第四章 向量组的线性相关性
1. 设v 1=(1, 1, 0)T , v 2=(0, 1, 1)T , v 3=(3, 4, 0)T , 求v 1-v 2及3v 1+2v 2-v 3. 解 v 1-v 2=(1, 1, 0)T -(0, 1, 1)T
T
=(1-0, 1-1, 0-1)
T
=(1, 0, -1) .
3v 1+2v 2-v 3=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).
2. 设3(a 1-a ) +2(a 2+a ) =5(a 3+a ) , 求a , 其中a 1=(2, 5, 1, 3)T , a 2=(10, 1, 5, 10), a 3=(4, 1, -1, 1). 解 由3(a 1-a ) +2(a 2+a ) =5(a 3+a ) 整理得 a =
T
T T
1
(3a 1+2a 2-5a 3) 6
1
3(2, 5, 1, 3) T +2(10, 1, 5, 10) T -5(4, 1, -1, 1) T ] 6
=
=(1, 2, 3, 4)T . 3. 已知向量组
A : a 1=(0, 1, 2, 3)T , a 2=(3, 0, 1, 2)T , a3=(2, 3, 0, 1)T ; B : b 1=(2, 1, 1, 2)T , b 2=(0, -2, 1, 1)T , b3=(4, 4, 1, 3)T , 证明B 组能由A 组线性表示, 但A 组不能由B 组线性表示. 证明 由
⎛0 1
(A , B ) =
2 3⎝⎛1r 0
~
0 0⎝
由
30122301
204⎫1-24⎪111⎪213⎪⎭⎛1
0~ 0 0⎝
r 031-24⎫32204⎪
1-6-15-7⎪2-8-17-9⎪⎭
031-24⎫
1-6-15-7⎪
041-35⎪00000⎪⎭
031-24⎫
1-6-15-7⎪0205-1525⎪041-35⎪⎭⎛1
r ~ 0
0 0⎝
知R (A ) =R (A , B ) =3, 所以B 组能由A 组线性表示.
⎛204⎫⎛102⎫⎛1 1-24⎪r 0-22⎪r 0
B = ~~111⎪ 01-1⎪ 0
213⎪ 01-1⎪ 0⎝⎭⎝⎭⎝
4. 已知向量组
A : a 1=(0, 1, 1), a 2=(1, 1, 0);
T
T
02⎫
1-1⎪
00⎪00⎪⎭
知R (B ) =2. 因为R (B ) ≠R (B , A ) , 所以A 组不能由B 组线性表示.
B : b 1=(-1, 0, 1), b 2=(1, 2, 1), b3=(3, 2, -1) , 证明A 组与B 组等价. 证明 由
T T T
⎛-11301⎫r ⎛-11301⎫r ⎛-11301⎫
(B , A ) = 02211⎪~ 02211⎪~ 02211⎪,
11-110⎪ 02211⎪ 00000⎪⎝⎭⎝⎭⎝⎭
知R (B ) =R (B , A ) =2. 显然在A 中有二阶非零子式, 故R (A ) ≥2, 又R (A ) ≤R (B , A ) =2, 所以R (A ) =2, 从而R (A ) =R (B ) =R (A , B ) . 因此A 组与B 组等价.
5. 已知R (a 1, a 2, a 3) =2, R (a 2, a 3, a 4) =3, 证明 (1) a1能由a 2, a 3线性表示; (2) a4不能由a 1, a 2, a 3线性表示.
证明 (1)由R (a 2, a 3, a 4) =3知a 2, a 3, a 4线性无关, 故a 2, a 3也线性无关. 又由R (a 1, a 2, a 3) =2知a 1, a 2, a 3线性相关, 故a 1能由a 2, a 3线性表示.
(2)假如a 4能由a 1, a 2, a 3线性表示, 则因为a 1能由a 2, a 3线性表示, 故a 4能由a 2, a 3线性表示, 从而a 2, a 3, a 4线性相关, 矛盾. 因此a 4不能由a 1, a 2, a 3线性表示.
6. 判定下列向量组是线性相关还是线性无关: (1) (-1, 3, 1)T , (2, 1, 0)T , (1, 4, 1)T ; (2) (2, 3, 0)T , (-1, 4, 0)T , (0, 0, 2)T .
解 (1)以所给向量为列向量的矩阵记为A . 因为
⎛-121⎫r ⎛-121⎫r ⎛-121⎫
A = 314⎪~ 077⎪~ 011⎪,
101⎪ 022⎪ 000⎪⎝⎭⎝⎭⎝⎭
所以R (A ) =2小于向量的个数, 从而所给向量组线性相关. (2)以所给向量为列向量的矩阵记为B . 因为
2-10
|B |=340=22≠0,
002
所以R (B ) =3等于向量的个数, 从而所给向量组线性相无关.
7. 问a 取什么值时下列向量组线性相关？ a 1=(a , 1, 1)T , a 2=(1, a , -1) T , a3=(1, -1, a ) T . 解 以所给向量为列向量的矩阵记为A . 由
a 11
|A |=1a -1=a (a -1)(a +1)
1-1a
知, 当a =-1、0、1时, R (A )
8. 设a 1, a 2线性无关, a 1+b , a 2+b 线性相关, 求向量b 用a 1, a 2线性表示的表示式. 解 因为a 1+b , a 2+b 线性相关, 故存在不全为零的数λ1, λ2使 λ1(a 1+b ) +λ2(a 2+b ) =0, 由此得 b =-
λλλλa 1-a 2=-a 1-(1-a 2,
λ1+λ2λ1+λ2λ1+λ2λ1+λ2
, 则
设c =-
λ1λ1+λ2
b =c a 1-(1+c ) a 2, c ∈R .
9. 设a 1, a 2线性相关, b 1, b 2也线性相关, 问a 1+b 1, a 2+b 2是否一定线性相关？试举例说明之.
解 不一定.
