How to prove my statement is correct about invertible matrices? 
*

*SO far i showed that if A matrix is left invertible (L) then in Ax = b, x has at most 1 solution.
I got that LAx = x = Lb, so x = Lb

*for right inverse (R) of A, in Ax = b, x has at least one solution. 
I got that x = Rb. 
In the book it says that x has 1 solution for case 1, and at least one solution for case 2.
Can someone explain my WHY? HOW?
 A: What you did for the left inverse is fine, but what you did for the right inverse doesn’t work: you know that $AR=I$, but you don’t know that $RA=I$, so you can’t say that $x=Rb$. (Actually, you may have seen that $ARb=Ib=b$ and realized on that basis that $Rb$ is a solution; if so, you’re right, but that doesn’t say anything one way or the other about whether it’s the only solution. In any case, you should justify it.)
Suppose that $A$ is an $m\times n$ matrix with a right inverse $R$. Then $R$ is $n\times m$, and $AR$ is $m\times m$. This implies that the $m$ rows of $A$ are linearly independent (why?), i.e., that the rank of $A$ is $m$, and hence that $m\le n$. From here there are several ways to argue that $Ax=b$ has at least one solution, depending on what you’ve already proved. One very straightforward way, however, is to consider what happens to the augmented matrix $\begin{bmatrix}A&b\end{bmatrix}$ when you row-reduce it.
A: Yes you are right. In 1., you SOLVED the equation, and thus you PROVED that $x=Lb$.
In 2., you can see easily that $x=Rb$ works, but this only proves that this is one solution. It could be the only one, or there could be others....
PS I don't know how you got the $x=Rb$, it is easy to guess or, to observe that $b=ARb$. Note that once you see that $Ax=ARb$, you cannot cancel the $A$ unless $A$ has a left inverse... 
