n1, n2, n3 是 Ax = b 的解，则 An1 = b, An2 = b, An3 = b，A(n1+n2) = 2bA(n3+2n2) = 3b则 A[3(n1+n2)-2(n3+2n2)] = 6b-6b = 0则 ξ2 = 3(n1+n2)-2(n3+2n2) 是 Ax = 0 的基础解系
