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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).

Thank you in advance for your input! If there's any need for clarification, please say so, I'll be glad to answer.

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    $\begingroup$ So $A(\gamma)$ is not invertible for every $\gamma$, e.g. $\gamma=0$, correct? Maybe you could write $A(\gamma)$ explicitly... $\endgroup$ – Miguel Jun 19 '17 at 10:50
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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 $\lambda x_0$.

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