How does one see whether to use a right- or left-side multiplication in a matrix equation?

Question

I understand that the main question is too generic. (But after having typed out the full question/answer I think this 'question' might be more of a confirmation :). I will try my best to form a series of questions along with some (hopefully) clear examples. I'm working with matrices and can understand calculating with them normally. But I'm having trouble with the equations where the matrices themselves are not given.

Basic left- vs right-sided multiplication examples (where X must be determined) are:

AX = B, where we left-multiply with A^-1 => A^-1AX = A^-1B. So we have X = A^-1B

And

XA = B, where we right-multiply with A^-1 => XAA^-1 = BA^-1. So we have X = BA^-1

These are simple examples, but I fail to understand more difficult ones. I will give an example with its answer(s) and hope somebody can answer my questions about the answer.

Question:

Suppose A,B,X are square matrices. A,X and A-AX are invertible. (A - AX)^-1 = X^-1B. Determine X.

Solution:

(A - AX)^-1 = X^-1B

(A - AX)(A - AX)^-1 = (A - AX) (X^-1B).

Q1: Why does the solution use left-multiplication? Why not right: (A-AX)^-1(A-AX)?

Q2: (has to do with Q1) If I'm not mistaken for all invertible A the following holds, right? AA^-1 = A^-1A = I

Continuation of answer:

I = (A - AX) (X^-1B)

I = AX^-1B - AB

AX^-1B = I + AB

(AX^-1B)^-1= (I + AB)^-1

B^-1XA^-1= (I + AB)^-1. Now B needs to be left-multiplied and A right-multiplied.

Q3: Is this because B is to the left of X and A to the right of X?

X= B(I + AB)^-1A

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Answer 1

Q1: Why does the solution use left-multiplication? Why not right?

Because this way, that $X$ on the right will hit the $X^{-1}$ on the right. If you had used right-multiplication, these would not meet and you have a mess on the right.

Q2: (has to do with Q1) If I'm not mistaken for all invertible $A$ the following holds, right? $AA^{-1} = A^{-}1A = I$

Yes, that is true.

Q3: Is this because $B$ is to the left of $X$ and $A$ to the right of $X$?

Well, it's because $B^{-1}$ is to the left of $X$, and $A^{-1}$ is to the right. The left side of that equation transitions from $B^{-1}XA^{-1}$ to $BB^{-1}XA^{-1}A$, to just $X$.