例如, 当a 1=(1, 2)T , a 2=(2, 4)T , b 1=(-1, -1) T , b 2=(0, 0)T 时, 有 a 1+b 1=(1, 2)T +b 1=(0, 1)T , a 2+b 2=(2, 4)T +(0, 0)T =(2, 4)T , 而a 1+b 1, a 2+b 2的对应分量不成比例, 是线性无关的.
10. 举例说明下列各命题是错误的:
(1)若向量组a 1, a 2, ⋅ ⋅ ⋅, a m 是线性相关的, 则a 1可由a 2, ⋅ ⋅ ⋅, a m 线性表示.
解 设a 1=e 1=(1, 0, 0, ⋅ ⋅ ⋅, 0), a 2=a 3= ⋅ ⋅ ⋅ =a m =0, 则a 1, a 2, ⋅ ⋅ ⋅, a m 线性相关, 但a 1不能由a 2, ⋅ ⋅ ⋅, a m 线性表示.
(2)若有不全为0的数λ1, λ2, ⋅ ⋅ ⋅, λm 使
λ1a 1+ ⋅ ⋅ ⋅ +λm a m +λ1b 1+ ⋅ ⋅ ⋅ +λm b m =0
成立, 则a 1, a 2, ⋅ ⋅ ⋅, a m 线性相关, b 1, b 2, ⋅ ⋅ ⋅, b m 亦线性相关. 解 有不全为零的数λ1, λ2, ⋅ ⋅ ⋅, λm 使
λ1a 1+ ⋅ ⋅ ⋅ +λm a m +λ1b 1+ ⋅ ⋅ ⋅ +λm b m =0,
原式可化为
λ1(a 1+b 1) + ⋅ ⋅ ⋅ +λm (a m +b m ) =0.
取a 1=e 1=-b 1, a 2=e 2=-b 2, ⋅ ⋅ ⋅, a m =e m =-b m , 其中e 1, e 2, ⋅ ⋅ ⋅, e m 为单位坐标向量, 则上式成立, 而a 1, a 2, ⋅ ⋅ ⋅, a m 和b 1, b 2, ⋅ ⋅ ⋅, b m 均线性无关.
(3)若只有当λ1, λ2, ⋅ ⋅ ⋅, λm 全为0时, 等式
λ1a 1+ ⋅ ⋅ ⋅ +λm a m +λ1b 1+ ⋅ ⋅ ⋅ +λm b m =0
才能成立, 则a 1, a 2, ⋅ ⋅ ⋅, a m 线性无关, b 1, b 2, ⋅ ⋅ ⋅, b m 亦线性无关. 解 由于只有当λ1, λ2, ⋅ ⋅ ⋅, λm 全为0时, 等式
由λ1a 1+ ⋅ ⋅ ⋅ +λm a m +λ1b 1+ ⋅ ⋅ ⋅ +λm b m =0
成立, 所以只有当λ1, λ2, ⋅ ⋅ ⋅, λm 全为0时, 等式
λ1(a 1+b 1) +λ2(a 2+b 2) + ⋅ ⋅ ⋅ +λm (a m +b m ) =0
成立. 因此a 1+b 1, a 2+b 2, ⋅ ⋅ ⋅, a m +b m 线性无关.
取a 1=a 2= ⋅ ⋅ ⋅ =a m =0, 取b 1, ⋅ ⋅ ⋅, b m 为线性无关组, 则它们满足以上条件, 但a 1, a 2, ⋅ ⋅ ⋅, a m 线性相关.
(4)若a 1, a 2, ⋅ ⋅ ⋅, a m 线性相关, b 1, b 2, ⋅ ⋅ ⋅, b m 亦线性相关, 则有不全为0的数, λ1, λ2, ⋅ ⋅ ⋅,
λm 使
λ1a 1+ ⋅ ⋅ ⋅ +λm a m =0, λ1b 1+ ⋅ ⋅ ⋅ +λm b m =0
同时成立.
解 a 1=(1, 0), a 2=(2, 0), b 1=(0, 3), b 2=(0, 4),
T
T
T
T
λ1a 1+λ2a 2 =0⇒λ1=-2λ2, λ1b 1+λ2b 2 =0⇒λ1=-(3/4)λ2,
⇒λ1=λ2=0, 与题设矛盾.
11. 设b 1=a 1+a 2, b 2=a 2+a 3, b3=a 3+a 4, b4=a 4+a 1, 证明向量组b 1, b 2, b 3, b 4线性相关. 证明 由已知条件得
a 1=b 1-a 2, a 2=b 2-a 3, a3=b 3-a 4, a4=b 4-a 1, 于是 a 1 =b 1-b 2+a 3
=b 1-b 2+b 3-a 4
=b 1-b 2+b 3-b 4+a 1, 从而 b 1-b 2+b 3-b 4=0,
这说明向量组b 1, b 2, b 3, b 4线性相关.
12. 设b 1=a 1, b 2=a 1+a 2, ⋅ ⋅ ⋅, b r =a 1+a 2+ ⋅ ⋅ ⋅ +a r , 且向量组a 1, a 2, ⋅ ⋅ ⋅ , a r 线性无关, 证明向量组b 1, b 2, ⋅ ⋅ ⋅ , b r 线性无关. 证明 已知的r 个等式可以写成
⎛1 0
(b 1, b 2, ⋅ ⋅ ⋅ , b r ) =(a 1, a 2, ⋅ ⋅ ⋅ , a r )
⋅⋅⋅ 0⎝
关.
13. 求下列向量组的秩, 并求一个最大无关组:
11⋅⋅⋅0⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅
1⎫1⎪, ⋅⋅⋅⎪1⎪⎭
上式记为B =AK . 因为|K |=1≠0, K 可逆, 所以R (B ) =R (A ) =r , 从而向量组b 1, b 2, ⋅ ⋅ ⋅ , b r 线性无
(1)a 1=(1, 2, -1, 4)T , a 2=(9, 100, 10, 4)T , a3=(-2, -4, 2, -8) T ; 解 由
9-2⎫⎛19-2⎫⎛1⎛1
r 2100-4⎪ 0820⎪r 0
(a 1, a 2, a 3) = ~~-1102⎪ 0190⎪ 0
4⎪ 0-320⎪ 04-8⎝⎭⎝⎭⎝
最大无关组.
