The probability limit of an inverse matrix all:
I am considering a question regarding the calculation of the probability limit for an inverse matrix.
Specifically, suppose we have a non-singular and squared matrix $M$ with dimension $2\times 2$, and let $a_{ij}$ be the $ij$th element in $M$. 
Assume that for all i and j = 1,2, the probability limit of $a_{ij}$ is $a_{ij}=a_{0,ij}+o(1)$, where the term $o(1)$ indicates that the remaining term goes to zero asymptotically. In this case we can state that $M=M_0+o(1)$, where the $ij$th element in $M_0$ is $a_{0,ij}$.
Now my question is the following. Let $M^{-1}$ be the inverse matrix of $M$. Given that $M=M_0+o(1)$, can we claim that $M^{-1}=M_0^{-1}+o(1)$? If this is true, then the calculation of $M^{-1}$ can be greatly simplified, especially at higher dimension.
I am not sure if this is a well-established fact or it is simply my claim. I would be appreciate if you can share with me your thought on this issue, or let me know any related paper/theory/lemma that you might know.
Thank you in advance. If the question is not clear please also let me know.
 A: I'm going to change your notation a little bit to avoid so many subscripts and to make things clearer. It seems that you're considering a sequence of random matrices
$$
M_n = \begin{bmatrix} a_n & b_n \\ c_n & d_n\end{bmatrix}
$$
where
$$
M_n \to_p M := \begin{bmatrix} a & b \\ c & d\end{bmatrix}
$$
and convergence is element-wise.
Your question seems to boil down to $M_n^{-1} \stackrel ?{\to_p} M^{-1}$.
Because $M_n$ is $2\times 2$ we know that
$$
M_n^{-1} = \frac{1}{\det M_n} \begin{bmatrix} d_n & -b_n \\ -c_n & a_n\end{bmatrix}
$$
and 
$$
M^{-1} = \frac{1}{\det M} \begin{bmatrix} d & -b \\ -c & a\end{bmatrix}.
$$
First off, consider $\det M_n = a_nd_n - b_n c_n$. Since all four of our random variable sequences converge to constants, we can use Slutsky's theorem to show that $\det M_n \to_p ad-bc = \det M$.
We also know (again by Slutsky) that $a_n / g_n \to_p a / g$ for some random sequence $g_n \to g$ provided $g_n, g \neq 0$ so $a_n / \det M_n \to a / \det M$ since we're assuming all $M_n$ are invertible along with $M$, so the determinants are never $0$. An analogous argument shows that this holds for all elements of $M_n^{-1}$.
In summary: Basically I'm using successive applications of Slutky's theorem to show that if $X_{ij} \to_p \mu_i$ then $\frac{\pm X_{ij}}{X_{1j}X_{4j} - X_{2j}X_{3j}} \to_p \frac{\pm \mu_i}{\mu_1\mu_4 - \mu_2\mu_3}$ provided the denominators are never $0$.
Another way to look at this is that the mapping $M_n \mapsto M_n^{-1}$ is continuous, provided the inverse exists, so by the continuous mapping theorem limits are preserved.
