0\\-a&1\end{array}\right).$$
Unless I completely misunderstood the motivation, I think this factorisation is just as useful.
