# Does the determinant of a substitution matrix need to be $\pm 1$?

The substitutional rule for the Fibonacci sequence is $$\sigma: L \rightarrow LS, S \rightarrow L$$, is: $$\sigma : \left ( \begin{array}{c} L \\ S\\ \end{array} \right ) \rightarrow \underbrace{\left ( \begin{array}{cc} 1 & 1 \\ 1 & 0 \\ \end{array} \right )}_{=\mathcal{S}} \left ( \begin{array}{c} L \\ S\\ \end{array} \right ) = \left ( \begin{array}{c} LS \\ S\\ \end{array} \right ),$$ and $$\mathrm{det}\, \mathcal{S} = -1$$.

The substitutional rule for the Octonacci (or Pell) sequence is $$\sigma: L \rightarrow LLS, S\rightarrow L$$, or: $$$$\sigma : \left ( \begin{array}{c} L \\ S\\ \end{array} \right ) \rightarrow \underbrace{\left ( \begin{array}{cc} 2 & 1 \\ 1 & 0 \\ \end{array} \right )}_{=\mathcal{S}} \left ( \begin{array}{c} L \\ S\\ \end{array} \right ) = \left ( \begin{array}{c} LLS \\ S\\ \end{array} \right ).$$$$

with $$\mathrm{det}\, \mathcal{S} = -1$$.

The substitution rule for the Ammmann-Beenker tiling is: $$\left ( \begin{array}{cc} 3 & 2 \\ 4 & 3 \end{array} \right )$$

whose determinant is $$1$$.

Do substitution matrices need to have determinant $$\pm 1$$?

What is the physical meaning?

No, for instance the Thue-Morse substitution $$a \mapsto ab, b \mapsto ba$$ has determinant $$0$$, and the period doubling substitution $$a \mapsto ab, b \mapsto aa$$ has determinant $$-2$$.
When a substitution matrix has determinant $$\pm 1$$, then we call the substitution unimodular. You can find plenty of literature on unimodular Pisot substitutions for instance, which is related to the long standing open question known as the Pisot conjecture or Pisot substitution conjecture.