1
$\begingroup$

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

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

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

$\endgroup$

closed as off-topic by Travis, Claude Leibovici, Shuchang, Swapnil Tripathi, Grigory M Jan 4 '15 at 8:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis, Claude Leibovici, Shuchang, Swapnil Tripathi, Grigory M
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 7
    $\begingroup$ 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
  • 3
    $\begingroup$ 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
7
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ That is a nice analogy! $\endgroup$ – Winther Jan 4 '15 at 7:18
  • 1
    $\begingroup$ Based on a true story, too! :-) $\endgroup$ – Asaf Karagila Jan 4 '15 at 7:50
3
$\begingroup$

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.

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

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ One also needs to know that inverses are unique, of course. $\endgroup$ – Casteels Jan 4 '15 at 7:13
  • 1
    $\begingroup$ @Casteels Yes. For the interested reader, that is answered (in somewhat more generality) here. $\endgroup$ – Winther Jan 4 '15 at 7:17

Not the answer you're looking for? Browse other questions tagged or ask your own question.