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?
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$\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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.
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