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 $B$?"
Let $$B=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$$ Then $$A^{-1}\ =\ \begin{pmatrix}1+a & b \\ c & 1+d\end{pmatrix}^{-1}$$ $$=\ \frac1{(1+a)(1+d)-bc} \begin{pmatrix}1+d & -b \\ -c & 1+a\end{pmatrix}$$ $$=\ \frac1{1+(a+d)+(ad-bc)}\left[\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\,+\,\begin{pmatrix}d & -b \\ -c & a\end{pmatrix}\right]$$ $$=\ \frac1{1+\mathrm{trace}(B)+\det(B)} \left[I+\det(B)\cdot B^{-1}\right]$$