= Row reduction with a parameterized right-hand side
{title2=$\tau=5$}
Work over the <rational numbers>, allowing unknowns in the <real numbers> when interpreting the geometry. Consider four equations in three unknowns:
$$
A=\begin{pmatrix}1&2&-1\\2&5&1\\3&7&0\\1&3&2\end{pmatrix},\qquad
b(\tau)=\begin{pmatrix}1\\4\\\tau\\3\end{pmatrix},\qquad Ax=b(\tau).
$$
The third row of $A$ is the sum of its first two rows, and the fourth is their difference, second minus first. The same relations must hold on the right: consistency requires $\tau=1+4=5$ and $3=4-1$. The latter condition already holds. <Gaussian elimination> makes this compatibility test explicit while exposing the remaining freedom.
Apply the <elementary row operations> in this order, using each row's current value:
$$
R_2\leftarrow R_2-2R_1,\quad R_3\leftarrow R_3-3R_1,\quad R_4\leftarrow R_4-R_1,
$$
then $R_3\leftarrow R_3-R_2$ and $R_4\leftarrow R_4-R_2$. The resulting <augmented matrix> is in <row echelon form> for every $\tau$:
$$
\left[\begin{array}{rrr|r}1&2&-1&1\\0&1&3&2\\0&0&0&\tau-5\\0&0&0&0\end{array}\right].
$$
When $\tau\ne5$, the third row has its <pivot element> in the augmented column. It says $0=\tau-5$, so the <linear system> has no solution. When $\tau=5$, that entire row vanishes and only two <pivot elements> remain. More equations than unknowns therefore do not force a unique solution: some equations may repeat information.
The final operation $R_1\leftarrow R_1-2R_2$ reduces the coefficient block to
$$
R=\begin{pmatrix}1&0&-7\\0&1&3\\0&0&0\\0&0&0\end{pmatrix},\qquad
Eb(\tau)=\begin{pmatrix}-3\\2\\\tau-5\\0\end{pmatrix}.
$$
The product of all six <elementary matrices> is
$$
E=\begin{pmatrix}5&-2&0&0\\-2&1&0&0\\-1&-1&1&0\\1&-1&0&1\end{pmatrix},\qquad EA=R.
$$
Direct <matrix multiplication> verifies both displayed transformations. Its <determinant> is one, so this transformation is reversible. In particular, the third row of $E$ extracts the discrepancy between the third equation and the sum of the first two.
\Image[/row-echelon-parameter-pivots.png]
{title=<Row echelon form> exposes compatibility; a right-hand-side <pivot element> distinguishes inconsistency from a <free variable in a linear system>}
For $\tau=5$, $[R\mid Eb]$ is the <reduced row echelon form> of the full <augmented matrix>. For $\tau\ne5$, only its coefficient block is reduced: scale the third row by $1/(\tau-5)$ and clear the entries above its augmented <pivot element>. The full <reduced row echelon form> is then
$$
\left[\begin{array}{rrr|r}1&0&-7&0\\0&1&3&0\\0&0&0&1\\0&0&0&0\end{array}\right].
$$
The last column is part of the <matrix> when testing the echelon conditions. Calling $[R\mid Eb]$ fully reduced before clearing that column would miss this distinction.
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