(2)a 1T =(1, 2, 1, 3), a 2T =(4, -1, -5, -6) , a3T =(1, -3, -4, -7) . 解 由
9-2⎫
10⎪
,
00⎪00⎪⎭
知R (a 1, a 2, a3) =2. 因为向量a 1与a 2的分量不成比例, 故a 1, a 2线性无关, 所以a 1, a 2是一个
41⎫⎛141⎫⎛141⎫⎛1
2-1-3⎪r 0-9-5⎪r 0-9-5⎪
, (a 1, a 2, a 3) = ~~1-5-4⎪ 0-9-5⎪ 000⎪
3-6-7⎪ 0-18-10⎪ 000⎪⎝⎭⎝⎭⎝⎭
知R (a 1T , a 2T , a3T ) =R (a 1, a 2, a3) =2. 因为向量a 1T 与a 2T 的分量不成比例, 故a 1T , a 2T 线性无关, 所以a 1, a 2是一个最大无关组.
14. 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:
T
T
⎛25
75 (1)
75 25⎝
[***********]13213448
⎫⎪⎪; ⎪⎭
解 因为
⎛25 75 75 25⎝⎛1 0 (2)
2 1⎝
⎛1 0 2 1⎝
[***********]13213448
⎫⎪⎪⎪⎭
r 2-3r 1r 3-3r 1r 4-r 1
~
⎛25 0 0 0⎝
31111
1743⎫23⎪35⎪35⎪⎭
r 4-r 3r 3-r 2
~
⎛25
0 0 0⎝
31100
1743⎫23⎪
,
13⎪00⎪⎭
所以第1、2、3列构成一个最大无关组.
1
201213025-14
1⎫-1⎪
. 3⎪-1⎪⎭
⎛1
0 0 0⎝
12-20
21-1-2
25-52
1⎫-1⎪1⎪-2⎪⎭
⎛1
0 0 0⎝
120021-20
25201⎫-1⎪
, -2⎪0⎪⎭
解 因为
1
201213025-14
1⎫-1⎪3⎪-1⎪⎭
r 3-2r 1r 4-r 1
~
r 3+r 2r 3r 4
~
所以第1、2、3列构成一个最大无关组.
15. 设向量组
(a , 3, 1)T , (2, b , 3)T , (1, 2, 1)T , (2, 3, 1)T
的秩为2, 求a , b .
解 设a 1=(a , 3, 1), a 2=(2, b , 3), a3=(1, 2, 1), a 4=(2, 3, 1). 因为
T
T
T
T
13⎫r ⎛1113⎫⎛12a 2⎫r ⎛11
(a 3, a 4, a 1, a 2) = 233b ⎪~ 01a -1-1⎪~ 01a -1-1⎪,
1113⎪ 01⎪ 002-a b -5⎪1b -6⎝⎭⎝⎭⎝⎭
而R (a 1, a 2, a 3, a 4) =2, 所以a =2, b =5.
16. 设a 1, a 2, ⋅ ⋅ ⋅, a n 是一组n 维向量, 已知n 维单位坐标向量e 1, e 2,⋅ ⋅ ⋅, e n 能由它们线性表示, 证明a 1, a 2, ⋅ ⋅ ⋅, a n 线性无关.
证法一 记A =(a 1, a 2, ⋅ ⋅ ⋅, a n ) , E =(e 1, e 2,⋅ ⋅ ⋅, e n ) . 由已知条件知, 存在矩阵K , 使
E =AK .
两边取行列式, 得
|E |=|A ||K |.
可见|A |≠0, 所以R (A ) =n , 从而a 1, a 2, ⋅ ⋅ ⋅, a n 线性无关.
证法二 因为e 1, e 2,⋅ ⋅ ⋅, e n 能由a 1, a 2, ⋅ ⋅ ⋅, a n 线性表示, 所以
R (e 1, e 2,⋅ ⋅ ⋅, e n ) ≤R (a 1, a 2, ⋅ ⋅ ⋅, a n ) ,
而R (e 1, e 2,⋅ ⋅ ⋅, e n ) =n , R (a 1, a 2, ⋅ ⋅ ⋅, a n ) ≤n , 所以R (a 1, a 2, ⋅ ⋅ ⋅, a n ) =n , 从而a 1, a 2, ⋅ ⋅ ⋅, a n 线性无关.
17. 设a 1, a 2, ⋅ ⋅ ⋅, a n 是一组n 维向量, 证明它们线性无关的充分必要条件是: 任一n 维向量都可由它们线性表示.
证明 必要性: 设a 为任一n 维向量. 因为a 1, a 2, ⋅ ⋅ ⋅, a n 线性无关, 而a 1, a 2, ⋅ ⋅ ⋅, a n , a 是n +1个n 维向量, 是线性相关的, 所以a 能由a 1, a 2, ⋅ ⋅ ⋅, a n 线性表示, 且表示式是唯一的.
充分性: 已知任一n 维向量都可由a 1, a 2, ⋅ ⋅ ⋅, a n 线性表示, 故单位坐标向量组e 1, e 2, ⋅ ⋅ ⋅, e n 能由a 1, a 2, ⋅ ⋅ ⋅, a n 线性表示, 于是有
n =R (e 1, e 2, ⋅ ⋅ ⋅, e n ) ≤R (a 1, a 2, ⋅ ⋅ ⋅, a n ) ≤n ,
即R (a 1, a 2, ⋅ ⋅ ⋅, a n ) =n , 所以a 1, a 2, ⋅ ⋅ ⋅, a n 线性无关.
18. 设向量组a 1, a 2, ⋅ ⋅ ⋅, a m 线性相关, 且a 1≠0, 证明存在某个向量a k (2≤k ≤m ) , 使a k 能由a 1, a 2, ⋅ ⋅ ⋅, a k -1线性表示.
