2?21??2????解 ?25???42?
??2?45?????1????5???1?21??初等行变换2??01??1??1??~??
(1??)(10??)(1??)(4??)??00??22??(1??)2(10??)?0 ???1且??10时，有唯一解. 当A?0,即
2(1??)(10??)(1??)(4??)当?0且?0，即??10时，无解.
22(1??)(10??)(1??)(4??)当?0且?0，即??1时，有无穷多解.
22?12?21???00? 此时，增广矩阵为?00?0000????x1???2??2??1?????????原方程组的解为?x2??k1?1??k2?0???0? (k1,k2?R)
?0??1??0??x????????3?
11．试利用矩阵的初等变换，求下列方阵的逆矩阵：
?3?20?1????321???21??02(1) ?315?; (2)
?1?2?3?2?.
?323??????0121????321100??32110???解 （1）?315010?~?0?14?11?323001??002?10???31??7?0???300?32022??2?~?0?1011?2?~?0?101?002?101??001?1???2???723?????100632??~?010?1?12?
1??001?10??22??0??0? 1??9?2??2?1?2?
1?02??36
23??7???632??故逆矩阵为??1?12?
1???10?2?2???3?20?11000???210100??02(2)
?1?2?3?20010? ???01?210001???1?2?3?20010???210001??01~?049510?30? ???02210100???0??1?2?3?2001??210001??01~?001110?3?4? ??00?2?1010?2????0??1?2?3?2001??210001??01~?001110?3?4? ???00?0121?6?10???1?200?1?1?2?2????010001`0?1?~?0010?1?136? ??000121?6?10?????100011?2?4???0?1??010001~?0010?1?136? ??000121?6?10?????11?2?4???0?1??01故逆矩阵为
??1?13?6??21?6?10????
37
?4?12．(1) 设A??2?3??0?(2) 设A??2??3?解
1?2??1?3????21?,B??22?,求X使AX?B;
?3?1?1?1????21???123??13?,B???2?31??,求X使XA?B.
??3?4???41?21?3?初等行变换?10???(1) ?AB???22122?~?01?31?13?1??00????102????1?X?AB???15?3?
?124???21??0?10????2?13?初等列变换?01?00?A???33?4???(2) ????~123?B????2?1?2?31???47??????????2?1?1??1?X?BA????474??．
??
0102??0?15?3? 1124??0??0?1?? ?1?4????
第四章 向量组的线性相关性
?(1,1,0)T,v2?(0,1,1)T,v3?(3,4,0)T,
求v1?v2及3v1?2v2?v3.
TT解 v1?v2?(1,1,0)?(0,1,1)
?(1?0,1?1,0?1)T?(1,0,?1)T
3v1?2v2?v3?3(1,1,0)T?2(0,1,1)T?(3,4,0)T
?(3?1?2?0?3,3?1?2?1?4,3?0?2?1?0)T ?(0,1,2)T
1．设v1
?a)?2(a2?a)?5(a3?a)其中a1?(2,5,1,3)T,
a2?(10,1,5,10)T,a3?(4,1,?1,1)T,求a
解 由3(a1?a)?2(a2?a)?5(a3?a)整理得
11a?(3a1?2a2?5a3)?[3(2,5,1,3)T?2(10,1,5,10)T?5(4,1,?1,1)T]
662．设3(a138
?(1,2,3,4)T
3．举例说明下列各命题是错误的:
(1)若向量组a1,a2,?,am是线性相关的,则a1可由a2,?am,线性表示. (2)若有不全为0的数?1,?2,?,?m使
?1a1????mam??1b1????mbm?0
成立,则a1,?,am线性相关,
b1,?,bm亦线性相关.
(3)若只有当?1,?2,?,?m全为0时,等式 才能成立,则a1,?,am线性无关,
?1a1????mam??1b1????mbm?0
b1,?,bm亦线性无关.
(4)若a1,?,am线性相关, b1,?,bm亦线性相关,则有不全为0的数, ?1,?2,?,?m使?1a1????mam?0,?1b1????mbm?0
同时成立.
?e1?(1,0,0,?,0)
a2?a3???am?0
满足a1,a2,?,am线性相关,但a1不能由a2,?,am,线性表示.
(2) 有不全为零的数?1,?2,?,?m使
?1a1????mam??1b1????mbm?0
解 (1) 设a1原式可化为
?1(a1?b1)????m(am?bm)?0
取a1?e1??b1,a2?e2??b2,?,am?em??bm 其中e1,?,em为单位向量,则上式成立,而
a1,?,am,b1,?,bm均线性相关
(3) 由?1a1????mam??1b1????mbm?0 (仅当?1????m?0) ?a1?b1,a2?b2,?,am?bm线性无关 取a1?a2???am?0 取b1,?,bm为线性无关组
满足以上条件,但不能说是a1,a2,?,am线性无关的.
TTTT(4) a1?(1,0) a2?(2,0) b1?(0,3) b2?(0,4) ?1a1??2a2?0??1??2?2??3? ??1??2?0与题设矛盾.
?1b1??2b2?0??1???2?4? 4．设b1?a1?a2,b2?a2?a3,b3?a3?a4,b4?a4?a1,证明向量组
b1,b2,b3,b4线性相关.
证明 设有x1,x2,x3,x4使得
x1b1?x2b2?x3b3?x4b4?0则
x1(a1?a2)?x2(a2?a3)?x3(a3?a4)?x4(a4?a1)?0
39
(x1?x4)a1?(x1?x2)a2?(x2?x3)a3?(x3?x4)a4?0 (1) 若a1,a2,a3,a4线性相关,则存在不全为零的数k1,k2,k3,k4, k1?x1?x4;k2?x1?x2;k3?x2?x3;k4?x3?x4;
由k1,k2,k3,k4不全为零,知x1,x2,x3,x4不全为零,即b1,b2,b3,b4线性相
关.
?x1?x4?0?1?x?x?0?1?1?2??(2) 若a1,a2,a3,a4线性无关,则??x2?x3?0?0?0?x?x?0?4?310011100?0知此齐次方程存在非零解 由
01100011则b1,b2,b3,b4线性相关.
综合得证. 5．设b1011000111??x1????0??x2??0
???0x3?????1??x4???a1,b2?a1?a2,?,br?a1?a2???ar,且向量组
a1,a2,?,ar线性无关,证明向量组b1,b2,?,br线性无关. 证明 设k1b1?k2b2???krbr?0则
(k1???kr)a1?(k2???kr)a2???(kp???kr)ap???krar?0
因向量组a1,a2,?,ar线性无关,故
?k1?k2???kr?0?1??1??k1??0????????k2???kr?0?01?1??k2??0????? ????????????????????????????????k0?010k?0???r????r1??101?1?1?0故方程组只有零解 因为
????0?01则k1?k2???kr?0所以b1,b2,?,br线性无关
6．利用初等行变换求下列矩阵的列向量组的一个最大无关组:
?25??75(1)
?75??25?
319494321753542043??1??132??0; (2)
?2?134???1?48??1201221??15?1?. ?3?13?04?1??40
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