由已知 (b1,b2,...,bs) = (a1,a2,...,as)K
K =
t1 0 ...t2
t2 t1 ...0
...
0 0 ...t1
|K| = t1^n + (-1)^(n-1) t2^n
所以 当 t1^n + (-1)^(n-1) t2^n ≠ 0 时 b1,b2,...,bs 线性无关,
故此时 b1,b2,...,bs 是方程组的基础解系