Consider the set of 5 vectors, $S$, from $\complex{4}$ \begin{equation*} S=\set{ \colvector{1 \\ 1 \\ 3 \\ 1},\, \colvector{2 \\ 1 \\ 2 \\ -1},\, \colvector{7 \\ 3 \\ 5 \\ -5},\, \colvector{1 \\ 1 \\ -1 \\ 2},\, \colvector{-1 \\ 0 \\ 9 \\ 0} } \end{equation*} and consider the infinite set of vectors $\spn{S}$ formed from all possible linear combinations of the elements of $S$. Here are four vectors we definitely know are elements of $\spn{S}$, since we will construct them in accordance with Definition SSCV, \begin{equation*} \vect{w}= (2)\colvector{1 \\ 1 \\ 3 \\ 1}+ (1)\colvector{2 \\ 1 \\ 2 \\ -1}+ (-1)\colvector{7 \\ 3 \\ 5 \\ -5}+ (2)\colvector{1 \\ 1 \\ -1 \\ 2}+ (3)\colvector{-1 \\ 0 \\ 9 \\ 0} = \colvector{-4\\2\\28\\10} \end{equation*} \begin{equation*} \vect{x}= (5)\colvector{1 \\ 1 \\ 3 \\ 1}+ (-6)\colvector{2 \\ 1 \\ 2 \\ -1}+ (-3)\colvector{7 \\ 3 \\ 5 \\ -5}+ (4)\colvector{1 \\ 1 \\ -1 \\ 2}+ (2)\colvector{-1 \\ 0 \\ 9 \\ 0} = \colvector{-26\\-6\\2\\34} \end{equation*} \begin{equation*} \vect{y}= (1)\colvector{1 \\ 1 \\ 3 \\ 1}+ (0)\colvector{2 \\ 1 \\ 2 \\ -1}+ (1)\colvector{7 \\ 3 \\ 5 \\ -5}+ (0)\colvector{1 \\ 1 \\ -1 \\ 2}+ (1)\colvector{-1 \\ 0 \\ 9 \\ 0} = \colvector{7\\4\\17\\-4} \end{equation*} \begin{equation*} \vect{z}= (0)\colvector{1 \\ 1 \\ 3 \\ 1}+ (0)\colvector{2 \\ 1 \\ 2 \\ -1}+ (0)\colvector{7 \\ 3 \\ 5 \\ -5}+ (0)\colvector{1 \\ 1 \\ -1 \\ 2}+ (0)\colvector{-1 \\ 0 \\ 9 \\ 0} = \colvector{0\\0\\0\\0} \end{equation*}
The purpose of a set is to collect objects with some common property, and to exclude objects without that property. So the most fundamental question about a set is if a given object is an element of the set or not. Let us learn more about $\spn{S}$ by investigating which vectors are elements of the set, and which are not.
First, is $\vect{u}=\colvector{-15\\-6\\19\\5}$ an element of $\spn{S}$? We are asking if there are scalars $\alpha_1,\,\alpha_2,\,\alpha_3,\,\alpha_4,\,\alpha_5$ such that \begin{equation*} \alpha_1\colvector{1 \\ 1 \\ 3 \\ 1}+ \alpha_2\colvector{2 \\ 1 \\ 2 \\ -1}+ \alpha_3\colvector{7 \\ 3 \\ 5 \\ -5}+ \alpha_4\colvector{1 \\ 1 \\ -1 \\ 2}+ \alpha_5\colvector{-1 \\ 0 \\ 9 \\ 0} =\vect{u} =\colvector{-15\\-6\\19\\5} \end{equation*}
Applying Theorem SLSLC we recognize the search for these scalars as a solution to a linear system of equations with augmented matrix \begin{equation*} \begin{bmatrix} 1 & 2 & 7 & 1 & -1 & -15 \\ 1 & 1 & 3 & 1 & 0 & -6 \\ 3 & 2 & 5 & -1 & 9 & 19 \\ 1 & -1 & -5 & 2 & 0 & 5 \end{bmatrix} \end{equation*} which row-reduces to \begin{equation*} \begin{bmatrix} \leading{1} & 0 & -1 & 0 & 3 & 10 \\ 0 & \leading{1} & 4 & 0 & -1 & -9 \\ 0 & 0 & 0 & \leading{1} & -2 & -7 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
At this point, we see that the system is consistent (Theorem RCLS), so we know there is a solution for the five scalars $\alpha_1,\,\alpha_2,\,\alpha_3,\,\alpha_4,\,\alpha_5$. This is enough evidence for us to say that $\vect{u}\in\spn{S}$. If we wished further evidence, we could compute an actual solution, say \begin{align*} \alpha_1&=2 & \alpha_2&=1 & \alpha_3&=-2 & \alpha_4&=-3 & \alpha_5&=2 \end{align*}
This particular solution allows us to write \begin{equation*} (2)\colvector{1 \\ 1 \\ 3 \\ 1}+ (1)\colvector{2 \\ 1 \\ 2 \\ -1}+ (-2)\colvector{7 \\ 3 \\ 5 \\ -5}+ (-3)\colvector{1 \\ 1 \\ -1 \\ 2}+ (2)\colvector{-1 \\ 0 \\ 9 \\ 0} =\vect{u} =\colvector{-15\\-6\\19\\5} \end{equation*} making it even more obvious that $\vect{u}\in\spn{S}$.
Let us do it again. Is $\vect{v}=\colvector{3\\1\\2\\-1}$ an element of $\spn{S}$? We are asking if there are scalars $\alpha_1,\,\alpha_2,\,\alpha_3,\,\alpha_4,\,\alpha_5$ such that \begin{equation*} \alpha_1\colvector{1 \\ 1 \\ 3 \\ 1}+ \alpha_2\colvector{2 \\ 1 \\ 2 \\ -1}+ \alpha_3\colvector{7 \\ 3 \\ 5 \\ -5}+ \alpha_4\colvector{1 \\ 1 \\ -1 \\ 2}+ \alpha_5\colvector{-1 \\ 0 \\ 9 \\ 0} =\vect{v} =\colvector{3\\1\\2\\-1} \end{equation*}
Applying Theorem SLSLC we recognize the search for these scalars as a solution to a linear system of equations with augmented matrix \begin{equation*} \begin{bmatrix} 1 & 2 & 7 & 1 & -1 & 3 \\ 1 & 1 & 3 & 1 & 0 & 1 \\ 3 & 2 & 5 & -1 & 9 & 2 \\ 1 & -1 & -5 & 2 & 0 & -1 \end{bmatrix} \end{equation*} which row-reduces to \begin{equation*} \begin{bmatrix} \leading{1} & 0 & -1 & 0 & 3 & 0 \\ 0 & \leading{1} & 4 & 0 & -1 & 0 \\ 0 & 0 & 0 & \leading{1} & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 & \leading{1} \end{bmatrix} \end{equation*}
At this point, we see that the system is inconsistent by Theorem RCLS, so we know there is not a solution for the five scalars $\alpha_1,\,\alpha_2,\,\alpha_3,\,\alpha_4,\,\alpha_5$. This is enough evidence for us to say that $\vect{v}\not\in\spn{S}$. End of story.