Product of inverse matrices $ (AB)^{-1}$ I am unsure how to go about doing this inverse product problem:
The question says to find the value of each matrix expression where A and B are the invertible 3 x 3 matrices such that 
$$A^{-1} = \left(\begin{array}{ccc}1& 2& 3\\ 2& 0& 1\\ 1& 1& -1\end{array}\right)
$$ and 
$$B^{-1}=\left(\begin{array}{ccc}2 &-1 &3\\ 0& 0 &4\\ 3& -2 & 1\end{array}\right)
$$
The actual question is to find $ (AB)^{-1}$. 
$ (AB)^{-1}$ is just $ A^{-1}B^{-1}$ and we already know matrices $ A^{-1}$ and $ B^{-1}$ so taking the product should give us the matrix 
$$\left(\begin{array}{ccc}11 &-7 &14\\ 7& -4 &7\\ -1& 1 & 6\end{array}\right)
$$
yet the answer is 
$$
\left(\begin{array}{ccc} 3 &7 &2 \\ 4& 4 &-4\\ 0 & 7 & 6 \end{array}\right)
$$
What am I not understanding about the problem or what am I doing wrong? Isn't this just matrix multiplication?

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:
$$
\begin{align*}
(AB)(AB)^{-1}&=I\\
(A^{-1}AB)(AB)^{-1}&=A^{-1}I\\
(IB)(AB)^{-1}&=A^{-1}\\
B(AB)^{-1}&=A^{-1}\\
B^{-1}B(AB)^{-1}&=B^{-1}A^{-1}\\
I(AB)^{-1}&=B^{-1}A^{-1}\\
(AB)^{-1}&=B^{-1}A^{-1}
\end{align*}
$$
