If u have a 4x3 matrix that is linearly independent, why can't you solve for Ax = b for EVERY b ?
Define $\displaystyle T:\mathbb{R}^3 \to \mathbb{R}^4$ by $\displaystyle T(\bold{x}) = A\bold{x}$.
It is straightforward to show that $\displaystyle T$ is a linear transformation.
Since the dimension of $\displaystyle \mathbb{R}^3$ over $\displaystyle \mathbb{R}$ is less than the dimension of $\displaystyle \mathbb{R}^4$ over $\displaystyle \mathbb{R}$ is means there is no surjective linear transformation. Thus, there exists $\displaystyle \bold{y}\in \mathbb{R}^4$ such that $\displaystyle A\bold{x} = \bold{y}$ is not solvable.
The point made before is that a n by m matrix, with independent columns, maps $\displaystyle R^n$ to an n dimensional subspace of $\displaystyle R^m$. If n< m, that is NOT all of $\displaystyle R^m$ and there will be vectors b not in that subspace so Ax= b would have no solution.
But if n= m, that is, if A is an n by n matrix with independent columns, then its image is all of $\displaystyle R^n$.