I would like to show that $(\mathbf{A} \mathbf{B})^{-1} = \mathbf{B}^{-1} \mathbf{A}^{-1}$, where $\mathbf{A}$ and $\mathbf{B}$ are $N \times N$ square matrices.
I think that this can be done as follows:
First, note that $(\mathbf{A}\mathbf{B})^{-1} (\mathbf{A} \mathbf{B}) = \mathbf{1}$ and also that $\mathbf{B}^{-1} \mathbf{A}^{-1} \mathbf{A} \mathbf{B} = \mathbf{B}^{-1} \mathbf{1} \mathbf{B} = \mathbf{B}^{-1} \mathbf{B} = \mathbf{1}$ (where $\mathbf{1}$ is the unit matrix). Thus
$(\mathbf{A} \mathbf{B})^{-1} \mathbf{A} \mathbf{B} = \mathbf{B}^{-1} \mathbf{A}^{-1} \mathbf{A} \mathbf{B}$
which implies that
$(\mathbf{A} \mathbf{B})^{-1} = \mathbf{B}^{-1} \mathbf{A}^{-1}$.
I am not sure that this is correct. It seems almost too easy! The correctness of my working is based on my assumption that $\mathbf{C} (\mathbf{A}\mathbf{B}) = (\mathbf{C} \mathbf{A}) \mathbf{B}$. Is this correct?
In normal linear algebra, this is the case. But is it the case in matrix multiplication? I am not sure whether the brackets result in a different order of multiplication and thus a different result.
Best Answer
Your reasoning is correct. However, it doesn't really make sense to ask whether $\mathbf C(\mathbf A\mathbf B)=\mathbf C\mathbf A\mathbf B$, since the right-hand side isn't defined unless either you already know that multiplication is associative or you've specified an order of evaluation (which is usual in programming languages but not in mathematics). The appropriate question is whether $\mathbf C(\mathbf A\mathbf B)=(\mathbf C\mathbf A)\mathbf B$, which is what you have in fact used in your derivation. This is called associativity; it holds for matrices; and it allows you to write $\mathbf C\mathbf A\mathbf B$ for either of the products.