= Solution
The initial <vector> records both initial <Fibonacci numbers>: $V_1=(U_2,U_1)^\top=(1,1)^\top$. For $n\ge2$,
$$
AV_{n-1}=\begin{pmatrix}1&1\\1&0\end{pmatrix}\begin{pmatrix}U_n\\U_{n-1}\end{pmatrix}=\begin{pmatrix}U_n+U_{n-1}\\U_n\end{pmatrix}=V_n.
$$
Conversely this <vector> recurrence forces the second component to be the preceding first component and forces their sum to be the next first component, recovering the <scalar> recurrence and its initial conditions. Iterating, or using <mathematical induction>, yields \b[$V_n=A^{n-1}(1,1)^\top$] for every $n\ge1$.
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