Linear Algebra - Complete solution for Ax = b Alright, I'm having some trouble understanding the "complete" solution for Ax = b.
For instance, suppose 
$$A = \pmatrix{ 1  & 2 & 2 & 2 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 7 & 10}$$
I can already see that $Ax = \pmatrix{1 \\ 5 \\ 6}$ is a solution for this system but after elimination we get 
$$A = \pmatrix{ 1  & 2 & 2 & 2 &b_1\\ 0 & 0 & 2 & 4 & b_2 - 2b_1 \\ 0 & 0 & 0 & 0 & b_3 - b_2 - b_1 0}$$
From there my textbook shows how to find a particular solution by setting all free variables to 0, yielding:
$$ x_1+ 2x_3 = 1$$ 
$$2x_3 = 3 $$
So that 
$$x-particular = \pmatrix{-2 \\ 0 \\ \frac{3}{2} \\ 0}$$
It then claims that the complete solution to Ax = b is given by 
$$x-complete + x_n$$
where $x_n$ is a "generic vector in the nullspace",and since $$x_n = c_1 * \pmatrix{-2 \\ 1 \\ 0 \\ 0} + c_2 * \pmatrix{2 \\ 0 \\ -2 \\ 1}$$
$$x-complete  = \pmatrix{-2 \\ 0 \\ \frac{3}{2} \\ 0} +   c_1 * \pmatrix{-2 \\ 1 \\ 0 \\ 0} + c_2 * \pmatrix{2 \\ 0 \\ -2 \\ 1}$$
What I don't understand is: 


*

*what is this "generic vector in the nullspace?

*Do we have to set the free variables to 0 and 1 respectively in order to find the nullspace? If not, how can this solution be considered complete if we're not describing all the possibles values the free variables could've taken?


Since $m>n$ the system $Ax=b$ has infinitely many or zero solutions depending upon the augmented RREF
$$A = \pmatrix{ 1  & 2 & 2 & 2 &b_1\\ 0 & 0 & 2 & 4 & b_2 - 2b_1 \\ 0 & 0 & 0 & 0 & b_3 - b_2 - b_1 }$$
Notably if $b_3-b_2-b_1\neq 0$ we have no solution otherwise the general solution is given by $x_P+x_H$ that is the sum of


*

*one particular soution to $Ax_P=b$

*the homogeneous solution to $Ax_H=0$

