Question

已知r1,r2,r3是四元线性方程组AX=B的解向量,R(A)=2

Answer 1

(1/3)(r1+2r2) = (1/3)(2,1,-1,3)^T 是 Ax=b 的特解

(r1+2r2)-(2r2+r3) = (1,0,-1,0)^T

(r1+2r2)+(r1-4r3) = (3,4,0,3)^T

线性无关,且由r(A)=2知是Ax=b的基础解系

所以 Ax=b 的通解为 (1/3)(2,1,-1,3)^T +k1(1,0,-1,0)^T+k2(3,4,0,3)^T