I'm studying Stang's linear algebra book. One of the questions is:
$$AB = C, \text{ find a formula for } A^{-1}$$
I solved this as:
$$(AB)^{-1} = C^{-1}$$
$$B^{-1}A^{-1} = C^{-1}$$
$$A^{-1} = BC^{-1}$$
This is correct according to answer at the back of the book, however I couldn't find a reference to say taking the inverse of both sides is allowable, or if instead I had to do it with some method of multiplication.
Best Answer
If you say matrices $A$ and $B$ are equal, i.e., $A = B$, then since they are equal, if one is invertible, the other better be invertible too since they are the same matrix, and they better have the same inverse (by a uniqueness argument I detailed below).
So $AB = C$ implies $(AB)^{-1} = C^{-1}$, since $AB$ is the same matrix as $C$ (which is what the equality says).
Note that the fact you can invert both sides doesn't just depend on the fact that both sides represent the same matrix $A$ (which is invertible). It also depends on the fact that if a matrix is invertible, its inverse is unique. An easy proof of this is here:
Let $A$ be an invertible matrix. Suppose $B$ and $C$ are both the inverses of $A$, i.e., $AB = BA = I$, and $AC = CA = I$.
Then $B = BI = B(AC) = (BA)C = IC = C$, so $B$ and $C$ are the same matrix. (I used here that matrix multiplication is associative, i.e., $B(AC) = (BA)C$).