If A is not invertible, then Ax=b has one of two problems. For almost all values of b, the system is overconstrained, and x has no exact solution. For few values of b (including b=0) the system is underconstrained, and x has infinitely many solutions.

#PivotColumns #IndependentColumns(A) Dim(ColumnSpace(A)) Dim(Im(A)) Rank A |
+ | #FreeColumns #DependentColumns(A) Dim(Null(A)) Dim(Ker(A)) Nullity A |
= | #Columns(A) n |