已知向量组a1 a2 a3线性无关

k1b1+k2b2+k3b3=0
k1(a1+a2)+k2(2a1+2a2+a3)+k3(a1-a2+2a3)=0
(k1+2k2+k3)a1+(k1+2k2-k3)a2+(k2+2k3)a3=0
=>
k1+2k2+k3=0 (1) and
k1+2k2-k3=0 (2) and
k2+2k3=0 (3)
(1)-(2) =>k3=0 (4)
from (4) and (3) => k2=0 (5)
from (1) and (4) and (5) => k1=0
=>向量组b1=a1+a2 b2=2a1+2a2+a3 b3=a1-a2+2a3也线性无关

证明：k1*b1+k2*b2+k3*b3=k1(a1+a2)+k2(2a1+2a2+a3)+k3(a1-a2+2a3)
=a1(k1+2k2+k3)+a2(k1+2k2-k3)+a3（k2+2k3)
因为a1 a2 a3线性无关，所以存在不全为0的数使得x1*a1+x2*a2+x3*a3=0成立
所以存在不全为零的数使得看看k1*b1+k2*b2+k3*b3=0成立，得证