The set [latex]{\bf L}[/latex] in [latex]\mathbb{R}^3[/latex] defined by the linear equations

[latex]\begin{align*} x_1 - 13x_2 + 4x_3 &= 2, \\ 3x_2 - x_3 &= 9 \end{align*}[/latex]

is an affine subspace of dimension [latex]1[/latex]. The corresponding linear subspace is defined by the linear equations obtained from the above by setting the constant terms to zero:

[latex]\begin{align*} x_1 - 13x_2 + 4x_3 &= 0, \\ 3x_2 - x_3 &= 0 \end{align*}[/latex]

We can solve for [latex]x_3[/latex] and get [latex]x_1 = x_2, \; x_3 = 3x_2[/latex]. We obtain a representation of the linear subspace as the set of vectors [latex]x \in \mathbb{R}^3[/latex] that have the form

[latex]\begin{align*} x_1 &:= \begin{pmatrix} 1 \\ 1 \\ 3 \end{pmatrix} t, \end{align*}[/latex]

for some scalar [latex]t=x_2[/latex]. Hence the linear subspace is the span of the vector [latex]u:=(1, 1, 3)[/latex], and is of dimension [latex]1[/latex].

We obtain a representation of the original affine set by finding a particular solution [latex]x^0[/latex], by setting say [latex]x_2 = 0[/latex] and solving for [latex]x_1, x_3[/latex]. We obtain

[latex]\begin{align*} x^0 &:= \begin{pmatrix} 38 \\ 0 \\ -9 \end{pmatrix}. \end{align*}[/latex]

The affine subspace [latex]{\bf L}[/latex] is thus the line [latex]x^0 + {\bf span}(u)[/latex], where [latex]x^0, u[/latex] are defined above.