Therefore, if $A+B$ is invertible/non-invertible then $I+BA^{-1}$ is invertible/non-invertible
Prove If $A$ is invertible then $A+B$ and $I+BA^{-1}$ are both invertible or both non-invertible
If $A+B$ is invertible, i.e. $A+B=C$ where $CC^{-1} = I$ then $(A+B)A^ {-1} = I +BA^{-1}$ but I get stuck here, I don't know how I show this leads to an invertible matrix, say D
Edit: $A+B=C \Rightarrow (A+B)A^{-1} = CA^{-1} = I + BA^{-1} = CA^{-1} \Rightarrow I + BA^{-1} - I = CA^{-1} - I \Rightarrow BA^{-1} = CA^{-1} - I$ $\Rightarrow BA^{-1}A = (CA^{-1} - I)A \Rightarrow B = C-A \Rightarrow B+A = C-A + A \Rightarrow B+A = C$
and if $A+B$ is not invertible then would the above argument still be used? I'm never counting on B or C to be invertible or non-invertible in that argument
Use the determinant condition: for any square matrix A, A is invertible iff det(A) is not equal to 0:
$$A+B=(I+BA^{-1})A$$
So
$$\det(A+B)=\det((I+BA^{-1})A)=\det(I+BA^{-1})\det(A)$$
Since $\det(A)\neq 0$,
$$\det(A+B)\neq 0\text{ iff } \det(I+BA^{-1})\neq 0$$