How to determine for which parameter values in a linear system, vector set spans $\mathbb{R^3}$ 
For vectors $v_1=(2,1,-1), v_2=(-m,-1,3), v_3=(-3,2,m+1), v_4=(1,2,1)$ determine for which values of $m$ the set {$v_1, v_2, v_3, v_4$} spans $\mathbb{R^3}$.

As far as I understand the problem is equivalent to finding whether one or more solutions exist for the matrix:
$$\begin{bmatrix}
2 &-m&-3&1 \\
1 &-1&2&2\\
-1&3&m+1&1
\end{bmatrix}
$$
that is if vector $\underline b = (b_1, b_2, b_3)$ is a linear combination of {$v_1, v_2, v_3, v_4$}. So am I correct that I need to solve for which $m$ the following matrix has solution/s:
$$\begin{bmatrix}
2 &-m&-3&1&b_1 \\
1 &-1&2&2&b_2\\
-1&3&m+1&1&b_3
\end{bmatrix}
$$
I'm particularly not sure if only one solution exists or infinitely many can exist for the span to exist.
 A: As a side note, the system is underdetermined — three equations with four unknowns — so it can't possibly have a unique solution. It either has no solutions or infinitely many solutions.
So when you say the problem is equivalent to finding whether one or more solutions exist, you're essentially correct. But for this question we don't care how many solutions exist, i.e we don't care to distinguish between "one" or "infinitely many". The real question is: do solutions (any number of them) exist or not?
Another side note is that the wording solutions exist for the matrix is meaningless. Matrices don't have solutions; equations do. Yes, matrices can represent systems of equations, but they can represent lots of other things too. So to talk about solutions, you should be talking about (systems of) equations — which, in particular, can be represented using matrices.
Now, to your actual question. Yes, you question is to determine whether for any vector $\mathbf{b}\in\mathbb{R}^3$ the system of equations represented by your second matrix as the augmented matrix has solutions (doesn't matter how many). But a somewhat easier approach would be not to use such an generic vector $\mathbf{b}$, so that you wouldn't have to do all the work in the last column. Instead, use the following theorem:

A set of $p$ vectors in $\mathbb{R}^n$ spans $\mathbb{R}^n$ if and only if the rank of the matrix with these vectors as columns (that's your first matrix) is $n$.

A: Consider the determinant of the matrix obtained by removing one column, e.g. the third one:
$$\det \left( \begin{matrix}2 & -m & 1 \\ 1 & -1 & 2 \\ -1 & 3 & 1 \end{matrix}  \right)=3m-12$$
Since this is nonzero for $m \neq 4$, the columns of this matrix (and thus also the columns of your original matrix) span $\mathbb{R}^3$.
As it turns out, $m=4$ is always a root of the determinant, no matter which of the columns you eliminate. Hence $v_1,\ldots,v_4$ do not span $\mathbb{R}^3$ for $m=4$. As you can see from the derivation, for $m \neq 4$ the linear combination of $v_1,\ldots,v_4$ for a point $v \in \mathbb{R}^3$ is not unique.
A: You’re on the right track: the given set of vectors spans $\mathbb R^3$ if every vector $\mathbf b\in\mathbb R^3$ can be written as a linear combination of the vectors. As you’ve already stated, this is equivalent to asking if the vector equation $A\mathbf x=\mathbf b$ has a solution for all $\mathbf b$, where $A$ is the matrix with the given vectors as its columns. For this to be true, the column space of $A$ must be all of $\mathbb R^3$, so we proceed to determine the dimension of the column space via row-reduction. Start by swapping the first two rows, then subtract twice the first row from the second and add the first row to the third to produce $$\begin{bmatrix}1&-1&2&2\\0&2-m&-7&-3\\0&2&m+3&3\end{bmatrix}.$$ (I started by swapping rows to avoid propagating $m$’s into other rows.) Next, swap the second and third rows so that we’re not dividing by $2-m$, which might be zero, add $(m-2)/2$ times the second row to the third and then multiply the third row by 2 to tidy things up a bit: $$\begin{bmatrix}1&-1&2&2\\0&2&m+3&3\\0&0&m^2+m-20&3m-12\end{bmatrix}.$$ We have pivots in the first two rows. For the column space to be three-dimensional, we also need a pivot in the third row. To put it another way, the column space will be less than all of $\mathbb R^3$ if the last row is all zeros. This occurs when $m$ is a solution to the system $$\begin{align}m^2+m-20&=0\\3m-12&=0.\end{align}$$ Therefore, the set of vectors spans $\mathbb R^3$ unless $m=4$.
