Motivation
It is not hard to show, by using the general solution, that
Proposition. If $(a_{n})_{n\in\Bbb{Z}}$ satisfies the recursive formula $ a_{n+2} = pa_{n+1} + qa_{n}$, then for any $n, i, j$ we have
$$ a_{n+i}a_{n+j} - a_{n+i+j}a_{n} = (-q)^{n}(a_{i}a_{j} - a_{i+j}a_{0}). \tag{*}$$
Indeed when $a_{n} = F_{n}$ and $i = j = 1$, (*) reduces to the Cassini's identity.
But by noting that the Cassini's identity is also proved by taking determinant of the identity
$$ \begin{pmatrix}1 & 1 \\ 1 & 0 \end{pmatrix}^{n} = \begin{pmatrix} F_{n+1} & F_{n} \\ F_{n} & F_{n-1} \end{pmatrix}, \tag{1} $$
I suspect that (*) may also be proved in a similar way which does not rely on the knowledge of the general solution.
Meditating on the identity (*), it is quite tempting to consider the matrix
$$ A_{n} = \begin{pmatrix} a_{n+i+j} & a_{n+i} \\ a_{n+j} & a_{n} \end{pmatrix} $$
as (*) is now read as $\det A_{n} = (-q)^{n}\det A_{0}$. And in view of the proof of the Cassini's identity, we would like to claim that
$$ A_{n} = \begin{pmatrix} p & q \\ 1 & 0 \end{pmatrix}^{n} A_{0}. $$
But this is not true except for some special case (such as $i = j = 1$). And neither I was able to find an alternative matrix relation that produces (*).
Question
So my question as follows: Is there any matrix relation that is analogous to (1) and produces (*)?