Matrix simplification

Starting with:

$(B^{-1} - A)^{-1} = (B^{-1} -B^{-1}B A)^{-1}$ $\implies (B^{-1} - A)^{-1} = \left(B^{-1} (I - BA)\right)^{-1}$

Using the fact that $(XY)^{-1} = Y^{-1}X^{-1}$ on the RHS gives:

$\implies (B^{-1} - A)^{-1} = (I - BA)^{-1}B$

Now:

$A^{-1} + (B^{-1} - A)^{-1} = A^{-1} + (I - BA)^{-1}B$ $\implies A^{-1} + (B^{-1} - A)^{-1} = A^{-1} + (I - BA)^{-1}BAA^{-1}$ $\implies A^{-1} + (B^{-1} - A)^{-1} = (I + (I - BA)^{-1}BA)A^{-1}$

Now, the identity matrix above can be written as: $I = (I - BA)^{-1}(I-BA)$ : $\implies A^{-1} + (B^{-1} - A)^{-1} = ( (I - BA)^{-1}(I-BA) + (I - BA)^{-1}BA)A^{-1}$ $\implies A^{-1} + (B^{-1} - A)^{-1} = (I - BA)^{-1}(I-BA +BA)A^{-1}$ $\implies A^{-1} + (B^{-1} - A)^{-1} = (I - BA)^{-1}A^{-1}$ $\implies (A^{-1} + (B^{-1} - A)^{-1})^{-1} = A(I-BA)$ $A - (A^{-1} + (B^{-1} - A)^{-1})^{-1} = A - A(I-BA)$ $\boxed{A - (A^{-1} + (B^{-1} - A)^{-1})^{-1} = ABA}$

$X = A - {({A}^{-1} + {({B}^{-1} - A)}^{-1})}^{-1}$

$A - X = {({A}^{-1} + {({B}^{-1} - A)}^{-1})}^{-1}$

$({A}^{-1} + {({B}^{-1} - A)}^{-1}) (A - X) = E$

$E - {A}^{-1} X + {({B}^{-1} - A)}^{-1} (A - X) = E$

${A}^{-1} X = {({B}^{-1} - A)}^{-1} (A - X)$

${A}^{-1} X = {({B}^{-1} - A)}^{-1} (A - X)$

$({B}^{-1} - A) {A}^{-1} X = A - X$

${B}^{-1} {A}^{-1} X - X = A - X$

${B}^{-1} {A}^{-1} X = A$

$X = \boxed{ABA}$

Since $A^{-1} + (B^{-1} - A)^{-1} = A^{-1} ( I + A ( B^{-1} - A)^{-1} )$ , then

$A - (A^{-1} + (B^{-1} - A)^{-1})^{-1} = A - (I + A (B^{-1} - A)^{-1})^{-1} A$

Taking $(B^{-1} - A)^{-1}$ as a common factor, the above becomes,

$= A - (B^{-1} - A) ((B^{-1} - A) + A)^{-1} A = A - (B^{-1} - A) B A = A - (I - A B) A = A - A + A BA = ABA$