Why can this underdetermined system in Ax=b be consistent? The question asks that, given a 5x7 matrix with rank of 4, is $A\vec{x}=\vec{b}$ consistent for all vectors $\vec{b}$ in $R^5$?
My attempt: If the system has rank = 4, this means that the rref has 4 leading 1's. With 5 rows, this means that the last row must be all zeros (because if there are non-zero entries in the 5th row, we could make it into a leading 1). The trouble is, my book says that the system may be inconsistent (which I agree with, given there would be an all-zero 5th row); however, if it is consistent, it will have 3 free parameters ($n-rank = 7-4 = 3$). I don't understand how these free parameters could exist -- if there are remaining digits in columns 5, 6, or 7 on row 5, wouldn't this make the rank of the system 5?
 A: Free parameters refer to the parameters that can take any value. 
To make things easy, assume the matrix $A$ is diagonal; the last row is $0$ because the rank is 4. If the system is consistent, which means that the last value of the vector $b$ is also $0$, we get that the 5th, 6th and 7th component of the vector $x$ must satisfy 
$$0\cdot x_5 + 0\cdot x_6 + 0\cdot x_7 = b_5 = 0$$
Which means that $x_5, x_6, x_7$ can take on any value.
In the general case if very similar: if the last row of the row echeleon form of the matrix $A$ is $0$, and the system is consistent, then it means that there are three parameters that can take any value and it wouldn't change anything.
A: The general condition for a non-homogeneous linear system to be consistent is that the rank of the matrix of the homogeneous linear system and the rank of the augmented matrix be the same.
Suppose you have a linear system of $m$ equations in $n$ unknowns ($m\le n$), represented by a $m\times n$ matrix $A$, and let $b$ be he column-vector of the right-hand-sides of the equations. 
If $A$ has the maximum rank $m$, then $\;\DeclareMathOperator{\rk}{rank}\rk(Ab)=m= \rk A$.
However if $\rk A<m$, there is no particular reason ensuring that $\;\rk (Ab)=\rk A$.
