Finding the vector form of the general solution Ax = 0 
Suppose that $x_1 = -1$, $x_2 = 2$, $x_3 = 4$, $x_4 = -3$ is a solution of a non-homogeneous linear system $A\mathbf{x} = \mathbf{b}$ and that the solution set of the homogeneous system $A\mathbf{x} =\mathbf{0}$ is given by the formulas: 
  $$\begin{align*}
x_1 &= -3r + 4s,\\
x_2 &= r - s,\\
x_3 &= r,\\
x_4 &= s.
\end{align*}$$
  Find the vector form of the general solutions of Ax = 0 and Ax = b

I ended up with something like: 
( -3 4)
( 1 -1)
( 1 0)
( 0 1) 

where I separated the $r$ and $s$ values, I haven't tried to actually solve though because I'm kinda confused about what I'm suppose to do with this. 
 A: You answer of the solutions to $A\vec{x}=0$ being equal to the span of $\begin{bmatrix}
-3\\ 
1\\ 
1\\ 
0
\end{bmatrix}$ and $
\begin{bmatrix}
4\\ 
-1\\ 
0\\ 
1
\end{bmatrix}$ is correct; now let's find the solutions to $A\vec{x}=\vec{b}$. If the kernel of a matrix is not equal to only the zero vector (as in our case, it actually has a dimension of two), then the matrix is non-invertible, and has infinitely many solutions for $\vec{b}$. By the property of linear transformations, if $A\vec{x}=0$, then $A(\vec{p}+\vec{x})=\vec{b}$ (where $\vec{p}$ is a vector such that $A\vec{p}=\vec{b}$) because $A(\vec{p}+\vec{x})=A\vec{p}+A\vec{x}=A\vec{p}+\vec{0}=\vec{b}+\vec{0}=\vec{b}$. Since you know that $
\begin{bmatrix}
-1\\ 
2\\ 
4\\ 
-3
\end{bmatrix}$ is a solution to $A\vec{x}=\vec{b}$, the set of solutions to this equations will consist of $\vec{p}$+any solution to $A\vec{x}=0$, which you found to be the span of $\begin{bmatrix}
-3\\ 
1\\ 
1\\ 
0
\end{bmatrix}$ and $
\begin{bmatrix}
4\\ 
-1\\ 
0\\ 
1
\end{bmatrix}$. Or, in other words, the set of $\left \{\vec{p}+\vec{x}| A\vec{x}=0 \right \}$ is a set of solutions to $A\vec{x}=\vec{b}$. Or, represented numerically, solutions to $A\vec{x}=\vec{b}$ are of form $
\underset{\text{A given solution to } A\vec{x}=\vec{b}}{\underbrace{\begin{bmatrix}
-1\\ 
2\\ 
4\\ 
-3
\end{bmatrix}}}$ $+$ $\underset{\text{span of Kernel } A} {\underbrace{r\begin{bmatrix}
-3\\ 
1\\ 
1\\ 
0
\end{bmatrix} + s
\begin{bmatrix}
4\\ 
-1\\ 
0\\ 
1
\end{bmatrix}}}$, as Arturo pointed out.
A: The key is to find a single solution to the non-homogeneous, which can be done by Gaussian elimination if such a solution exists. The general solution is then this solution plus any linear combination of solutions to the homogeneous equation.
A: You are close, but you aren't writing it right.
First, let's look at the solution to the homogeneous system. Writing out $r$ and $s$ explicitly in all the equation, we get:
$$\begin{align*}
x_1 & = -3r + 4s\
x_2 & = 1r -1 s\
x_3 & = 1r + 0s\
x_4 &= 0r + 1s.
\end{align*}
To write it as a sum of vectors, the vectors should have four coordinates; they are column vectors, not row vectors:
$$\left(\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right) = \left(\begin{array}{r}-3\\1\\1\\0\end{array}\right)r + \left(\begin{array}
{r}4\\-1\\0\\1\end{array}\right)s.$$
So you see, the vectors go down, not across.
Now, the particular solution you have can be written analogously:
$$\left(\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right) = \left(\begin{array}{r}-1\\2\\4\\-3\end{array}\right).$$
The solutions to the system $A\mathbf{x}=\mathbf{b}$ are then equal to the particular solution plus the general solution to the homogeneous system. So putting it all together, you get:
$$\left(\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right) = \left(\begin{array}{r}-1\\2\\4\\-3\end{array}\right) + \left(\begin{array}{r}-3\\1\\1\\0\end{array}\right)r + \left(\begin{array}{r}4\\-1\\0\\1\end{array}\right)s.$$
