Commutative property of matrix multiplication (or lack thereof) Assuming $A$ and $B$ are invertible matrices and are of proper dimensions to be multiplied (say, $2\times2$), is the following expression correct for all examples of matrices $A$ and $B$?
$$(A^{-1}B)(AB^{-1}) = A^{-1}BAB^{-1} = A^{-1}AB^{-1}B = I^2 = I$$
My understanding is that for matrices $A$ and $B$, $AB$ doesn't necessarily equal $BA$ as matrix multiplication is not commutative.
I'm trying to simplify the below expression:
$$(AB)^{-1}(AC^{-1})(D^{-1}C^{-1})^{-1}D^{-1}$$
Nothing is given about the matrices $A$, $B$, $C$, or $D$ beyond that they are invertible and of correct dimensions such that any matrix multiplication is possible. My process is as below:
$$\begin{align}(AB)^{-1}(AC^{-1})(D^{-1}C^{-1})^{-1}D^{-1} &= (A^{-1}B^{-1})(AC^{-1})(DC)D^{-1}\\
&= A^{-1}B^{-1}AC^{-1}DCD^{-1}\\
&= B^{-1}A^{-1}AC^{-1}CDD^{-1}\\
&= B^{-1}I^3\\
&= B^{-1}\end{align}$$
You'll notice the error I've made here: $(AB)^{-1} = (B^{-1}A^{-1})$, not $(A^{-1}B^{-1})$, but in the end it doesn't change the answer. The answer in the textbook is indeed $B^{-1}$. According to the above:
$$(B^{-1}A^{-1}) = (A^{-1}B^{-1})$$
But matrices are not commutative? Why does the algebra suggest they are?
 A: In general you won't have any commutative property with matrices, $AB \neq BA$. And you won't be able to simplify $(A^{-1}B)(AB^{-1})$. It is in general the final form of this calculus.
For instance $$A=\left(\begin{matrix}
1&2 \\ 3&4
\end{matrix} \right) \qquad B=\left(\begin{matrix}
5&6 \\ 7&8
\end{matrix} \right)$$ $$AB=\left(\begin{matrix}
19&22 \\ 43&50
\end{matrix} \right) \qquad BA=\left(\begin{matrix}
23&34 \\ 31&46
\end{matrix} \right)$$ $$AB \neq BA$$  $$(A^{-1}B)(AB^{-1}) = \left(\begin{matrix}
-17&10 \\ 22&-13
\end{matrix} \right)$$
To help you remember this non commutative property remind that matrices are a representation of linear functions and that the matrix product corresponds to the functional composition which is intuitively noncommutative.
In your example : $(AB)^{-1}(AC^{-1})(D^{−1}C^{−1})^{−1}D^{−1}=B^{-1}A^{-1}AC^{-1}CDD^{-1}=B^{-1}$
Getting a good answer coming from a wrong calculus does not validate any hypothesis. Your "According to the above" is logically incorrect.
The algebra suggests nothing here.
A: Focusing on just the relevant part of your calculation: you wrote
$$ (AB)^{-1} = A^{-1}B^{-1} = B^{-1}A^{-1} $$
That's how we usually write it, but a more explicit form of the reasoning actually being presented here is:


*

*We know that $(AB)^{-1} = A^{-1}B^{-1}$.

*We know that $A^{-1}B^{-1} = B^{-1}A^{-1}$.

*Therefore $(AB)^{-1} = B^{-1}A^{-1}$.


(3) would indeed follow from (1) and (2), if they were true; but they are not.  This is a valid argument with false premises; as such, it does not prove the conclusion.  (The conclusion happens to be true anyway.)
So, your question "Why does the algebra suggest [that matrices commute]?" is taking a true conclusion to suggest true premises, which doesn't work.
tl;dr: Even when two wrongs make a right, they're still wrong.
A: That expression is not always correct. For example, let
$$A=\begin{bmatrix}
 0  & 1  \\
 -1 & 0
\end{bmatrix},\quad
B=\begin{bmatrix}
 0  & 1  \\
 1 &  0
\end{bmatrix}
$$
We then get
$$(A^{-1}B)(AB^{-1})=\begin{bmatrix}
 -1  & 0  \\
  0 &  -1
\end{bmatrix}
$$
As others have explained, you have no reason to assume that general matrices will commute.
