If $A B = I$ then $B A = I$ and $rank (A) = rank (A) = n$

We start by proving $rank (B) = n$ if $A B = I$ .
We define $x$ such that $B x = 0$ and then we basically “convert” $x$ into $0$ .
Hence, $x = 0$ is the only solution to $B x = 0$ .
So now thinking in terms of augmented matrix and systems of equations, we can see that if both $b, x$ are $0$ , our augmented matrix has no free parameters ⇒ rank is same as the number of columns (n).
Next, we prove $B A = I$ by defining $y = B x$ . By System-Rank Theorem, we are guaranteed to find a solution.
Then similarly, $rank (A) = n$ .

If $B, C$ are both inverses of $A$ then $B = C$

If A is invertible and we have it on both sides of an equation, we can cancel it out.

A B = A C or B A = C A

This works only for invertible matrices as technically “cancellation” means we are multiplying both sides by something that cancels out the effect. Here we are cancelling out by multiplying both sides by $A^{- 1}$ which cancels out the $A$ on each side. If inverse does not exist this wouldn’t work.