Solution
= Solution
Use <mathematical induction> on $n$. For $n=0$, the identity <matrix> gives $(1,2)^T=(L_1,L_0)^T$. If $A^n(1,2)^T=(L_{n+1},L_n)^T$, applying $A$ and using part (a) gives $A^{n+1}(1,2)^T=(L_{n+2},L_{n+1})^T$. Therefore
$$
\boxed{A^n\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix}L_{n+1}\\L_n\end{pmatrix}\quad(n\geq0).}
$$
The base case is essential: the starting vector represents consecutive <Lucas numbers>, not two arbitrary initial values.