I have a $4(\text{rows}) \times 5(\text{col})$ matrix. Lets call it $A$. I want to solve $AX = b$ where $b$ is $4 \times 1$ vector. Verified with Matlab that rank(A) is $4$. So all $b$ must be in the range space of the operator $A$ and not just in a 'least-squared' sense.
However if i use X = pinv(A)*b, i get a 'solution' that does not satisfy $AX = b$.
What am i missing?