= Reading solutions from reduced row echelon form
{title2=$x=(-3,2,0)^T+t(7,-3,1)^T$}
For <row reduction with a parameterized right-hand side> at $\tau=5$, the nonzero reduced equations are
$$
x_1-7x_3=-3,\qquad x_2+3x_3=2.
$$
Column 3 has no <pivot element>, so $x_3$ is a <free variable in a linear system>. Taking $x_3=t$ gives every solution:
$$
x=x_0+tv,\qquad x_0=\begin{pmatrix}-3\\2\\0\end{pmatrix},\quad v=\begin{pmatrix}7\\-3\\1\end{pmatrix},\quad t\in\mathbb R.
$$
There is no further restriction on $t$. Independent substitution into the original <linear system> gives
$$
Ax_0=\begin{pmatrix}1\\4\\5\\3\end{pmatrix},\qquad Av=\begin{pmatrix}7-6-1\\14-15+1\\21-21\\7-9+2\end{pmatrix}=0.
$$
Conversely, the two reduced equations force exactly these coordinate formulas, so no solutions have been omitted. The same computation with zero right-hand side gives the <kernel> $\ker A=\operatorname{span}\{v\}$. Its <dimension> is one. The <rank-nullity theorem> appears here as $2+1=3$: two <pivot columns> and one freely chosen coordinate account for the three unknowns.<Interactive Linear Algebra by Dan Margalit and others/The Rank Theorem>[Interactive Linear Algebra by Dan Margalit and others; §2.9, Rank Theorem and preceding pivot-column/free-variable count]{ref}
\Image[/row-echelon-affine-solution-line.png]
{title=One <free variable in a linear system> traces an <affine solution space of a linear equation> parallel to its <kernel>}
In general a consistent <linear system> has solution set $x_0+\ker A$. Subtracting $Ax_0=b$ from $Ax=b$ proves this description in both directions. The particular solution places the <affine subspace>, and the <kernel> supplies its directions.<Interactive Linear Algebra by Dan Margalit and others/Solution Sets>[Interactive Linear Algebra by Dan Margalit and others; §2.4.2, Key Observation and its proof: translate homogeneous solutions by any particular solution]{ref} Over the <real numbers> this example is an entire line; over the <rational numbers> the same formula uses rational $t$. A <free variable in a linear system> over a <finite field> instead has only finitely many choices.
The <pivot columns> identify a different object: the original <column space>. Here they select
$$
a_1=(1,2,3,1)^T,\qquad a_2=(2,5,7,3)^T,
$$
which form its <basis>, while $a_3=-7a_1+3a_2$. The corresponding reduced columns are $(1,0,0,0)^T$ and $(0,1,0,0)^T$; their <linear span> is a different <plane>. For example, every original output satisfies $y_3=y_1+y_2$, but $(1,0,0,0)^T$ does not. Since $E$ is an <invertible matrix>, column <linear relations> survive $EA=R$ even though the output <vectors> move. This is why choosing original columns at the reduced pivot positions works.
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