The question started with how to compute the matrix inverse with a scale separated orders.
A related question could be found here: Compute matrix inverse at scale separated orders
This is, in $2\times 2$ matrix example, how to compute the matrix inverse of $$ \begin{pmatrix} a_{11} 10^{10}& a_{12} 10^{30} \newline a_{21} 10^{20}& a_{22} 10^{40} \end{pmatrix} $$ where the entry $a_{ij}\sim O(1)$ while being multiplied by some scale separation parameters of order $(\frac{1}{\epsilon})^N$ for some integer $N$. i.e. is it possible to compute the matrix inverse directly from $a_{ij}$ and then put in the orderings.(I've been studying scale separation and singular values for quite a while, so those were a bit standard.)
Initially, this might sound like some non trivial task, because $$ \begin{pmatrix} 1 & 1 \newline 1 & 1 \end{pmatrix} $$ were non invertible matrixs, but $$ \begin{pmatrix} 1\cdot 10^{10}& 1\cdot 10^{30} \newline 1\cdot 10^{20}& 1\cdot 10^{40} \end{pmatrix} $$ were invertible matrix.
However, consider the following matrix written with sympy code
x=10;
M = Matrix(( [1*(x)**1, 2*(x)**2, 3*(x)**5], [3*(x)**5, 6*(x)**5, 2*(x)**1], [2*(x)**1, 0*(x)**2, 1*(x)**2] ))
M.inv().evalf()
$$ \left[\begin{matrix}-1.66417044797627 \cdot 10^{-5} & 5.54723482658756 \cdot 10^{-9} & 0.0499251123298411\\8.32074129518481 \cdot 10^{-6} & 1.66389308623494 \cdot 10^{-6} & -0.0249625566641717\\3.32834089595254 \cdot 10^{-6} & -1.10944696531751 \cdot 10^{-9} & 1.49775340317864 \cdot 10^{-5}\end{matrix}\right] $$
at $x=100$
$$ \left[\begin{matrix}-1.66663916712041 \cdot 10^{-9} & 5.55546389040138 \cdot 10^{-16} & 0.00499991750136123\\8.33319583449097 \cdot 10^{-10} & 1.66663888934722 \cdot 10^{-11} & -0.00249995875068062\\3.33327833424083 \cdot 10^{-11} & -1.11109277808028 \cdot 10^{-17} & 1.64997277544921 \cdot 10^{-9}\end{matrix}\right] $$
At $x=1000$
$$ \left[\begin{matrix}-1.66666638916671 \cdot 10^{-13} & 5.55555463055571 \cdot 10^{-23} & 0.000499999916750014\\8.33333194583355 \cdot 10^{-14} & 1.66666638888894 \cdot 10^{-16} & -0.000249999958375007\\3.33333277833343 \cdot 10^{-16} & -1.11111092611114 \cdot 10^{-25} & 1.66499972277755 \cdot 10^{-13}\end{matrix}\right] $$
At $x=100000$
$$ \left[\begin{matrix}-1.66666666663889 \cdot 10^{-21} & 5.55555555546296 \cdot 10^{-37} & 4.99999999991667 \cdot 10^{-6}\\8.33333333319445 \cdot 10^{-22} & 1.66666666663889 \cdot 10^{-26} & -2.49999999995833 \cdot 10^{-6}\\3.33333333327778 \cdot 10^{-26} & -1.11111111109259 \cdot 10^{-41} & 1.66664999997222 \cdot 10^{-21}\end{matrix}\right] $$
That all it takes to compute the matrix inverse of the scale separated order was to compute the the result at $(\frac{1}{\epsilon})^1$!!
Further, consider
x=10;
M = Matrix(( [1*(x)**1, 1*(x)**2 ], [1*(x)**5, 1*(x)**5]))
M.inv().evalf()
$$\left[\begin{matrix}-0.0111111111111111 & 1.11111111111111 \cdot 10^{-5}\\0.0111111111111111 & -1.11111111111111 \cdot 10^{-6}\end{matrix}\right] $$
at $x=100$
$$\left[\begin{matrix}-0.000101010101010101 & 1.01010101010101 \cdot 10^{-10}\\0.000101010101010101 & -1.01010101010101 \cdot 10^{-12}\end{matrix}\right] $$
if this seemed a bit disappointing, $x=1000$
$$\left[\begin{matrix}-1.001001001001 \cdot 10^{-6} & 1.001001001001 \cdot 10^{-15}\\1.001001001001 \cdot 10^{-6} & -1.001001001001 \cdot 10^{-18}\end{matrix}\right] $$
and at, $x=10000$
$$\left[\begin{matrix}-1.000100010001 \cdot 10^{-8} & 1.000100010001 \cdot 10^{-20}\\1.000100010001 \cdot 10^{-8} & -1.000100010001 \cdot 10^{-24}\end{matrix}\right] $$
A pattern started to emerge!
Another example
x=10;
M = Matrix(( [1*(x)**1, 1*(x)**2 ], [2*(x)**5, 2*(x)**5]))
M.inv().evalf()
$$\left[\begin{matrix}-0.0111111111111111 & 5.55555555555556 \cdot 10^{-6}\\0.0111111111111111 & -5.55555555555556 \cdot 10^{-7}\end{matrix}\right]$$
At $x=100$
$$\left[\begin{matrix}-0.000101010101010101 & 5.05050505050505 \cdot 10^{-11}\\0.000101010101010101 & -5.05050505050505 \cdot 10^{-13}\end{matrix}\right]$$
At $x=1000$
$$\left[\begin{matrix}-1.001001001001 \cdot 10^{-6} & 5.00500500500501 \cdot 10^{-16}\\1.001001001001 \cdot 10^{-6} & -5.005005005005 \cdot 10^{-19}\end{matrix}\right]$$
That all the non invertible matrix were special types of zeros marked by the "non uniform convergence" of scale separated order. Further, as observed, the matrix entry $a_{ij}$ converged to another set of non invertable matrix when $(\frac{1}{\epsilon})\rightarrow \infty $ a closed topological system of zeros!
How to compute the matrix inverse of the scale separated orders, such as $$ \begin{pmatrix} a_{11} 10^{10}& a_{12} 10^{30} \newline a_{21} 10^{20}& a_{22} 10^{40} \end{pmatrix} $$, how to keep track of the scale separated order parameter $(\frac{1}{\epsilon})^N$, is there a number algebra which include the non invertible matrix as the structure of zeros?