# Matrix Inversion vs. Limit

First: I am a physicist, so please excuse my possibly flawed notation in some points.

I've got a set of physical conditions dependent of some perturbative parameter $\gamma$. This parameter describes a special case

I can write my conditions in the form $A(\gamma)x = b$ with an invertible real $3\times3$-matrix $A$ and real vectors $x,b$. This of course is solved for $x$ by inversion, so $x = A^{-1}b$.

Now I want to consider the limit $\gamma\rightarrow0$ to the unperturbed case. Because one of the conditions goes to $0=0$, $\lim_{\gamma\rightarrow0} a_{3i}=0=\lim_{\gamma\rightarrow0} a_{i3}$. Physically, the thing that makes sense to me is that the limit exists for $x_1, x_2$ but not for one component $x_3$. $x_1$ and $x_2$ should go to the solution one would get for an inversion of the submatrix $$\begin{pmatrix}a_{11}&a_{12}\\ a_{21} & a_{22}\end{pmatrix},$$ so the result one would get from just leaving out the $0=0$ equation. This would result to the perturbed case converging to the unperturbed in two components. However, when I perform the inversion and then the limit, the result is different from doing the limit and then inverting the submatrix.

Is there a way to "deal" with this? Or has there been a mistake on my side? Essentially, this comes down to the question whether there is a way to "commute" limit and matrix inversion for a matrix that becomes singular in the limit (or rather somehow extracting the inverse of the submatrix out of the inverted matrix).

• So $A(\gamma)$ is not invertible for every $\gamma$, e.g. $\gamma=0$, correct? Maybe you could write $A(\gamma)$ explicitly... Jun 19, 2017 at 10:50
If $A(\gamma)$ is not invertible then there is either zero of infinitively many solution to the equation $A(\gamma)x = b$. If it is not invertible, then there exists $x_0$ such that $A(\gamma)x_0 = 0$. In that case, it is easy to check that if $y$ is a solution then $y + \lambda x_0$ is also a solution for all $\lambda \in \mathbb{R}$.
I guess that when you (numerically?) solve you system before or after the limit $\gamma \rightarrow 0$, you find two different solutions because you are in the case where you have an infinite number of solutions. They are the same up to $\lambda x_0$.