linear algebramatrices
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?
One of my best college math professor always said:
Make a drawing first.
Although, he couldn't have made this one on the blackboard.
We usually see matrices as linear transformations. The inverse of $A$, when it exists, means simply "reversing" what $A$ does as a function. The transpose originates in a different point of view.
So we have vector spaces $X,Y$, and $A:X\to Y$ is linear. For many reasons, we often look at the linear functionals on the space; that way we get the dual $$ X^*=\{f:X\to\mathbb R:\ f\ \text{ is linear}\}, $$ and correspondingly $Y^*$. Now the map $A$ induces a natural map $A^*:Y^*\to X^*$, by $$ (A^*g)(x)=g(Ax). $$ In the particular case where $X=\mathbb R^n$, $Y=\mathbb R^m$, one can check that $X^*=X$ and $Y^*=Y$, in the sense that all linear functionals $f:\mathbb R^n\to\mathbb R$ are of the form $f(x)=y^Tx$ for some fixed $y\in\mathbb R^n$. In this situation $A$ is an $m\times n$ matrix, and the matrix of $A^*$ is the transpose of $A$.
Best Answer
For the inverse, you can just apply the definition and compute:
$(ab)(b^{-1}a^{-1})=a(bb^{-1})a^{-1}=aa^{-1}=id$
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).