linear algebra - If an invertible matrix $A$ commutes with $C$, show that $A^{-1}$ commutes with $C$

This is taken from a problem in Introduction to Linear Algebra by Strang:

If an invertible matrix $A$ commutes with $C$, show that $A^{-1}$ commutes with $C$

This is what I tried to do, take note of the specific order in which I've multiplied the various matrices, as matrix multiplication is not commutative. (i.e.. left multiplication $\neq$ right multiplication)

Because A commutes with C, this means $AC = CA$

$$\begin{equation} \label{eq1} \begin{split} AC & = CA \\ \implies & IAC & = CAI \\ \implies & AA^{-1}(AC) & = (CA)A^{-1}A \ \ \ (AA^{-1} = I = A^{-1}A)\\ \implies & A(A^{-1}(AC)) &= ((CA)A^{-1})A \\ \implies & A^{-1}((AA^{-1})(AC)) &= A^{-1}((CA)A^{-1})A) \ \ \ \text{Left multiply by } A^{-1} \\ \implies & A^{-1}A^{-1}((AA^{-1})(AC))&= A^{-1}A^{-1}(CA) \ \ \ \text{Left multiply by } A^{-1} \\ \implies & A^{-1}(C)&= A^{-1}A^{-1}(CA) \\ \end{split} \end{equation}$$

But the problem comes up in the last step, as it not yet been proven that $A^{-1}$ commutes with $CA$, and it would not constitute a proof to assume that $A^{-1}$ commutes with $CA$, when we are trying to find out if $A^{-1}$ commutes with $C$ in the first place. If $A^{-1}$ did commute with $CA$ then we could proceed as follows.

$$\begin{equation} \begin{split} \text{Assuming } A^{-1} \ \text{commutes with } CA \\ \implies A^{-1}(C)&= (CA)A^{-1}A^{-1} \\ \implies A^{-1}(C)&= (C)(AA^{-1})A^{-1} \\ \implies A^{-1}(C)&= (C)A^{-1} \\ q.e.d \\ \end{split} \end{equation}$$

For reference this was the official solution, which I'm quite sure makes the assumption that $A^{-1}$ commutes with $CA$.

If $AC = CA$ , multiply left and right by $A^{-1}$ to find $CA^{-1} = A^{-1}C$

The above only works if left multiplication by $A^{-1}$ is equivalent to right multiplication by $A^{-1}$, which only occurs if it is assumed that $A^{-1}$ commutes with $CA$

But I'm not satisfied with the above proof method in assuming $A^{-1}$ commutes with $CA$, nor the official solution, which I'm sure is making the same assumption. Is there any possible way to complete the proof without making this assumption?

Answer

Suppose that $AC=CA$. Multiplying on the right by $A^{-1}$, we get $ACA^{-1}=C$, and then multiplying on the left by $A^{-1}$, we get $CA^{-1}=A^{-1}C$.