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?

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 $A \cdot A^{-1} = A^{-1}\cdot A = I$); and $B^{-1}$ is the inverse of $B$ (i.e $B\cdot B^{-1} = B^{-1} \cdot B = I$).

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