Given that $A,B,X$ are square matrices with same dimensions, find all solutions for $X$ of the equation $$(X+B)^{-1}=A+BX^{-1}$$
It is also mentioned that $A,B,X,A+B,B+X,X+A$ are all regular. The first thing I've tried is to group terms which contains $X$ on the LHS and other stuff to the RHS. After multiplying by $X+B$ from the right side I got
$$I=AX+AB+B+BX^{-1}B$$
The problem here is because I have both $X$ and $X^{-1}$ in the linear terms, so I cannot separate it from other terms.
Then I tried to get rid of $X^{-1}$, so I multiplied with $B^{-1}$ from the left
$$B^{-1}-B^{-1}AB-I=B^{-1}AX+X^{-1}B$$
and then by $X$ from the left side
$$X\left(B^{-1}-B^{-1}AB-I\right)=XB^{-1}AX+B$$
How to proceed? Maybe this is a very easy problem which can be solved using some formula similar to quadratic equation, but I am new to matrices, so any help will be appreciated.
Well, yours is not actually an easy problem.
A possible way (probably not the best) to put your equation in a more tractable form could be as follows. From $$ \left( {{\bf X + B}} \right)^{\, - \,{\bf 1}} = {\bf A} + {\bf B}\,{\bf X}^{\, - \,{\bf 1}} $$ the LHS can be rewritten as $$ \begin{array}{l} \left( {{\bf X + B}} \right)^{\, - \,{\bf 1}} = \left( {{\bf X + B}\,{\bf X}^{\, - \,{\bf 1}} {\bf X}} \right)^{\, - \,{\bf 1}} = \left( {\left( {{\bf I + B}\,{\bf X}^{\, - \,{\bf 1}} } \right){\bf X}} \right)^{\, - \,{\bf 1}} = \\ = {\bf X}^{\, - \,{\bf 1}} \left( {{\bf I + B}\,{\bf X}^{\, - \,{\bf 1}} } \right)^{\, - \,{\bf 1}} \\ \end{array} $$ which gives: $$ {\bf X}^{\, - \,{\bf 1}} \left( {{\bf I + B}\,{\bf X}^{\, - \,{\bf 1}} } \right)^{\, - \,{\bf 1}} = {\bf A} + {\bf B}\,{\bf X}^{\, - \,{\bf 1}} = \left( {{\bf A} - {\bf I}} \right) + \left( {{\bf I + B}\,{\bf X}^{\, - \,{\bf 1}} } \right) $$ Making the substitution $$ \left\{ \begin{array}{l} {\bf Y} = {\bf I + B}\,{\bf X}^{\, - \,{\bf 1}} \\ {\bf X}^{\, - \,{\bf 1}} = {\bf B}^{\, - \,{\bf 1}} \left( {{\bf Y} - {\bf I}} \right)\, \\ \end{array} \right. $$ we arrive at: $$ {\bf B}^{\, - \,{\bf 1}} \left( {{\bf Y} - {\bf I}} \right) = \left( {{\bf A} - {\bf I}} \right){\bf Y} + {\bf Y}^{\,{\bf 2}} $$ i.e.