How can a system $A\vec{x}=\vec{b}$ have more than one particular solution? How can a system $A\vec{x}=\vec{b}$ have more than one particular solution? The matrix $A$ is an arbitrary $m \times n$ matrix. I understand that if there is an $m \times n$ matrix where $n > m$, that there will be a null space solution (1 or more) which will occupy a line, plane, etc. but how can there be more than one particular solution of such a system? 
This problem comes from Introduction to Lin Alg - Strang Ch.3.4 Problem 13b. Thanks.
 A: It actually doesn't matter whether $n>m$. What does matter is that there is a nontrivial null space. This is because if $z$ is in the null space and $Ax=b$ then 
$$A(x+z)=Ax+Az=b+0=b$$
so $x+z$ is another solution to the equation.  Ultimately the problem of whether there is a nontrivial null space depends on the matrix, not on $n$ and $m$. That said, there will always be a nontrivial null space if $n>m$.
A: Maybe it helps if you explore a simpler case. Let $A = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$. This matrix could form the system of equations
$$\begin{align*}
x_1 + x_2 &= c_1 \\
2x_1 + 2x_2 &= c_2
\end{align*}$$
Now, suppose that $c_2 = 2c_1$. From this, both equations tell us that $x_1 + x_2 = c_1$. The best we could do is say
$$x_2 = -x_1 + c_1.$$
As you might be able to tell, this represents a line -- any point along that line satisfies the system of equations. So in fact there is more than one solution, but more specifically there are infinitely many solutions, and all solutions lie along the line $x_2 = -x_1 + c_1$.
A: Ian answered before I do, so instead I'm going to help you visualise this:
In $\mathbb{R}^3$, imagine two planes that intersect. The line in which they intersect in is the solution space, which contains many (in fact, infinitely many) particular solutions. So it is totally possible to have more than one particular solution.
A: If $A$ has maximum rank, that is, we have a left inverse, then we can do $$Ax = b \implies x = A^{-1}b$$
and the system is solved. Take, for concreteness: $$\begin{cases} x+y+z = 1 \\ 2x+2y+2z = 2\end{cases}$$
Does the matrix $$A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \end{pmatrix}$$ have a left inverse? What are the solutions of this problem?
A: that happens only if $Ax = 0$ has nontrivial solutions. reason is you can always add any particular solution a solution of $Ax = 0$
take for example $x + y = 2$ one particular solution is $x = 1, y = 1.$ the homogeneous solution are $x = t, y = -t$ where $t$ is any number. now you can see that $x = 1+t, y = 1-t$ is also a solution of $x + y = 2$.
A: This is what I saw from the official solution of 18.06SC Linear Algebra taught by Gilbert Strang:

The system $Ax = b$ can have more than one particular solution if $x_n\in N(A)$ is in the nullspace of $A$ and $x_p$ is one particular solution, then $x_p + x_n$ is also a particular solution.

I think it's more a definition kind of question. On 18.06 Strang calculates the particular solution by letting all free variables being $0$s, which is the reason why it is "particular".
But from my understanding, the particular solution is really just a solution that gives $Ax = b$, it doesn't have to have all free variables being $0$. Because even with a $x_p$ that have non-zero free variables, $x_{complete}$ is still $x_p + x_n$.
