If we have two matrices $A$ and $B$ then the following property is true.


I can't understand how the property is true. Can anyone give me a intuitive proof for the property?


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    $\begingroup$ That formula is not true. Did you mean to write $(AB)^{-1}=B^{-1}A^{-1}$, when $A$ and $B$ are two invertible matrices of the same dimension? $\endgroup$ – Casteels Jan 4 '15 at 6:04
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    $\begingroup$ This is not in general true; rather, if $A, B$ are invertible, then so is $AB$, and $(AB)^{-1} = B^{-1} A^{-1}$. $\endgroup$ – Travis Jan 4 '15 at 6:04

If to get from my home to the university I need to take a bus to Jerusalem, and then an internal bus; on the way back I first need to take an internal bus, and then a bus from Jerusalem.

If you think about $AB$ as applying $B$ and then applying $A$, reversing it means first undoing $A$ and then undoing $B$.

(Of course, as noted on this page, you need to know that $A$ and $B$ are invertible to begin with, but the intuition is there.)

  • $\begingroup$ That is a nice analogy! $\endgroup$ – Winther Jan 4 '15 at 7:18
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    $\begingroup$ Based on a true story, too! :-) $\endgroup$ – Asaf Karagila Jan 4 '15 at 7:50

This is false in general.

A counter example:

Consider $A = \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}, B = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$

Then $AB=\begin{bmatrix}1&2\\0&2\end{bmatrix}$ and $(AB)^{-1}=\begin{bmatrix}1&-1\\0&\frac{1}{2}\end{bmatrix}\neq A^{-1}B^{-1}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}1&0\\0&\frac{1}{2}\end{bmatrix}=\begin{bmatrix}1&-\frac{1}{2}\\0&\frac{1}{2}\end{bmatrix}$

$(AB)^{-1}=B^{-1}A^{-1}$ is the general result.



If $A$ and $B$ are invertible then

$$(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AA^{-1} = I$$

and it follows that (the definition of the inverse of $X$ is $XX^{-1} = I$)

$$(AB)^{-1} = B^{-1}A^{-1}$$

The order of $A$ and $B$ have to be reversed in the equation for the inverse of $AB$ in order for the defining relation $(AB)(AB)^{-1} = I$ to be satisfied.

  • $\begingroup$ One also needs to know that inverses are unique, of course. $\endgroup$ – Casteels Jan 4 '15 at 7:13
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    $\begingroup$ @Casteels Yes. For the interested reader, that is answered (in somewhat more generality) here. $\endgroup$ – Winther Jan 4 '15 at 7:17

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