Why is $(AB)^{-1}=A^{-1}B^{-1}?$ If we have two matrices $A$ and $B$ then the following property is true.
$$(AB)^{-1}=A^{-1}B^{-1}.$$
I can't understand how the property is true. Can anyone give me a intuitive proof for the property?
 A: If to get from my home to the university I need to take a bus to Jerusalem, and then an internal bus; on the way back I first need to take an internal bus, and then a bus from Jerusalem.
If you think about $AB$ as applying $B$ and then applying $A$, reversing it means first undoing $A$ and then undoing $B$.
(Of course, as noted on this page, you need to know that $A$ and $B$ are invertible to begin with, but the intuition is there.)
A: If $A$ and $B$ are invertible then
$$(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AA^{-1} = I$$
and it follows that (the definition of the inverse of $X$ is $XX^{-1} = I$)
$$(AB)^{-1} = B^{-1}A^{-1}$$
The order of $A$ and $B$ have to be reversed in the equation for the inverse of $AB$ in order for the defining relation $(AB)(AB)^{-1} = I$ to be satisfied.
A: This is false in general. 
A counter example:
Consider $A = \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}, B = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$
Then $AB=\begin{bmatrix}1&2\\0&2\end{bmatrix}$ and $(AB)^{-1}=\begin{bmatrix}1&-1\\0&\frac{1}{2}\end{bmatrix}\neq A^{-1}B^{-1}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}1&0\\0&\frac{1}{2}\end{bmatrix}=\begin{bmatrix}1&-\frac{1}{2}\\0&\frac{1}{2}\end{bmatrix}$
$(AB)^{-1}=B^{-1}A^{-1}$ is the general result.
$B^{-1}A^{-1}=\begin{bmatrix}1&-1\\0&\frac{1}{2}\end{bmatrix}=(AB)^{-1}$
