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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?

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7 Answers 7

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

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    $\begingroup$ 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 $\endgroup$
    – 7raiden7
    Jan 25, 2017 at 8:27
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    $\begingroup$ Tomáš Zato: matrix multiplication is function composition. $(f \circ g)^{-1} = g^{-1} \circ f^{-1}$ $\endgroup$ Jun 14, 2018 at 22:21
  • $\begingroup$ 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 $$(AB)^{-1} = B^{-1}A^{-1}.$$ $\endgroup$ Jan 5 at 10:35
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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 $A^{-1}B^{-1}$.

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    $\begingroup$ This is better than the accepted answer because it actually explains the WHY. $\endgroup$ Feb 11, 2015 at 13:44
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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: $$ \begin{align*} (AB)(AB)^{-1}&=I\\ (A^{-1}AB)(AB)^{-1}&=A^{-1}I\\ (IB)(AB)^{-1}&=A^{-1}\\ B(AB)^{-1}&=A^{-1}\\ B^{-1}B(AB)^{-1}&=B^{-1}A^{-1}\\ I(AB)^{-1}&=B^{-1}A^{-1}\\ (AB)^{-1}&=B^{-1}A^{-1} \end{align*} $$

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

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

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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.

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$(A^{-1}B^{-1})$ 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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  • $\begingroup$ Welcome to MSE. Your answer adds nothing new to the already existing answers. $\endgroup$ Apr 24 at 21:37
  • $\begingroup$ @JoséCarlosSantos Please direct me to the answer that contains a proof (or a link to one) which is directly related to the mistake the original poster made in the question. $\endgroup$ Apr 25 at 0:02
  • $\begingroup$ The mistake lies in assuming that $(AB)^{-1}=A^{-1}B^{-1}$. Every answer explains that this is false in general and that what we actually have is $(AB)^{-1}=B^{-1}A^{-1}$. It follows from this the matrix multiplication is not commutative. $\endgroup$ Apr 25 at 5:59

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