设向量组A:a1,a2,a3线性无关,向量b1能由向量组A线性表示,向量b2不能由向量组A线性表示,k为任意常数,
答案:2 悬赏:0
解决时间 2021-02-08 00:46
- 提问者网友:清羽墨安
- 2021-02-07 17:34
设向量组A:a1,a2,a3线性无关,向量b1能由向量组A线性表示,向量b2不能由向量组A线性表示,k为任意常数,
最佳答案
- 二级知识专家网友:零负荷的放任
- 2021-02-07 17:54
(1)
k1a1+k2a2+k3a3+k4(kb1+b2)=0
k1a1+k2a2+k3a3+k4(k(k1'a1+k2'a2+k3'a3)+b2)=0
(k1+kk1'k4)a1 +(k2+kk2'k4)a2+(k3+kk3'k4)a3+ k4b2=0
=>
(k1+kk1'k4) = 0 (1) and
(k2+kk2'k4) = 0 (2) and
(k3+kk3'k4) = 0 (3) and
k4=0 (4)
from (1),(2),(3),(4)
=> k1=k2=k3=k4=0
向量组a1,a2,a3,kb1+b2线性无关
(2)
k1a1+k2a2+k3a3+k4(b1+kb2)=0
k1a1+k2a2+k3a3+k4((k1'a1+k2'a2+k3'a3)+kk4b2)=0
(k1+k1'k4)a1 +(k2+k2'k4)a2+(k3+k3'k4)a3+ kk4b2=0
=>
(k1+k1'k4) = 0 (1) and
(k2+k2'k4) = 0 (2) and
(k3+k3'k4) = 0 (3) and
kk4=0 (4)
if k =0, 向量组a1,a2,a3,b1+kb2线性相关
if k不等于0,向量组a1,a2,a3,b1+kb2线性无关
k1a1+k2a2+k3a3+k4(kb1+b2)=0
k1a1+k2a2+k3a3+k4(k(k1'a1+k2'a2+k3'a3)+b2)=0
(k1+kk1'k4)a1 +(k2+kk2'k4)a2+(k3+kk3'k4)a3+ k4b2=0
=>
(k1+kk1'k4) = 0 (1) and
(k2+kk2'k4) = 0 (2) and
(k3+kk3'k4) = 0 (3) and
k4=0 (4)
from (1),(2),(3),(4)
=> k1=k2=k3=k4=0
向量组a1,a2,a3,kb1+b2线性无关
(2)
k1a1+k2a2+k3a3+k4(b1+kb2)=0
k1a1+k2a2+k3a3+k4((k1'a1+k2'a2+k3'a3)+kk4b2)=0
(k1+k1'k4)a1 +(k2+k2'k4)a2+(k3+k3'k4)a3+ kk4b2=0
=>
(k1+k1'k4) = 0 (1) and
(k2+k2'k4) = 0 (2) and
(k3+k3'k4) = 0 (3) and
kk4=0 (4)
if k =0, 向量组a1,a2,a3,b1+kb2线性相关
if k不等于0,向量组a1,a2,a3,b1+kb2线性无关
全部回答
- 1楼网友:你好陌生人
- 2021-02-07 18:40
反证法:
假设
a1,a2………am,lb1+b2线性相关
则
存在x1,...,xm使得
lb1+b2=x1a1+...+xmam.............(1)
又已知
向量b1能由向量组a线性表示
则
存在y1,...,ym使得
b1=y1a1+...+ymam.........(2)
(1)-(2)*l得
b2=(x1-ly1)a1+...+(xm-lym)am
得
b2能由向量组a线性表示
与已知矛盾
故,m+1个向量a1,a2………am,lb1+b2必线性无关
我要举报
如以上问答内容为低俗、色情、不良、暴力、侵权、涉及违法等信息,可以点下面链接进行举报!
大家都在看
推荐资讯