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), \(W=\spn{S}\text{.}\) So by Theorem SSS we know that \(W\) is a subspace of \(M_{32}\text{.}\) 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 \(W\text{?}\) To answer this, we want to determine if \(\vect{y}\) can be written as a linear combination of the five matrices in \(S\text{.}\) 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}\text{.}
\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\text{.}
\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}\text{.}
\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 \(\alpha_1=2\text{,}\) \(\alpha_2=-1\text{,}\) \(\alpha_3=0\text{,}\) \(\alpha_4=1\text{.}\) So the scalars \(\alpha_1=2\text{,}\) \(\alpha_2=-1\text{,}\) \(\alpha_3=0\text{,}\) \(\alpha_4=1\text{,}\) \(\alpha_5=0\) will provide a linear combination of the elements of \(S\) that equals \(\vect{y}\text{,}\) 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 \(W=\spn{S}\text{.}\)

Second, is
\begin{equation*}
\vect{x}=\begin{bmatrix}
2 & 1 \\ 3 & 1 \\ 4 & -2
\end{bmatrix}
\end{equation*}
in \(W\text{?}\) To answer this, we want to determine if \(\vect{x}\) can be written as a linear combination of the five matrices in \(S\text{.}\) 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}\text{.}
\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\text{.}
\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}\text{.}
\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 \(W\text{,}\) and so we conclude that \(\vect{x}\not\in W\text{.}\)