证明 因为a 1, a 2, ⋅ ⋅ ⋅, a m 线性相关, 所以存在不全为零的数λ1, λ2, ⋅ ⋅ ⋅, λm , 使
λ1a 1+λ2a 2+ ⋅ ⋅ ⋅ +λm a m =0,
而且λ2, λ3,⋅ ⋅ ⋅, λm 不全为零. 这是因为, 如若不然, 则λ1a 1=0, 由a 1≠0知λ1=0, 矛盾. 因此存在k (2≤k ≤m ) , 使
λk ≠0, λk +1=λk +2= ⋅ ⋅ ⋅ =λm =0,
于是
λ1a 1+λ2a 2+ ⋅ ⋅ ⋅ +λk a k =0,
a k =-(1/λk )(λ1a 1+λ2a 2+ ⋅ ⋅ ⋅ +λk -1a k -1) ,
即a k 能由a 1, a 2, ⋅ ⋅ ⋅, a k -1线性表示.
19. 设向量组B : b 1, ⋅ ⋅ ⋅, b r 能由向量组A : a 1, ⋅ ⋅ ⋅, a s 线性表示为
(b 1, ⋅ ⋅ ⋅, b r ) =(a 1, ⋅ ⋅ ⋅, a s ) K , 其中K 为s ⨯r 矩阵, 且A 组线性无关. 证明B 组线性无关的充分必要条件是矩阵K 的秩R (K ) =r .
证明 令B =(b 1, ⋅ ⋅ ⋅, b r ) , A =(a 1, ⋅ ⋅ ⋅, a s ) , 则有B =AK . 必要性: 设向量组B 线性无关.
由向量组B 线性无关及矩阵秩的性质, 有 r =R (B ) =R (AK ) ≤min{R (A ) , R (K )}≤R (K ) , 及 R (K ) ≤min{r , s }≤r . 因此R (K ) =r .
充分性: 因为R (K ) =r , 所以存在可逆矩阵C , 使KC (b 1, ⋅ ⋅ ⋅, b r ) C =( a1, ⋅ ⋅ ⋅, a s ) KC =(a 1, ⋅ ⋅ ⋅, a r ) .
因为C 可逆, 所以R (b 1, ⋅ ⋅ ⋅, b r ) =R (a 1, ⋅ ⋅ ⋅, a r ) =r , 从而b 1, ⋅ ⋅ ⋅, b r 线性无关.
20. 设
⎛E ⎫
= r ⎪为K 的标准形. 于是 ⎝O ⎭
⎧β1= α2+α3+ ⋅ ⋅ ⋅ +αn ⎪β2=α1 +α3+ ⋅ ⋅ ⋅ +αn
⎨ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ , ⎪
⎩βn =α1+α2+α3+ ⋅ ⋅ ⋅ +αn -1
证明向量组α1, α2, ⋅ ⋅ ⋅, αn 与向量组β1, β2, ⋅ ⋅ ⋅, βn 等价. 证明 将已知关系写成
⎛0 1
(β1, β2, ⋅ ⋅ ⋅ , βn ) =(α1, α2, ⋅ ⋅ ⋅ , αn ) 1
⋅⋅⋅ ⎝1
将上式记为B =AK . 因为
101⋅⋅⋅1110⋅⋅⋅1⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅
1⎫1⎪1⎪, ⋅⋅⋅⎪⎪0⎭
01|K |=1
⋅⋅⋅1101⋅⋅⋅1110⋅⋅⋅1⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅11
1=(-1) n -1(n -1) ≠0, ⋅⋅⋅0
所以K 可逆, 故有A =BK -1. 由B =AK 和A =BK -1可知向量组α1, α2, ⋅ ⋅ ⋅, αn 与向量组β1, β2, ⋅ ⋅ ⋅, βn 可相互线性表示. 因此向量组α1, α2, ⋅ ⋅ ⋅, αn 与向量组β1, β2, ⋅ ⋅ ⋅, βn 等价.
21. 已知3阶矩阵A 与3维列向量x 满足A 3x =3A x -A 2x , 且向量组x , A x , A 2x 线性无关.
(1)记P =(x , A x , A 2x ) , 求3阶矩阵B , 使AP =PB ; 解 因为
AP =A (x , A x , A 2x ) =(A x , A 2x , A 3x ) =(A x , A 2x , 3A x -A 2x )
⎛000⎫
=(x , A x , A x ) 103⎪,
01-1⎪⎝⎭
2
⎛000⎫
所以B = 103⎪.
01-1⎪⎝⎭
(2)求|A |.
解 由A 3x =3A x -A 2x , 得A (3x -A x -A 2x ) =0. 因为x , A x , A 2x 线性无关, 故3x -A x -A 2x ≠0, 即方程A x =0有非零解, 所以R (A )
x -8x 2+10x 3+2x 4=0⎧⎪1
(1)⎨2x 1+4x 2+5x 3-x 4=0;
⎪⎩3x 1+8x 2+6x 3-2x 4=0
解 对系数矩阵进行初等行变换, 有
40⎫⎛1-8102⎫r ⎛10
A = 245-1⎪ ~ 01-3/4-1/4⎪,
386-2⎪ 00⎪00⎝⎭⎝⎭
于是得
⎧x 1=-4x 3
⎨x =(3/4) x +(1/4) x . ⎩234
T
T
T
T
取(x 3, x 4) =(4, 0), 得(x 1, x 2) =(-16, 3); 取(x 3, x 4) T =(0, 4)T , 得(x 1, x 2) T =(0, 1)T . 因此方程组的基础解系为
ξ1=(-16, 3, 4, 0), ξ2=(0, 1, 0, 4).
T
T
2x -3x 2-2x 3+x 4=0⎧⎪1
(2)⎨3x 1+5x 2+4x 3-2x 4=0.
