1. Represent the vectors in X as rows in a matrix:
\begin{pmatrix} 1 & 2 \\ -1 & 2 \\ 5 & 3 \\ \end{pmatrix}
2. Use Gaussian elimination to row reduce the matrix to row echelon form:
\begin{pmatrix} 1 & 2 \\ -1 & 2 \\ 5 & 3 \\ \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 2 \\ 0 & 4 \\ 0 & -7 \\ \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 2 \\ 0 & 1 \\ 0 & -7 \\ \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{pmatrix}
3. The pivot positions indicate the independent vectors:
(1, 2) and (5, 3) .
4. Verify that (\begin{pmatrix} -1 & 2 \end{pmatrix}) is a linear combination of (\begin{pmatrix} 1 & 2 \end{pmatrix}) and (\begin{pmatrix} 5 & 3 \end{pmatrix}):
We have already row-reduced the matrix and seen that (\begin{pmatrix} -1 & 2 \end{pmatrix}) did not contribute a new pivot position.
Therefore, the basis for S(X) is {(1, 2), (5, 3)}.