Uniqueness of Inverse

Question

I am having trouble understanding the logic for a few steps in the following. I'll point the steps out at the end.

If B and C are inverses of a square matrix A, then B = C.

Proof: Since B is an inverse of A, we have BA = I and AB = I.

Also, since C is an inverse of A, we have CA = I and AC = I.

So AB = I $\implies$ CAB = CI $\implies$ IB = C $\implies$ B = C.

So my confusions is in the last set of steps. How do we get from:

CAB = CI $\implies$ IB = C

And

IB = C $\implies$ B =C

The above steps are not obvious to me.

Also why is the proof not:

CA = I $\implies$ CA = BA, and by cancelation of matrices we have C = B?

Answer 1

you have $AB=I$, then multiply for the left side of this equation for $C$ you have: $$C(AB)=CI$$ for the property associative of multiplication of matrix we have $C(AB)=(CA)B$,we know that too $CI=C$ and $CA=I$ $$(CA)B=CI<=>IB=C<=>B=C$$

Answer 2

Your simple proof just needs a little more justification. To expand upon your proof, you can say

Since $CA=I=BA$, we have $CAB=BAB$, which simplifies to $C=B$.