Product of inverse matrices $ (AB)^{-1}$

Actually the inverse of matrix product does not work in that way. Suppose that we have two invertible matrices, $A$ and $B$. Then it holds: $$ (AB)^{-1}=B^{-1}A^{-1}, $$ and, in general: $$ \left(\prod_{k=0}^NA_k\right)^{-1}=\prod_{k=0}^NA^{-1}_{N-k} $$

Note that the matrix multiplication is not commutative, i.e, you'll not always have: $AB = BA$.

Now, say the matrix $A$ has the inverse $A^{-1}$ (i.e $A.A^{-1} = A^{-1}.A = I$); and $B^{-1}$ is the inverse of $B$ (i.e $B.B^{-1} = B^{-1}.B = I$).

Claim

$B^{-1}A^{-1}$ is the inverse of $AB$. So basically, what I need to prove is: $(B^{-1}A^{-1})(AB) = (AB)(B^{-1}A^{-1}) = I$.

Note that, although matrix multiplication is not commutative, it is however, associative. So:

• $(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = (B^{-1}I)B = B^{-1}B=I$

• $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = A^{-1}IA = (A^{-1}I)A = A^{-1}A=I$

So, the inverse if $AB$ is indeed $B^{-1}A^{-1}$, and NOT $A^{-1}B^{-1}$.

Not really. Matrices do not follow exponential laws. In fact, $(AB)^{-1}=B^{-1}A^{-1}$. Here is the proof:

Let $I$ be a 3 by 3 identity matrix. If $A$ and $B$ are 3 by 3 invertible matrices, then: $$ \begin{align*} (AB)(AB)^{-1}&=I\\ (A^{-1}AB)(AB)^{-1}&=A^{-1}I\\ (IB)(AB)^{-1}&=A^{-1}\\ B(AB)^{-1}&=A^{-1}\\ B^{-1}B(AB)^{-1}&=B^{-1}A^{-1}\\ I(AB)^{-1}&=B^{-1}A^{-1}\\ (AB)^{-1}&=B^{-1}A^{-1} \end{align*} $$