Inverse of a Product of Matrices: Why is $(AB)^{-1} = B^{-1} \cdot A^{-1}$? I'm having a bit of difficulty conceptualizing a rule for the inverse of a product of matrices and I'd appreciate any input on it.
Suppose I let:
$A^{-1} = \begin{bmatrix}1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B^{-1} = \begin{bmatrix}5 & 6 \\ 7 & 8 \end{bmatrix}$
From what I understand of the rule:
$(AB)^{-1} = B^{-1} \cdot A^{-1}$
However while I understand matrix multiplication isn't necessarily commutative, I'm unclear on why this must be the case.
It intuitively seems that simplifying the exponent would give me:
$(AB)^{-1} = A^{-1} \cdot B^{-1}$
So why isn't this the case?
 A: $$
\Big( AB \Big) \Big( B^{-1} A^{-1}\Big) = A\Big(BB^{-1}\Big)A^{-1}
$$
After you do the obvious cancelation, then another cancelation becomes obvious if it wasn't already.
A: The answer is more obvious when you think of a matrix as representing a linear map.
If  $$ V_1 \stackrel{A}{\longrightarrow} V_2 \stackrel{B}{\longrightarrow} V_3$$ represents the maps $A$ and $B$, their composite is (unfortunately) $BA$, and their inverses go in the "opposite direction":
$$V_1 \stackrel{A^{-1}}{\longleftarrow} V_2 \stackrel{B^{-1}}{\longleftarrow} V_3$$
So the natural map to look at is $A^{-1}B^{-1}$.
A: This has nothing to do with matrices or commutativity.
Suppose you have two invertible functions $f,g$ with appropriate domains and you want to find $x$ when given $y=f(g(x))  = (f \circ g)(x)$. That is, we want to compute $(f \circ g)^{-1} (y)$.
First apply $f^{-1}$ to get $f^{-1}(y) = f^{-1}(f(g(x))) = g(x)$, and then apply $g^{-1}$ to get $g^{-1}(f^{-1}(y)) = g^{-1}(g(x)) = x$.
That is, $x=(f \circ g)^{-1} (y) = g^{-1}(f^{-1}(y)) = (g^{-1} \circ f^{-1})(y)$,
hence
$(f \circ g)^{-1} = g^{-1} \circ f^{-1}$.
A: Check by definition that $(AB)^{-1}=B^{-1}A^{-1}$:
$$(AB)(B^{-1}A^{-1})=ABB^{-1}A^{-1}=A(BB^{-1})A^{-1}=AIA^{-1}=AA^{-1}=I$$
$$(B^{-1}A^{-1})AB=B^{-1}A^{-1}AB=B^{-1}(A^{-1})AB=B^{-1}IB=B^{-1}B=I$$
A: Because if you put on your socks and then your shoes, in that order, and now you want to take them off again, you have to take off your shoes before you take off your socks.
(In any group, $(AB)^{-1} = B^{-1}A^{-1}$.)
A: You know that $A^{-1}$ and $B^{-1}$ both exist. Then, you want to find $X$ ($(AB)^{-1}$) such that: $$ABX = I.$$
So, multiplying by $A^{-1}$ on the left: $$A^{-1}ABX = A^{-1}I \implies BX = A^{-1}.$$ Now we multiply by $B^{-1}$ on the left: $$B^{-1}BX = B^{-1}A^{-1} \implies X = B^{-1}A^{-1}.$$
If you solve $XAB = I$, you'll get the same thing.
A: The definition of the inverse is that
$$
CC^{-1} = C^{-1}C = I.
$$
Try multiplying $AB$ by $B^{-1}A^{-1}$ vs. $A^{-1}B^{-1}$ (on both the right and the left), and see what you get.
A: Like you said, matrix multiplication isn't commutative. So $ABA^{-1}B^{-1} \not = I$, since the matrices don't commute in general. However, $ABB^{-1}A^{-1} = I$, so $(AB)^{-1} = B^{-1}A^{-1}$.
A: 
Lets ${\cal I}$ the inverse of $AB$. Then,

$$
\left(\,AB\,\right){\cal I}={\bf 1}\ \Longrightarrow\ B{\cal I}=A^{-1}{\bf 1}=A^{-1}
\ \Longrightarrow\ \color{#66f}{\large{\cal I}=B^{-1}A^{-1}}
$$

Similarly,

$$
{\cal I}\left(\,AB\,\right)={\bf 1}\ \Longrightarrow\ {\cal I}A={\bf 1}B^{-1}=B^{-1}
\ \Longrightarrow\ \color{#66f}{\large{\cal I}=B^{-1}A^{-1}}
$$