⎪⎩8x 1+7x 2+6x 3-3x 4=0
解 对系数矩阵进行初等行变换, 有
⎛2-3-21⎫r ⎛102/19-1/19⎫
4-2⎪ ~ 0114/19-7/19⎪, A = 35
876-3⎪ 0000⎪⎝⎭⎝⎭
于是得
⎧x 1=-(2/19) x 3+(1/19) x 4
⎨x =-(14/19) x +(7/19) x . ⎩234
取(x 3, x 4) T =(19, 0)T , 得(x 1, x 2) T =(-2, 14)T ; 取(x 3, x 4) T =(0, 19)T , 得(x 1, x 2) T =(1, 7)T . 因此方程组的基础解系为
ξ1=(-2, 14, 19, 0)T , ξ2=(1, 7, 0, 19)T .
(3)nx 1 +(n -1) x 2+ ⋅ ⋅ ⋅ +2x n -1+x n =0. 解 原方程组即为
x n =-nx 1-(n -1) x 2- ⋅ ⋅ ⋅ -2x n -1.
取x 1=1, x 2=x 3= ⋅ ⋅ ⋅ =x n -1=0, 得x n =-n ;
取x 2=1, x 1=x 3=x 4= ⋅ ⋅ ⋅ =x n -1=0, 得x n =-(n -1) =-n +1; ⋅ ⋅ ⋅ ;
取x n -1=1, x 1=x 2= ⋅ ⋅ ⋅ =x n -2=0, 得x n =-2. 因此方程组的基础解系为 ξ1=(1, 0, 0, ⋅ ⋅ ⋅, 0, -n ) T , ξ2=(0, 1, 0, ⋅ ⋅ ⋅, 0, -n +1) T , ⋅ ⋅ ⋅,
ξn -1=(0, 0, 0, ⋅ ⋅ ⋅, 1, -2) T .
23. 设A = R (B ) =2.
解 显然B 的两个列向量应是方程组AB =0的两个线性无关的解. 因为 A =
r
⎛2-213⎫, 求一个4⨯2矩阵B , 使AB =0, 且
⎪
9-528⎝⎭
1/8⎫⎛2-213⎫ ~ ⎛10-1/8
⎪ 01-5/8-11/8⎪,
9-528⎝⎭⎝⎭
所以与方程组AB =0同解方程组为
⎧x 1=(1/8) x 3-(1/8) x 4
⎨x =(5/8) x +(11/8) x . ⎩234
取(x 3, x 4) T =(8, 0)T , 得(x 1, x 2) T =(1, 5)T ; 取(x 3, x 4) T =(0, 8)T , 得(x 1, x 2) T =(-1, 11)T . 方程组AB =0的基础解系为
ξ1=(1, 5, 8, 0)T , ξ2=(-1, 11, 0, 8)T .
⎛1
5
因此所求矩阵为B =
8 0⎝
-1⎫11⎪
. 0⎪8⎪⎭
24. 求一个齐次线性方程组, 使它的基础解系为
ξ1=(0, 1, 2, 3)T , ξ2=(3, 2, 1, 0)T .
解 显然原方程组的通解为
⎛x 1⎫⎧x 1=3k 2⎛0⎫⎛3⎫ x ⎪ 1⎪ 2⎪⎪x 2=k 1+2k 2
2
, 即=k +k ⎪1 ⎪2 ⎪⎨x =2k +k , (k 1, k 2∈R ) ,
21x 312
⎪⎪3 3⎪ 0⎪
⎝⎭⎝⎭⎩x 4=3k 1⎝x 4⎭
消去k 1, k 2得
⎧2x 1-3x 2+x 4=0
⎨x -3x +2x =0, ⎩134
此即所求的齐次线性方程组.
25. 设四元齐次线性方程组 I :
⎧x 1+x 2=0
⎨x -x =0 , II: ⎩24⎧x 1-x 2+x 3=0
⎨x -x +x =0. ⎩234
求: (1)方程I 与II 的基础解系; (2) I与II 的公共解. 解 (1)由方程I 得⎨
⎧x 1=-x 4
.
x =x ⎩24
取(x 3, x 4) T =(1, 0)T , 得(x 1, x 2) T =(0, 0)T ; 取(x 3, x 4) T =(0, 1)T , 得(x 1, x 2) T =(-1, 1)T . 因此方程I 的基础解系为
ξ1=(0, 0, 1, 0)T , ξ2=(-1, 1, 0, 1)T . 由方程II 得⎨
⎧x 1=-x 4
.
x =x -x ⎩234
T
T
T
取(x 3, x 4) T =(1, 0)T , 得(x 1, x 2) T =(0, 1)T ; 取(x 3, x 4) =(0, 1), 得(x 1, x 2) =(-1, -1) . 因此方程II 的基础解系为
ξ1=(0, 1, 1, 0)T , ξ2=(-1, -1, 0, 1)T . (2) I与II 的公共解就是方程
T
⎧x 1+x 2=0⎪x -x 4=0
III : ⎨2
x 1-x 2+x 3=0⎪
⎩x 2-x 3+x 4=0
的解. 因为方程组III 的系数矩阵
⎛1100⎫⎛1 010-1⎪r 0
A = ~
1-110⎪ 0 01-11⎪ 0⎝⎭⎝
所以与方程组III 同解的方程组为
100
01⎫0-1⎪
,
1-2⎪00⎪⎭
x =-x 4⎧⎪1
⎨x 2=x 4.
⎪⎩x 3=2x 4
取x 4=1, 得(x 1, x 2, x 3) =(-1, 1, 2), 方程组III 的基础解系为 ξ=(-1, 1, 2, 1)T .
因此I 与II 的公共解为x =c (-1, 1, 2, 1)T , c ∈R .
26. 设n 阶矩阵A 满足A =A , E 为n 阶单位矩阵, 证明
R (A ) +R (A -E ) =n .
证明 因为A (A -E ) =A 2-A =A -A =0, 所以R (A ) +R (A -E ) ≤n . 又R (A -E ) =R (E -A ) , 可知
R (A ) +R (A -E ) =R (A ) +R (E -A ) ≥R (A +E -A ) =R (E ) =n ,
由此R (A ) +R (A -E ) =n .
