<s>[INST] 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?
 [/INST] $(A^{-1}B^{-1})$ is not always equal to $(AB)^{-1}$.
Analogous to matrix transpose $(AB)^T = B^TA^T$, we have $(AB)^{-1} = B^{-1}A^{-1}$.
Further, matrix multiplication is not commutative. Here is a proof to show this, but we can see this fact from a simple counterexample involving two square matrices $A$ and $B$.
$A = \begin{pmatrix}1 & 2\\\ -1 & 0\end{pmatrix}$ $B = \begin{pmatrix}1 & -1\\\ 0 & 1\end{pmatrix}$
$AB = \begin{pmatrix}1 & 1\\\ -1 & 1\end{pmatrix}$, $BA = \begin{pmatrix}2 & 2\\\ -1 & 0\end{pmatrix}$
$AB \neq BA$
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