Find a Basis for the Subspace spanned by Five Vectors

We apply the leading 1 method.
Let A$A$ be the matrix whose column vectors are vectors in the set S$S$:

A=⎡⎣⎢⎢⎢122−1131115−15114−12702⎤⎦⎥⎥⎥.$$A=\left[\begin{array}{ccccc}1& 1& 1& 1& 2\\ 2& 3& 5& 1& 7\\ 2& 1& -1& 4& 0\\ -1& 1& 5& -1& 2\end{array}\right].$$

Applying the elementary row operations to A$A$, we obtain

A=⎡⎣⎢⎢⎢122−1131115−15114−12702⎤⎦⎥⎥⎥−→−−−−R4+R1R2−2R1R3−2R1⎡⎣⎢⎢⎢100011−1213−361−12023−44⎤⎦⎥⎥⎥−→−−−R4−2R2R1−R2R3+R2⎡⎣⎢⎢⎢10000100−23002−112−13−1−2⎤⎦⎥⎥⎥−→−−−−R4−2R3R1−2R3R2+R3⎡⎣⎢⎢⎢10000100−2300001012−10⎤⎦⎥⎥⎥=rref(A).$$\begin{array}{r}A=\left[\begin{array}{ccccc}1& 1& 1& 1& 2\\ 2& 3& 5& 1& 7\\ 2& 1& -1& 4& 0\\ -1& 1& 5& -1& 2\end{array}\right]\underset{R_4+R_1}{\overset{\begin{array}{c}{R}_2-2R_1\\ R_3-2R_1\end{array}}{\to }}\left[\begin{array}{ccccc}1& 1& 1& 1& 2\\ 0& 1& 3& -1& 3\\ 0& -1& -3& 2& -4\\ 0& 2& 6& 0& 4\end{array}\right]\\ \underset{R_4-2R_2}{\overset{\begin{array}{c}{R}_1-R_2\\ R_3+R_2\end{array}}{\to }}\left[\begin{array}{ccccc}1& 0& -2& 2& -1\\ 0& 1& 3& -1& 3\\ 0& 0& 0& 1& -1\\ 0& 0& 0& 2& -2\end{array}\right]\underset{R_4-2R_3}{\overset{\begin{array}{c}{R}_1-2R_3\\ R_2+R_3\end{array}}{\to }}\left[\begin{array}{ccccc}1& 0& -2& 0& 1\\ 0& 1& 3& 0& 2\\ 0& 0& 0& 1& -1\\ 0& 0& 0& 0& 0\end{array}\right]=\mathrm{rref}(A).\end{array}$$

Observe that the first, second, and fourth column vectors of rref(A)$\mathrm{rref}(A)$ contain the leading 1 entries.
Hence, the first, second, and fourth column vectors of A$A$ form a basis of Span(S)$\mathrm{Span}(S)$.
Namely,

⎧⎩⎨⎪⎪⎪⎪⎪⎪⎡⎣⎢⎢⎢122−1⎤⎦⎥⎥⎥,⎡⎣⎢⎢⎢1311⎤⎦⎥⎥⎥,⎡⎣⎢⎢⎢114−1⎤⎦⎥⎥⎥⎫⎭⎬⎪⎪⎪⎪⎪⎪$$\left\{\left[\begin{array}{c}1\\ 2\\ 2\\ -1\end{array}\right],\left[\begin{array}{c}1\\ 3\\ 1\\ 1\end{array}\right],\left[\begin{array}{c}1\\ 1\\ 4\\ -1\end{array}\right]\right\}$$

is a basis for Span(S)$\mathrm{Span}(S)$.