27. 设A 为n 阶矩阵(n ≥2) , A *为A 的伴随阵, 证明
2T
T
⎧n 当R (A ) =n
⎪
R (A *)=⎨1 当R (A ) =n -1.
⎪⎩0 当R (A ) ≤n -2
证明 当R (A ) =n 时, |A |≠0, 故有 |AA *|=||A |E |=|A |≠0, |A *|≠0, 所以R (A *)=n .
当R (A ) =n -1时, |A |=0, 故有 AA *=|A |E =0,
即A *的列向量都是方程组A x =0的解. 因为R (A ) =n -1, 所以方程组A x =0的基础解系中只含一个解向量, 即基础解系的秩为1. 因此R (A *)=1.
当R (A ) ≤n -2时, A 中每个元素的代数余子式都为0, 故A *=O , 从而R (A *)=0. 28. 求下列非齐次方程组的一个解及对应的齐次线性方程组的基础解系:
x +x 2=5⎧⎪1
(1)⎨2x 1+x 2+x 3+2x 4=1;
⎪⎩5x 1+3x 2+2x 3+2x 4=3
解 对增广矩阵进行初等行变换, 有
⎛11005⎫r ⎛1010-8⎫
B = 21121⎪ ~ 01-1013⎪.
53223⎪ 00012⎪⎝⎭⎝⎭
与所给方程组同解的方程为
x =-x 3-8⎧⎪1
⎨x 2= x 3+13. ⎪⎩x 4= 2
当x 3=0时, 得所给方程组的一个解η=(-8, 13, 0, 2). 与对应的齐次方程组同解的方程为
T
x =-x 3⎧⎪1
⎨x 2= x 3. ⎪⎩x 4=0
当x 3=1时, 得对应的齐次方程组的基础解系ξ=(-1, 1, 1, 0).
T
x -5x 2+2x 3-3x 4=11⎧⎪1
(2)⎨5x 1+3x 2+6x 3-x 4=-1.
⎪⎩2x 1+4x 2+2x 3+x 4=-6
解 对增广矩阵进行初等行变换, 有
⎛1-52-311⎫r ⎛109/7-1/21⎫ B = 536-1-1⎪ ~ 01-1/71/2-2⎪.
2421-6⎪ 000⎪00⎝⎭⎝⎭
与所给方程组同解的方程为
⎧x 1=-(9/7) x 3+(1/2) x 4+1⎨x =(1/7)x -(1/2) x -2. ⎩234
当x 3=x 4=0时, 得所给方程组的一个解
η=(1, -2, 0, 0)T .
与对应的齐次方程组同解的方程为
⎧x 1=-(9/7) x 3+(1/2) x 4⎨x =(1/7)x -(1/2) x . ⎩234
分别取(x 3, x 4) =(1, 0), (0, 1), 得对应的齐次方程组的基础解系
T
T
T
ξ1=(-9, 1, 7, 0)T . ξ2=(1, -1, 0, 2)T .
29. 设四元非齐次线性方程组的系数矩阵的秩为3, 已知η1, η2, η3是它的三个解向量. 且
η1=(2, 3, 4, 5)T , η2+η3=(1, 2, 3, 4)T ,
求该方程组的通解.
解 由于方程组中未知数的个数是4, 系数矩阵的秩为3, 所以对应的齐次线性方程组的基础解系含有一个向量, 且由于η1, η2, η3均为方程组的解, 由非齐次线性方程组解的结构性质得
2η1-(η2+η3) =(η1-η2) +(η1-η3) = (3, 4, 5, 6)
为其基础解系向量, 故此方程组的通解:
x =k (3, 4, 5, 6)T +(2, 3, 4, 5)T , (k ∈R ) .
30. 设有向量组A : a 1=(α, 2, 10), a 2=(-2, 1, 5), a3=(-1, 1, 4), 及b =(1, β, -1) , 问α, β为何值时
(1)向量b 不能由向量组A 线性表示;
(2)向量b 能由向量组A 线性表示, 且表示式唯一;
(3)向量b 能由向量组A 线性表示, 且表示式不唯一, 并求一般表示式.
T
T
T
T
T
⎛-1-2α1⎫
12β⎪ 解 (a 3, a 2, a 1, b ) = 1
4510-1⎪⎝⎭1⎫⎛-1-2α
~ 0-11+αβ+1⎪. 004+α-3β⎪⎝⎭
r
(1)当α=-4, β≠0时, R (A ) ≠R (A , b ) , 此时向量b 不能由向量组A 线性表示.
(2)当α≠-4时, R (A ) =R (A , b ) =3, 此时向量组a 1, a 2, a 3线性无关, 而向量组a 1, a 2, a 3, b 线
性相关, 故向量b 能由向量组A 线性表示, 且表示式唯一.
(3)当α=-4, β=0时, R (A ) =R (A , b ) =2, 此时向量b 能由向量组A 线性表示, 且表示式不唯一.
当α=-4, β=0时,
⎛-1-2-41⎫
(a 3, a 2, a 1, b ) = 1120⎪
4510-1⎪⎝⎭
方程组(a 3, a 2, a 1) x =b 的解为
⎛10-21⎫
~ 013-1⎪, 0000⎪⎝⎭
r
⎛x 1⎫⎛2⎫⎛1⎫⎛2c +1⎫
x 2⎪=c -3⎪+ -1⎪= -3c -1⎪, c ∈R .
⎪ 1⎪ 0⎪ c ⎪
⎭⎝⎭⎝⎭⎝x 3⎭⎝
因此 b =(2c +1) a 3+(-3c -1) a 2+c a 1, 即 b = ca 1+(-3c -1) a 2+(2c +1) a 3, c∈R .
