$$\{ (-1, 1, -2, 2) + \alpha(1, 2, -3, 4) + \beta(1, -2, 3, -4) \mid \alpha, \beta \in \mathbb{R}\}$$
The given solution set suggests the system of equations will have 2 pivots and 2 free variables.
From my understanding, the (-1, 1, -2, 2) is going to be the right side of the non-homogenous system (the b in Ax = b). It means the matrix will have to have 4 rows, and I can't figure out how to get the 2 free variables in there if the number of variables = rows?
The given set of vectors is a subset of $\mathbb R^4$, so you know that the coefficient matrix $A$ must have four columns (not rows—these vectors are the $\mathbf x$ in $A\mathbf x=\mathbf b$). Now, the solution set of $A\mathbf x=\mathbf b$ is a coset of $A$’s null space, so we surmise that $(-1,1,-2,2)^T$ is a particular solution and that $\mathcal N(A)$ is spanned by $\{(1,2,-3,4)^T, (1,-2,3,-4)^T\}$. These vectors are obviously linearly independent, therefore $\operatorname{rank}(A)=2$. The row space of a matrix is the orthogonal complement of its null space, so we need two linearly independent vectors that are orthogonal to the two that we were given. Finding them is a straightforward matter of solving the homogeneous system $$(1,2,-3,4)\cdot\mathbf a = 0 \\ (1,-2,3,-4)\cdot\mathbf a = 0.$$ A simple row-reduction yields $(0,3,2,0)^T$ and $(0,2,0,-1)^T$. We take these vectors as the rows of $A$. (It can have more rows than this, but any additional rows must be linear combinations of these two to maintain the correct rank.) To find $\mathbf b$, simply apply $A$ to the particular solution. Putting this all together produces the system $$3x_2+2x_3 = -1 \\ 2x_2-x_4=0.$$ I’ll leave it to you to verify that the given set of vectors indeed satisfies this system.
In the comments you ask an interesting question: why aren’t there any zero entries in the two vectors that generate the null space? Those zeros that you were expecting to see are artifacts of the solution method that you’re used to. The basis of a vector space isn’t unique, so there are many ways to describe the null space. Indeed, if we go through the usual process for solving the system of equations derived above, we end up with the row-reduced augmented matrix $$\left[\begin{array}{rrrr|r}0&1&0&-\frac12&0 \\ 0&0&1&\frac34&-\frac12\end{array}\right]$$ which gives the null space basis $\{(1,0,0,0)^T,(0,1/2,-3/4,1)^T\}$. You can verify that these vectors span the same space as the two given in the problem. The particular solution that one might derive from this reduced matrix is also different from the one given in the problem, too, by the way. This shouldn’t be surprising, either: any particular solution will do to describe the system’s solution set.