Span

Given vectors ${\mathbf{v}}_1,{\mathbf{v}}_2,...,{\mathbf{v}}_{\mathbf{p}}\in {\mathbb{R}}^n$. The set of all linear combinations of ${\mathbf{v}}_1,{\mathbf{v}}_2,...,{\mathbf{v}}_{\mathbf{p}}$ is called the span of ${\mathbf{v}}_1,{\mathbf{v}}_2,...,{\mathbf{v}}_{\mathbf{p}}$.

Thus, $\text{span}\{{\mathbf{v}}_1,{\mathbf{v}}_2,\dots ,{\mathbf{v}}_{\mathbf{p}}\}$ can be represented like this: $c_1{\mathbf{v}}_1+c_2{\mathbf{v}}_2+\cdots +c_p{\mathbf{v}}_{\mathbf{p}}$ Where $c_1,c_2,...,c_p$ are scalars.

$c_1\left[\begin{array}{c}1\\ -2\\ -3\end{array}\right]+c_2\left[\begin{array}{c}2\\ 5\\ 6\end{array}\right]=\left[\begin{array}{c}7\\ 4\\ 15\end{array}\right]$ $\left[\begin{array}{c}{c}_1\\ -2c_1\\ -3c_1\end{array}\right]+\left[\begin{array}{c}2c_2\\ 5c_2\\ 6c_2\end{array}\right]=\left[\begin{array}{c}7\\ 4\\ 15\end{array}\right]$ $\left[\begin{array}{ccc}1& 2& 7\\ -2& 5& 4\\ -3& 6& 15\end{array}\right]$ $\left[\begin{array}{ccc}1& 2& 7\\ -2& 5& 4\\ -1& 2& 5\end{array}\right]$ $\left[\begin{array}{ccc}1& 2& 7\\ 0& 1& -6\\ -1& 2& 5\end{array}\right]$ $\left[\begin{array}{ccc}1& 2& 7\\ 0& 1& -6\\ 0& 1& 3\end{array}\right]$ $\left[\begin{array}{ccc}1& 0& 19\\ 0& 1& -6\\ 0& 1& 3\end{array}\right]$ $\left[\begin{array}{ccc}1& 0& 19\\ 0& 1& -6\\ 0& 0& 9\end{array}\right]$

Row₃ / 9

$\left[\begin{array}{ccc}1& 0& 19\\ 0& 1& -6\\ 0& 0& 1\end{array}\right]$

So $y$ is not in span $\{{\mathbf{v}}_1,{\mathbf{v}}_2\}$ Look at Row Operations