The set of all $3\times 2$ matrices forms a vector space when we use the operations of matrix addition (Definition MA) and scalar matrix multiplication (Definition MSM), as was shown in Example VSM. Consider the subset \begin{equation*} S=\set{ \begin{bmatrix} 3 & 1 \\ 4 & 2 \\ 5 & -5 \end{bmatrix},\, \begin{bmatrix} 1 & 1 \\ 2 &-1 \\ 14 & -1 \end{bmatrix},\, \begin{bmatrix} 3 & -1 \\ -1&2 \\ -19 & -11 \end{bmatrix},\, \begin{bmatrix} 4 & 2 \\ 1 & -2 \\ 14 & -2 \end{bmatrix},\, \begin{bmatrix} 3 & 1 \\ -4 & 0 \\ -17 & 7 \end{bmatrix} } \end{equation*} and define a new subset of vectors $W$ in $M_{32}$ using the span (Definition SS), $W=\spn{S}$. So by Theorem SSS we know that $W$ is a subspace of $M_{32}$. While $W$ is an infinite set, and this is a precise description, it would still be worthwhile to investigate whether or not $W$ contains certain elements.
First, is \begin{equation*} \vect{y}=\begin{bmatrix} 9 & 3 \\ 7 & 3 \\ 10 & -11 \end{bmatrix} \end{equation*} in $W$? To answer this, we want to determine if $\vect{y}$ can be written as a linear combination of the five matrices in $S$. Can we find scalars, $\alpha_1,\,\alpha_2,\,\alpha_3,\,\alpha_4,\,\alpha_5$ so that \begin{align*} &\begin{bmatrix} 9 & 3 \\ 7&3 \\ 10 & -11 \end{bmatrix}\\ &= \alpha_1 \begin{bmatrix} 3 & 1 \\ 4 & 2 \\ 5 & -5 \end{bmatrix} +\alpha_2 \begin{bmatrix} 1 & 1 \\ 2 & -1 \\ 14 & -1 \end{bmatrix} +\alpha_3 \begin{bmatrix} 3 & -1 \\ -1 & 2 \\ -19 & -11 \end{bmatrix} +\alpha_4 \begin{bmatrix} 4 & 2 \\ 1 & -2 \\ 14 & -2 \end{bmatrix} +\alpha_5 \begin{bmatrix} 3 & 1 \\ -4 & 0 \\ -17 & 7 \end{bmatrix}\\ &= \begin{bmatrix} 3\alpha_1 +\alpha_2 +3\alpha_3 +4\alpha_4 +3\alpha_5 & \alpha_1 +\alpha_2 -\alpha_3 +2\alpha_4 +\alpha_5\\ 4\alpha_1 +2\alpha_2 -\alpha_3 +\alpha_4 -4\alpha_5& 2\alpha_1 -\alpha_2 +2\alpha_3 -2\alpha_4 \\ 5\alpha_1 +14\alpha_2 -19\alpha_3 +14\alpha_4 -17\alpha_5& -5\alpha_1 -\alpha_2 -11\alpha_3 -2\alpha_4 +7\alpha_5 \end{bmatrix} \end{align*}
Using our definition of matrix equality (Definition ME) we can translate this statement into six equations in the five unknowns, \begin{align*} 3\alpha_1 +\alpha_2 +3\alpha_3 +4\alpha_4 +3\alpha_5& =9\\ \alpha_1 +\alpha_2 -\alpha_3 +2\alpha_4 +\alpha_5& =3\\ 4\alpha_1 +2\alpha_2 -\alpha_3 +\alpha_4 -4\alpha_5& =7\\ 2\alpha_1 -\alpha_2 +2\alpha_3 -2\alpha_4 & =3\\ 5\alpha_1 +14\alpha_2 -19\alpha_3 +14\alpha_4 -17\alpha_5& =10\\ -5\alpha_1 -\alpha_2 -11\alpha_3 -2\alpha_4 +7\alpha_5&=-11 \end{align*}
This is a linear system of equations, which we can represent with an augmented matrix and row-reduce in search of solutions. The matrix that is row-equivalent to the augmented matrix is \begin{equation*} \begin{bmatrix} \leading{1} & 0 & 0 & 0 & \frac{5}{8} & 2\\ 0 & \leading{1} & 0 & 0 & \frac{-19}{4} & -1\\ 0 & 0 & \leading{1} & 0 & \frac{-7}{8} & 0\\ 0 & 0 & 0 & \leading{1} & \frac{17}{8} & 1\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
So we recognize that the system is consistent since the final column is not a pivot column (Theorem RCLS), and compute $n-r=5-4=1$ free variables (Theorem FVCS). While there are infinitely many solutions, we are only in pursuit of a single solution, so let us choose the free variable $\alpha_5=0$ for simplicity's sake. Then we easily see that $\alpha_1=2$, $\alpha_2=-1$, $\alpha_3=0$, $\alpha_4=1$. So the scalars $\alpha_1=2$, $\alpha_2=-1$, $\alpha_3=0$, $\alpha_4=1$, $\alpha_5=0$ will provide a linear combination of the elements of $S$ that equals $\vect{y}$, as we can verify by checking, \begin{align*} \begin{bmatrix} 9 & 3 \\ 7 & 3 \\ 10 & -11 \end{bmatrix} = 2 \begin{bmatrix} 3 & 1 \\ 4 & 2 \\ 5 & -5 \end{bmatrix} +(-1) \begin{bmatrix} 1 & 1 \\ 2 & -1 \\ 14 & -1 \end{bmatrix} +(1) \begin{bmatrix} 4 & 2 \\ 1 & -2 \\ 14 & -2 \end{bmatrix} \end{align*} So with one particular linear combination in hand, we are convinced that $\vect{y}$ deserves to be a member of $W=\spn{S}$.
