Formal expression for $A^{-1}$, given $A=1+B$ I have a 2x2 real matrix $A$, with $\det A\not=0$ and I define it as
 $$A=1+B$$ where 1 is the identity and B another regular and real matrix.
I need to express the inverse $A^{-1}$ in terms of B: I have used $A^{-1}A=1$ in order to derive a recurrence relation that yelds
$$A^{-1}=1-B+B^{2}-B^{3}+B^{4}-...$$ 
which formally is a geometric series in B.
My questions are:  


*

*"When this series is properly defined?"

*"Does exist a constraint on $B$ in order to deal with a convergent series?"

*"If for some $B$ the series diverge, what is the expression for $A^{-1}$ in terms of $B$?"

 A: You have conditions on the spectral radius $\rho$ :


*

*if $\rho(B) < 1$ then your serie converge

*if $\rho(B) > 1$ then your serie diverge


And if $\rho(B) = 1$ it can converge or diverge 
The convergence is proved by knowing that $\rho(B) = \inf_N N(B)$ over all the operator norms
The divergence is simple : just take an engeinvector with eigenvalue $\lambda > \rho(B) > 1$ and apply this to the partial sum
A: Let
$$B=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$$
Then
$$A^{-1}\ =\ \begin{pmatrix}1+a & b \\ c & 1+d\end{pmatrix}^{-1}$$
$$=\ \frac1{(1+a)(1+d)-bc} \begin{pmatrix}1+d & -b \\ -c & 1+a\end{pmatrix}$$
$$=\ \frac1{1+(a+d)+(ad-bc)}\left[\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\,+\,\begin{pmatrix}d & -b \\ -c & a\end{pmatrix}\right]$$
$$=\ \frac1{1+\mathrm{trace}(B)+\det(B)} \left[I+\det(B)\cdot B^{-1}\right]$$
A: If you introduce a norm $\| \cdot \|$ on the space of matrices such that the addition and multiplication of matrices become continuous in that norm, then a sufficient condition for that series to be convergent is $\| B \| < 1$.
One such norm is the operator norm: if $A : \Bbb R^n \to \Bbb R^n$ is a matrix, then one defines
$$\| A \| = \sup _{\| v \| = 1} \| Av \|$$
which is perfect for your needs.
