$|ab-cd|=1$ for infinite chessboard

Lemma 1. If in a square matrix we replace the first row with the sum between the first and second row, its determinant stays the same.

Lemma 2. If $\frac{p_n}{q_n}$ and $\frac{p_{n+1}}{q_{n+1}}$ are two consecutive convergents of the continued fraction of $\alpha$, $$\det\begin{pmatrix}p_n & p_{n+1} \\ q_n & q_{n+1}\end{pmatrix}\in\{-1,+1\}$$

From now on, let rows be black diagonals in the original chessboard.

Lemma 3. By Lemma 1,

$$\begin{array}{l} \ldots\verb \76543234567\\ldots\\ \ldots\verb \65432123456\\ldots\\\ldots\verb \11111111111\\ldots \\\ldots \verb \65432123456\\ldots\\ \ldots\verb \76543234567\\ldots\\ \end{array}$$

works. Just to be clear, the previous and next rows are given by: $$ \ldots \verb \13 11 9 7 5 3 5 7 9 11 13 \ \ldots$$

but the repeating pattern $$\begin{array}{l} \ldots\verb \11111111111\\ldots\\ \ldots\verb \21212121212\\ldots\\\ldots\verb \11111111111\\ldots \\\ldots \verb \21212121212\\ldots\\ \ldots\verb \11111111111\\ldots\\ \end{array}$$ works just as fine.

Another interesting chance is given by placing the Fibonacci number $F_{i+j}$ at the position $(i,j)$ - the extension of the Fibonacci sequence to the negative indices is given, as usual, by the relation $F_n=F_{n+2}-F_{n+1}$, so that $F_{-n} = (-1)^{n+1} F_n$. This works by Lemma 2 or Lemma 1.