Whenever there is a transpose or inverse of a group of matrices, I just reverse their order. For eg: $(ABC)^{-1} = C^{-1}B^{-1}A^{-1}$ and $(ABC)^{T} = C^{T}B^{T}A^{T}$

But usually, I am taking this reverse "rule" for granted without really knowing why I have to reverse their order whenever there is an inverse or transpose.

What is the reason for reversing their order?

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    $\begingroup$ Well, if you don't reverse the factors the equalities are simply not true! $\endgroup$ – Mariano Suárez-Álvarez Jul 4 '11 at 23:28
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    $\begingroup$ In the morning, you put on your socks then your shoes. But when you come home, you have to take off your shoes before your socks! $\endgroup$ – Corey Jul 4 '11 at 23:34

For the inverse, you can just apply the definition and compute:


and similarly $(b^{-1}a^{-1})(ab)=id$. Hence $b^{-1}a^{-1}$ is an (and by uniqueness of inverses, the) inverse of $ab$.

A popular way to 'explain' this, is by saying "the inverse of 'putting on socks and then shoes', is 'taking off shoes and then socks'".

For the transpose, you can also apply the definition: if $A$ has matrix elements $(a_{ij})_{ij}$, then $A^T$ has matrix elements $(a_{ji})_{ij}$. Then compute the matrix elements of $(AB)^T$ and of $B^TA^T$ and see that they are equal.

A more conceptual way however is the following: taking the dual is a contravariant functor. That is, a linear map $A:V\to W$ has dual map $A^*:W^*\to V^*$ between the dual spaces (in the other direction!), and this is compatible with composition: $(A\circ B)^*=B^*\circ A^*$. The definition is $A^*=-\circ A$, so just precompose the linear functional with $A$; the composition property is then immediate. Now if $A$ has matrix $a_{ij}$ with repsect to some bases of $V,W$, then the matrix of $A^*$ w.r.t. the corresponding dual bases of $V^*,W^*$ is $a_{ji}$ (this is an easy check by writing out the definitions).

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    $\begingroup$ For transposes, consider rectangular matrices: $AB$ may be defined but not $BA$, if they are not square. Hence, $(AB)^T$ cannot be $A^T B^T$ because they do not "match", while if $AB$ is defined so is $B^T A^T$. Of course, this is not a proof that $(AB)^T=B^T A^T$ but it does help to remember the right equation. $\endgroup$ – lhf Jul 5 '11 at 1:06
  • $\begingroup$ Nice interpretation, it is fresh to me. $\endgroup$ – Sunni Jul 5 '11 at 1:19

For inverses, there is a "common sense" interpretation. To undo a multistep procedure, you undo each step in the reverse order in which you did it. Example

Procedure: Dress your feet
1.  Put on socks.
2.  Put on shoes.
3.  Tie shoes.

To undo:
1.  Untie shoes
2.  Remove shoes
3.  Remove socks.

This principle is applicable to composition of functions in general. My students find the explanation plausible and palatable.

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  • $\begingroup$ But it's not wrong per se if they aren't written in reverse, right? It's just that you will always wind up moving the matrices around anyway to proceed with multiplication? $\endgroup$ – compguy24 Jun 8 '15 at 15:43

(AB)^-1=B^-1 A^-1 multiply AB on both sides and dividing by AB the value will not change so.. AB (AB)^-1= B^-1 A^-1 (AB) (AB)^-1= B^-1 A^-1 (AB)/AB (AB)^-1= B^-1 A^-1

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    $\begingroup$ How do you divide by a matrix? Does $A/B$ mean $AB^{-1}$ or $B^{-1}A$ or something else? $\endgroup$ – robjohn May 22 '14 at 15:40
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    $\begingroup$ Please use proper formatting. $\endgroup$ – user88595 May 22 '14 at 15:47

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