Factor: $A^{2025}-3A^{2024}+A^{2023}=A^{2023}(A^2-3A+I)$.

$|A|=10-9=1$, so $|A^{2023}|=1$. Hence $\det (A^{2023}(A^2-3A+I))=|A^2-3A+I|$.

$A^2=\left[\begin{array}{c}1321\\ 2134\end{array}\right]$.

$A^2-3A+I=\left[\begin{array}{c}13-6+121-9\\ 21-934-15+1\end{array}\right]=\left[\begin{array}{c}812\\ 1220\end{array}\right]$.

Determinant $=8\cdot 20-12\cdot 12=160-144=16$.