31. 设a =(a 1, a 2, a 3) T , b =(b 1, b 2, b 3) T , c=(c 1, c 2, c 3) T , 证明三直线 l 1: a 1x +b 1y +c 1=0,
l 2: a 2x +b 2y +c 2=0, (a i +b i ≠0, i =1, 2, 3) l 3: a 3x +b 3y +c 3=0,
相交于一点的充分必要条件为: 向量组a , b 线性无关, 且向量组a , b , c 线性相关. 证明 三直线相交于一点的充分必要条件为方程组
2
2
a x +b 1y +c 1=0a x +b 1y =-c 1⎧⎧⎪1⎪1
⎨a 2x +b 2y +c 2=0, 即⎨a 2x +b 2y =-c 2 ⎪⎪⎩a 3x +b 3y +c 3=0⎩a 3x +b 3y =-c 3
有唯一解. 上述方程组可写为x a +y b =-c . 因此三直线相交于一点的充分必要条件为c 能由a , b 唯一线性表示, 而c 能由a , b 唯一线性表示的充分必要条件为向量组a , b 线性无关, 且向量组a , b , c 线性相关.
32. 设矩阵A =(a 1, a 2, a 3, a 4) , 其中a 2, a 3, a 4线性无关, a 1=2a 2- a 3. 向量b =a 1+a 2+a 3+a 4, 求方程A x =b 的通解.
解 由b =a 1+a 2+a 3+a 4知η=(1, 1, 1, 1)T 是方程A x =b 的一个解. 由a 1=2a 2- a 3得a 1-2a 2+a 3=0, 知ξ=(1, -2, 1, 0)T 是A x =0的一个解.
由a 2, a 3, a 4线性无关知R (A ) =3, 故方程A x =b 所对应的齐次方程A x =0的基础解系中含
一个解向量. 因此ξ=(1, -2, 1, 0)T 是方程A x =0的基础解系. 方程A x =b 的通解为
x =c (1, -2, 1, 0)T +(1, 1, 1, 1)T , c ∈R .
33. 设η*是非齐次线性方程组A x =b 的一个解, ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r ,是对应的齐次线性方程组的一个基础解系, 证明:
(1)η*, ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性无关; (2)η*, η*+ξ1, η*+ξ2, ⋅ ⋅ ⋅, η*+ξn -r 线性无关.
证明 (1)反证法, 假设η*, ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性相关. 因为ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性无关, 而
η*, ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性相关, 所以η*可由ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性表示, 且表示式是唯一的, 这说
明η*也是齐次线性方程组的解, 矛盾.
(2)显然向量组η*, η*+ξ1, η*+ξ2, ⋅ ⋅ ⋅, η*+ξn -r 与向量组η*, ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 可以相互表示, 故这两个向量组等价, 而由(1)知向量组η*, ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性无关, 所以向量组η*, η*+ξ1,
η*+ξ2, ⋅ ⋅ ⋅, η*+ξn -r 也线性无关.
34. 设η1, η2, ⋅ ⋅ ⋅, ηs 是非齐次线性方程组A x =b 的s 个解, k 1, k 2, ⋅ ⋅ ⋅, k s 为实数, 满足k 1+k 2+ ⋅ ⋅ ⋅ +k s =1. 证明
x =k 1η1+k 2η2+ ⋅ ⋅ ⋅ +k s ηs
也是它的解.
证明 因为η1, η2, ⋅ ⋅ ⋅, ηs 都是方程组A x =b 的解, 所以 A ηi =b (i =1, 2, ⋅ ⋅ ⋅, s ) ,
从而 A (k 1η1+k 2η2+ ⋅ ⋅ ⋅ +k s ηs ) =k 1A η1+k 2A η2+ ⋅ ⋅ ⋅ +k s A ηs =(k 1+k 2+ ⋅ ⋅ ⋅ +k s ) b =b . 因此x =k 1η1+k 2η2+ ⋅ ⋅ ⋅ +k s ηs 也是方程的解.
35. 设非齐次线性方程组A x =b 的系数矩阵的秩为r , η1, η2, ⋅ ⋅ ⋅, ηn -r +1是它的n -r +1个线性无关的解. 试证它的任一解可表示为
x =k 1η1+k 2η2+ ⋅ ⋅ ⋅ +k n -r +1ηn -r +1, (其中k 1+k 2+ ⋅ ⋅ ⋅ +k n -r +1=1).
证明 因为η1, η2, ⋅ ⋅ ⋅, ηn -r +1均为A x =b 的解, 所以ξ1=η2-η1, ξ2=η3-η1, ⋅ ⋅ ⋅, ξn -r =η
n -r +1-1均为
η
A x =b 的解.
用反证法证: ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性无关.
设它们线性相关, 则存在不全为零的数λ1, λ2, ⋅ ⋅ ⋅, λn -r , 使得 λ1ξ1+ λ2ξ2+ ⋅ ⋅ ⋅ + λ n-r ξ n-r =0,
即 λ1(η2-η1) + λ2(η3-η1) + ⋅ ⋅ ⋅ + λ n-r (ηn -r +1-η1) =0, 亦即 -(λ1+λ2+ ⋅ ⋅ ⋅ +λn -r ) η1+λ1η2+λ2η3+ ⋅ ⋅ ⋅ +λ n-r ηn -r +1=0, 由η1, η2, ⋅ ⋅ ⋅, ηn -r +1线性无关知
-(λ1+λ2+ ⋅ ⋅ ⋅ +λn -r ) =λ1=λ2= ⋅ ⋅ ⋅ =λn -r =0,
矛盾. 因此ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性无关. ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 为A x =b 的一个基础解系.
设x 为A x =b 的任意解, 则x -η1为A x =0的解, 故x -η1可由ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性表出, 设 x -η1=k 2ξ1+k 3ξ2+ ⋅ ⋅ ⋅ +k n -r +1ξn -r
=k 2(η2-η1) +k 3(η3-η1) + ⋅ ⋅ ⋅ +k n -r +1(ηn -r +1-η1) , x =η1(1-k 2-k 3 ⋅ ⋅ ⋅ -k n -r +1) +k 2η2+k 3η3+ ⋅ ⋅ ⋅ +k n-r +1ηn -r +1. 令k 1=1-k 2-k 3 ⋅ ⋅ ⋅ -k n -r +1, 则k 1+k 2+k 3 ⋅ ⋅ ⋅ -k n -r +1=1, 于是 x =k 1η1+k 2η2+ ⋅ ⋅ ⋅ +k n -r +1ηn -r +1.