Second, is \begin{equation*} \vect{x}=\begin{bmatrix} 2 & 1 \\ 3 & 1 \\ 4 & -2 \end{bmatrix} \end{equation*} in $W$? To answer this, we want to determine if $\vect{x}$ can be written as a linear combination of the five matrices in $S$. Can we find scalars, $\alpha_1,\,\alpha_2,\,\alpha_3,\,\alpha_4,\,\alpha_5$ so that \begin{align*} &\begin{bmatrix} 2 & 1 \\ 3 & 1 \\ 4 & -2 \end{bmatrix}\\ &= \alpha_1 \begin{bmatrix} 3 & 1 \\ 4 & 2 \\ 5 & -5 \end{bmatrix} +\alpha_2 \begin{bmatrix} 1 & 1 \\ 2 & -1 \\ 14 & -1 \end{bmatrix} +\alpha_3 \begin{bmatrix} 3 & -1 \\ -1 & 2 \\ -19 & -11 \end{bmatrix} +\alpha_4 \begin{bmatrix} 4 & 2 \\ 1 & -2 \\ 14 & -2 \end{bmatrix} +\alpha_5 \begin{bmatrix} 3 & 1 \\ -4 & 0 \\ -17 & 7 \end{bmatrix}\\ &= \begin{bmatrix} 3\alpha_1 +\alpha_2 +3\alpha_3 +4\alpha_4 +3\alpha_5 & \alpha_1 +\alpha_2 -\alpha_3 +2\alpha_4 +\alpha_5\\ 4\alpha_1 +2\alpha_2 -\alpha_3 +\alpha_4 -4\alpha_5& 2\alpha_1 -\alpha_2 +2\alpha_3 -2\alpha_4 \\ 5\alpha_1 +14\alpha_2 -19\alpha_3 +14\alpha_4 -17\alpha_5& -5\alpha_1 -\alpha_2 -11\alpha_3 -2\alpha_4 +7\alpha_5 \end{bmatrix} \end{align*} Using our definition of matrix equality (Definition ME) we can translate this statement into six equations in the five unknowns, \begin{align*} 3\alpha_1 +\alpha_2 +3\alpha_3 +4\alpha_4 +3\alpha_5& =2\\ \alpha_1 +\alpha_2 -\alpha_3 +2\alpha_4 +\alpha_5& =1\\ 4\alpha_1 +2\alpha_2 -\alpha_3 +\alpha_4 -4\alpha_5& =3\\ 2\alpha_1 -\alpha_2 +2\alpha_3 -2\alpha_4 & =1\\ 5\alpha_1 +14\alpha_2 -19\alpha_3 +14\alpha_4 -17\alpha_5& =4\\ -5\alpha_1 -\alpha_2 -11\alpha_3 -2\alpha_4 +7\alpha_5&=-2 \end{align*} This is a linear system of equations, which we can represent with an augmented matrix and row-reduce in search of solutions. The matrix that is row-equivalent to the augmented matrix is \begin{equation*} \begin{bmatrix} \leading{1} & 0 & 0 & 0 & \frac{5}{8} & 0\\ 0 & \leading{1} & 0 & 0 & -\frac{38}{8} & 0\\ 0 & 0 & \leading{1} & 0 & -\frac{7}{8} & 0\\ 0 & 0 & 0 & \leading{1} & -\frac{17}{8} & 0\\ 0 & 0 & 0 & 0 & 0 & \leading{1}\\ 0 & 0 & 0 & 0 & 0 & 0\ \end{bmatrix} \end{equation*} Since the last column is a pivot column, Theorem RCLS tells us that the system is inconsistent. Therefore, there are no values for the scalars that will place $\vect{x}$ in $W$, and so we conclude that $\vect{x}\not\in W$.