Matrices satisfying $(A-B)C=BA^{-1}$ where $A$ is Non singular If $A$,$B$,$C$ are Square matrices  satisfying $$(A-B)C=BA^{-1},$$ where $A$ is nonsingular.
Then which is true among these?


*

*$C(A-B)=BA^{-1}$

*$(A-B)C=A^{-1}B$

*$C(A-B)=A^{-1}B$

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


My try:
Since $A$ is invertible,
$$(A-B)CA=B$$ $\implies$
$$ACA=B(CA+I)$$
$$ACA-ACB=BCA-ACB+B$$ $\implies$
$$AC(A-B)=BCA-ACB+B.$$
Now pre multiplying with $A^{-1}$ we get
$$A^{-1}AC(A-B)=A^{-1}BCA-CB+A^{-1}B$$ $\implies$
$$C(A-B)=A^{-1}BCA-CB+A^{-1}B.$$
Any help here?
 A: (1) and (2) are incorrect.  Take $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, $C = \begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix}$.  Verify that the given equation is satisfied.
$$(A-B)C = \begin{pmatrix} ? & -1 \\ ? & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \\
BA^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$
but
$$C(A-B) = \begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} -1 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ -1 & 0 \end{pmatrix} \ne BA^{-1}$$
$$A^{-1}B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ -1 & 0 \end{pmatrix} \ne (A-B)C.$$
(4) is correct.  To show this, consider
$$(A-B)(C+A^{-1}) = (A-B)C + I - BA^{-1} = [(A-B)C - BA^{-1}] + I = I.$$
Use (4) to prove (3).  From (4), we have $C = (A-B)^{-1} - A^{-1}$.
$$C(A-B) = [(A-B)^{-1} - A^{-1}] (A-B) = I - A^{-1} (A-B) = A^{-1}B$$
