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?
On
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.
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...