Solution S.C15 Chris Black
For $\vect{b}$ to be an element of $W=\spn{S}$ there must be a linear combination of the vectors in $S$ that equals $\vect{b}$ (Definition SSCV). The existence of such scalars is equivalent to the linear system $\linearsystem{A}{\vect{b}}$ being consistent, where $A$ is the matrix whose columns are the vectors from $S$ (Theorem SLSLC). \begin{align*} \begin{bmatrix} 3 & 1 & 1 & 2 & 4\\ 2 & 0 & 1 & 1 & 3\\ 3 & 3 & 0 & 3 & 1 \end{bmatrix} &\rref \begin{bmatrix} \leading{1} & 0 & 1/2 & 1/2 & 0\\ 0 & \leading{1} & -1/2 & 1/2 & 0\\ 0 & 0 & 0 & 0 & \leading{1} \end{bmatrix} \end{align*} So by Theorem RCLS the system is inconsistent, which indicates that $\vect{b}$ is not an element of the subspace $W$.