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?
$\endgroup$7 Answers
$\begingroup$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: $$ \left(\prod_{k=0}^NA_k\right)^{-1}=\prod_{k=0}^NA^{-1}_{N-k} $$
$\endgroup$ 3 $\begingroup$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 $A^{-1}B^{-1}$.
$\endgroup$ 1 $\begingroup$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: $$ \begin{align*} (AB)(AB)^{-1}&=I\\ (A^{-1}AB)(AB)^{-1}&=A^{-1}I\\ (IB)(AB)^{-1}&=A^{-1}\\ B(AB)^{-1}&=A^{-1}\\ B^{-1}B(AB)^{-1}&=B^{-1}A^{-1}\\ I(AB)^{-1}&=B^{-1}A^{-1}\\ (AB)^{-1}&=B^{-1}A^{-1} \end{align*} $$
$\endgroup$ $\begingroup$$(AB)^{-1}$ is not equal to $A^{-1}B^{-1}$, but it is equal to $B^{-1}A^{-1}$.
$\endgroup$ 0 $\begingroup$Intuitively, think of matrices as linear operators. To reverse a composition of operators you have to re-run it backwards. So if you want the inverse of the operator $A(B(x))$ on vector $x$, you need to first reverse $A$ and then reverse $B$, so that $A^{-1}(A(B(x))) = B(x)$, so $B^{-1}(A^{-1}(A(B(x)))) = B^{-1}Bx = x$
$\endgroup$ $\begingroup$I have some personal opinions which might perfect it (many students made mistakes about this), in your case it works fine $$(AB)^{-1}=B^{-1}A^{-1}$$ Because you already have the fact that $A,B$ are both square matrix and invertible, but suppose $A$ is $m\times n$ matrix, and $B$ is $n\times m$ matrix, then $AB$ is $m\times m$ matrix which might be invertible, but in this case we don't even have square matrix so we could never have such fomular.
$\endgroup$ $\begingroup$$(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 $B$.
$A = \begin{pmatrix}1 & 2\\\ -1 & 0\end{pmatrix}$ $B = \begin{pmatrix}1 & -1\\\ 0 & 1\end{pmatrix}$
$AB = \begin{pmatrix}1 & 1\\\ -1 & 1\end{pmatrix}$, $BA = \begin{pmatrix}2 & 2\\\ -1 & 0\end{pmatrix}$
$AB \neq BA$
$\endgroup$ 3
