Question

这种题答案不唯一吧，哪位大神能做出来的答案是唯一的一个答案么？

Answer 1

a1+2a2-a3 = a1+a2+a3+a4, 则 2a3+a4 = a2；
a1+a2+a3+a4 = a1+3a2+a3+2a4， 则 a4 = -2a2
得a3 = (3/2)a2, β = a1+(1/2)a2.
Ax = β, a1x1 + a2x2 + (3/2)a2x3 - 2a2x4 = a1 + (1/2)a2,
x1 = 1, x2+(3/2)x3-2x4 = 1/2, 即 2x2 = 1 - 3x3 +4x4
取 x3 = x4 = 1, 得特解 (x2, x3, x4)^T = (1, 1, 1)^T,
导出组 2x2 = - 3x3 +4x4 ，
取 x3 = -2，x4 = 0, 得基础解系 (x2, x3, x4)^T = (3, -2, 0)^T,
取 x3 = 0，x4 = 1, 得基础解系 (x2, x3, x4)^T = (2, 0, 1)^T,
通解 (x2, x3, x4)^T = (1, 1, 1)^T+k(3, -2, 0)^T+c (2, 0, 1)^T
= (1+3k+2c, 1-2k, 1+c)^T
方程组Ax = β 的通解是 x = (1，1+3k+2c, 1-2k, 1+c)^T，
其中， k， c 为任意常数。