7. If $a_{1}=a_{2}=1, a_{n}=4 a_{n-1}-a_{n-2}$, then $a_{n} a_{n-2}-a_{n-1}^{2}=$
7. 2 .

From the known equation, we can transform it into $a_{n}+a_{n-2}=4 a_{n-1}, 4 a_{n-2}=a_{n-1}+a_{n-3}$. Multiplying the two equations yields $a_{n} a_{n-2}+a_{n-2}^{2}=a_{n-1}^{2}+a_{n-1} a_{n-3}$, thus $a_{n} a_{n-2}-a_{n-1}^{2}=a_{n-1} a_{n-3}-a_{n-2}^{2}$, indicating that $a_{n} a_{n-2}-a_{n-1}^{2}=a_{3} a_{1}-a_{2}^{2}=2$ is a constant.