Rank-reducibility of Latin squares Consider the following Latin square with rank $N= 9$:
$$
\begin{bmatrix}
  5 & 3 & 1 & 2 & 4 & 7 & 6 & 8 & 9 \\
  3 & 7 & 9 & 6 & 8 & 4 & 5 & 2 & 1 \\
  8 & 5 & 4 & 9 & 1 & 2 & 7 & 3 & 6 \\
  9 & 2 & 3 & 8 & 7 & 6 & 1 & 4 & 5 \\
  7 & 4 & 5 & 3 & 9 & 1 & 2 & 6 & 8 \\
  6 & 1 & 2 & 5 & 3 & 9 & 8 & 7 & 4 \\
  2 & 9 & 6 & 7 & 5 & 8 & 4 & 1 & 3 \\
  4 & 8 & 7 & 1 & 6 & 5 & 3 & 9 & 2 \\
  1 & 6 & 8 & 4 & 2 & 3 & 9 & 5 & 7 \\
\end{bmatrix}$$
This LS is rank-reducible, as it has the property that
we can obtain a Latin square $L'$ with rank $N'=8$ directly from it.
The value $9$ in $L_{4,1}$ is the key to this reduction. For every row $R$, if $L_{R,C} = 9$ then
$L_{R,1} = L_{4,C}$.
This allows us to remove row 4 and column 1, change each $L_{R,C} = 9$ to $L_{R,1}$ in the remaining
rows, and the result will be a Latin square of rank $N=8$:
$$
\begin{bmatrix}
  3 & 1 & 2 & 4 & 7 & 6 & 8 & 5 \\
  7 & 3 & 6 & 8 & 4 & 5 & 2 & 1 \\
  5 & 4 & 8 & 1 & 2 & 7 & 3 & 6 \\
  4 & 5 & 3 & 7 & 1 & 2 & 6 & 8 \\
  1 & 2 & 5 & 3 & 6 & 8 & 7 & 4 \\
  2 & 6 & 7 & 5 & 8 & 4 & 1 & 3 \\
  8 & 7 & 1 & 6 & 5 & 3 & 4 & 2 \\
  6 & 8 & 4 & 2 & 3 & 1 & 5 & 7 \\
\end{bmatrix}$$
If the pivot value in $V=L_{4,1}$ was any other value than 9, we would just swap the values $V$ and $9$ throughout the square before
proceeding with the reduction.
I wonder, then, are there any other cases which would allow a similarly direct construction of a reduced-rank Latin square?
 A: You're essentially doing transversal prolongation in reverse.
A transversal is a set of entries with one representative from each row, each column, and each symbol.  We can extend a Latin square of order $n$ with a transversal to a Latin square of order $n+1$ by (a) replacing the transversal with a new symbol, (b) adding the corresponding symbols to the new margins, and (c) placing the new symbol in the intersection of the new margins.
Here's a simple example:
$\begin{bmatrix}
3 & 1 & \color{red}{\mathbf 2} \\
\color{red}{\mathbf 1} & 2 & 3 \\
2 & \color{red}{\mathbf 3} & 1 \\
\end{bmatrix}
\longmapsto
\begin{bmatrix}
3 & 1 & \color{blue}{\mathbf \infty} & \color{red}{\mathbf 2} \\
\color{blue}{\mathbf \infty} & 2 & 3 & \color{red}{\mathbf 1} \\
2 & \color{blue}{\mathbf \infty} & 1 & \color{red}{\mathbf 3} \\
\color{red}{\mathbf 1} & \color{red}{\mathbf 3} & \color{red}{\mathbf 2} & \color{blue}{\mathbf \infty} \\
\end{bmatrix}
$
In your example, you've started off with a prolonged Latin square (with its rows and columns permuted), and have discovered the Latin square it was prolonged from.
This construction generalizes to $k$-plexes ($k$ representatives of each row, column, and symbol) where $k \leqslant n/2$.
