Question 1
If I define $A(z) = \sum_{n \ge 0} a_n \frac{z^n}{n!} \tag 1$
(where $a_n$ are $3\times 3$ constant matrices indexed with n),
then can we re-write $\sum_{n \ge 1} a_{n-1} \frac{z^n}{n!} \tag 2 $ and $\sum_{n \ge 2} a_{n-2} \frac{z^n}{n!} \tag 3 $ in terms of A(z)?
Question 2
Can we write $\psi(z)=\sum_{n \ge 0} na_{n} z^n \tag 4 $
in terms of any any existing infinite series such that we can also re write
$\sum_{n \ge 1} a_{n-1} z^n \tag 5 $ and $\sum_{n \ge 2} a_{n-2} z^n \tag 6 $ in terms of $\psi(z)$?
NB :: Thease all are part of my attempts to solve $na_n=a_{n-1}+a_{n-2}$ without using ODE. Means by using power series or any other infinte series
The nine entries in your matrix are independent, so it is slightly simpler to think of nine separate equations.
Differentiate the first equation (2). Then let $m=n-1$ and relabel the result. Unfortunately, this turns it into a differential equation.
Let $\xi(z)$ be equation (3). Then $\xi^{\prime}(z)$ can be rearranged in terms of $\psi(z)$.
$B(z)=\sum_{n\geq1}a_{n-1}{z^n\over n!}\\B\,^{\prime}(z)=\sum_{n\geq1}a_{n-1}{z^{n-1}\over (n-1)!}=\sum_{m\geq0}a_m{z^m\over m!}=B(z)$