= Solution
The pivot columns are columns 1, 2 and 3 of the original matrix. A <basis> for its <column space> is therefore
$$
\boxed{\left\{\begin{pmatrix}2\\1\\1\\0\\1\end{pmatrix},\begin{pmatrix}2\\1\\2\\0\\0\end{pmatrix},\begin{pmatrix}1\\0\\1\\1\\0\end{pmatrix}\right\}.}
$$
Use the original columns, rather than the columns of the row-reduced matrix: <elementary row operations> preserve column dependence relations, but can change the column space as a subspace of $\mathbb R^5$.
Back to article page
