# 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: $$\left(\prod_{k=0}^NA_k\right)^{-1}=\prod_{k=0}^NA^{-1}_{N-k}$$

• For the sake of simplicity, let's assume $\prod_{k=0}^{N-1}A_i=A$ and $A_N=B$. You can easily verify that both A and B are invertible. Now you are looking for a matrix $C$ such that $C\cdot (AB) = I$. For the associative property lhs is equal to $(CA)B$. Since B is invertible, there exists a matrix $D$ such that $DB=I$. Since this matrix exists and is unique, it holds that $CA=D=B^{-1}$. Also, A is invertible, and we can multiply both lhs and rhs by $A^{-1}$ on the right. It finally holds: $C=B^{-1}A^{-1}$. By induction you prove the rest. Let me know if you need help with that Jan 25 '17 at 8:27
• Tomáš Zato: matrix multiplication is function composition. $(f \circ g)^{-1} = g^{-1} \circ f^{-1}$ Jun 14 '18 at 22:21
• And this makes so much sense intuitively. First you applied the transformation $B$ and then applied $A$. Now, if you wanna go back, you first have to invert $A$ and then invert $B$. Thus first you have to apply $A^{-1}$ and then $B^{-1}$. Thus $$(AB)^{-1} = B^{-1}A^{-1}.$$ Jan 5 at 10:35

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 $$A^{-1}B^{-1}$$.

• This is better than the accepted answer because it actually explains the WHY. Feb 11 '15 at 13:44

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*}

$(AB)^{-1}$ is not equal to $A^{-1}B^{-1}$, but it is equal to $B^{-1}A^{-1}$.

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.

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 $$B^{-1}(A^{-1}(A(B(x)))) = B^{-1}Bx = x$$