Distributing a matrix Since matrix mutiplication is not commutative, the two ways in which you can factorize matrices makes a difference in which side the factor goes on. 
In particular, if I want to distribute
$$((I - A) + A)(I - A)^{-1},$$
would it become
$$(I - A)(I - A)^{-1} + A(I - A)^{-1} $$
OR would it be
$$(I - A)^{-1}(I - A) + (I - A)^{-1}A?$$
How do I know which side it goes on? I think the first one is correct.
 A: In general, this is what we call "right distributivity" - I usually hear the context for this in the sense of ring axioms. Let's sojourn into this a bit - though if you're not familiar with abstract algebra, this won't be particularly enlightening, and you might be better off skipping to the very end. 

Let $(R,+,\cdot,0,1)$ be a ring; then we call left-distributivity and define it by
$$a \cdot (b+c) = a\cdot b + a \cdot c$$
Similarly, right-distributivity is given by
$$(b+c)\cdot a = b\cdot a + c\cdot a$$
Note: we are not guaranteed that $a\cdot b = b\cdot a$ unless $R$ is a commutative ring. 
In the context of matrices over rings, for which I reference Wikipedia, you can define $M_n(R)$ as the $n\times n$ matrices over a ring $R$ (i.e. its elements come from the ring, and the addition and multiplication of elements are shared). Notably, we have that $M_n(R)$ is a commutative ring if and only if $R$ is a commutative ring and $n=1$ (so basically effectively no different from working in the ring in question). 

So what does this mean? This means, in your case, you probably do not have $AB=BA$ (of course, I imagine you know this). And thus in the context of the distributivity thigns above, you would have
$$(B+C)A = BA + CA$$
Your example has $B = I-A$ and $C=A$. And thus, your first example is correct: if you are distributing something on the right side, and cannot ensure commutativity, you should multiply that element by everything in the brackets on the right side.
A: Your first answer is correct.  There are two distributive laws for matrices,
$$A(B+C)=AB+AC\quad\hbox{and}\quad (A+B)C=AC+BC\ ,$$
but not $A(B+C)=BA+CA$ or $(A+B)C=AC+CB$ or.....