36. 设
V 1={x =(x 1, x 2, ⋅ ⋅ ⋅, x n ) T | x 1, ⋅ ⋅ ⋅, x n ∈R 满足x 1+x 2+ ⋅ ⋅ ⋅ +x n =0}, V 2={x =(x 1, x 2, ⋅ ⋅ ⋅, x n ) | x 1, ⋅ ⋅ ⋅, x n ∈R 满足x 1+x 2+ ⋅ ⋅ ⋅ +x n =1},
问V 1, V 2是不是向量空间？为什么？ 解 V 1是向量空间, 因为任取
α=(a 1, a 2, ⋅ ⋅ ⋅, a n ) T ∈V 1, β=(b 1, b 2, ⋅ ⋅ ⋅, b n ) T ∈V 1, λ∈∈R , 有 a 1+a 2+ ⋅ ⋅ ⋅ +a n =0, b 1+b 2+ ⋅ ⋅ ⋅ +b n =0,
从而 (a 1+b 1) +(a 2+b 2) + ⋅ ⋅ ⋅ +(a n +b n ) =(a 1+a 2+ ⋅ ⋅ ⋅ +a n ) +(b 1+b 2+ ⋅ ⋅ ⋅ +b n ) =0, λa 1+λa 2+ ⋅ ⋅ ⋅ +λa n =λ(a 1+a 2+ ⋅ ⋅ ⋅ +a n ) =0, 所以 α+β=(a 1+b 1, a 2+b 2, ⋅ ⋅ ⋅, a n +b n ) T ∈V 1, λα=(λa 1, λa 2, ⋅ ⋅ ⋅, λa n ) T ∈V 1. V 2不是向量空间, 因为任取
α=(a 1, a 2, ⋅ ⋅ ⋅, a n ) T ∈V 1, β=(b 1, b 2, ⋅ ⋅ ⋅, b n ) T ∈V 1, 有 a 1+a 2+ ⋅ ⋅ ⋅ +a n =1, b 1+b 2+ ⋅ ⋅ ⋅ +b n =1,
从而 (a 1+b 1) +(a 2+b 2) + ⋅ ⋅ ⋅ +(a n +b n ) =(a 1+a 2+ ⋅ ⋅ ⋅ +a n ) +(b 1+b 2+ ⋅ ⋅ ⋅ +b n ) =2,
T
所以 α+β=(a 1+b 1, a 2+b 2, ⋅ ⋅ ⋅, a n +b n ) T ∉V 1.
37. 试证: 由a 1=(0, 1, 1), a 2=(1, 0, 1), a3=(1, 1, 0)所生成的向量空间就是R . 证明 设A =(a 1, a2, a 3) , 由
T
T
T
3
011
|A |=101=-2≠0,
110
知R (A ) =3, 故a 1, a2, a 3线性无关, 所以a 1, a2, a 3是三维空间R 3的一组基, 因此由a 1, a2, a 3所生成的向量空间就是R .
38. 由a 1=(1, 1, 0, 0) T , a 2=(1, 0, 1, 1) T 所生成的向量空间记作V 1, 由b 1=(2, -1, 3, 3) T , b 2=(0, 1, -1, -1) 所生成的向量空间记作V 2, 试证V 1=V 2. 证明 设A =(a 1, a2) , B =(b 1, b2) . 显然R (A ) =R (B ) =2, 又由
T
3
⎛1 1
(A , B ) =
0 0⎝
10112-133
0⎫⎛11⎪r 0 ~ -1⎪ 0
0-1⎪⎭⎝
1
-1002-300
0⎫1⎪, 0⎪0⎪⎭
知R (A , B ) =2, 所以R (A ) =R (B ) =R (A , B ) , 从而向量组a 1, a2与向量组b 1, b2等价. 因为向量组a 1, a2与向量组b 1, b2等价, 所以这两个向量组所生成的向量空间相同, 即V 1=V 2.
39. 验证a 1=(1, -1, 0) T , a 2=(2, 1, 3) T , a 3=(3, 1, 2) T 为R 3的一个基, 并把v 1=(5, 0, 7) T , v 2=(-9, -8, -13) T 用这个基线性表示. 解 设A =(a 1, a2, a3) . 由
123
|(a 1, a 2, a 3) |=-111=-6≠0,
032
知R (A ) =3, 故a 1, a2, a3线性无关, 所以a 1, a2, a3为R 的一个基. 设x 1a 1+x 2a 2+x 3a 3=v 1, 则
3
x +2x 2+3x 3=5⎧⎪1
⎨-x 1+x 2+x 3=0, ⎪⎩3x 2+2x 3=7
解之得x 1=2, x 2=3, x 3=-1, 故线性表示为v 1=2a 1+3a 2-a 3. 设x 1a 1+x 2a 2+x 3a 3=v 2, 则
x +2x 2+3x 3=-9⎧⎪1
⎨-x 1+x 2+x 3=-8, ⎪⎩3x 2+2x 3=-13
解之得x 1=3, x 2=-3, x 3=-2, 故线性表示为v 2=3a 1-3a 2-2a 3.
40. 已知R 的两个基为
a 1=(1, 1, 1), a 2=(1, 0, -1) , a 3=(1, 0, 1), b 1=(1, 2, 1)T , b 2=(2, 3, 4)T , b 3=(3, 4, 3)T . 求由基a 1, a 2, a 3到基b 1, b 2, b 3的过渡矩阵P . 解 设e 1, e 2, e 3是三维单位坐标向量组, 则
T
T
T
3
⎛111⎫
(a 1, a 2, a 3) =(e 1, e 2, e 3) 100⎪,
1-11⎪⎝⎭
⎛111⎫
(e 1, e 2, e 3) =(a 1, a 2, a 3) 100⎪,
1-11⎪⎝⎭⎛123⎫
于是 (b 1, b 2, b 3) =(e 1, e 2, e 3) 234⎪
143⎪⎝⎭
-1
⎛111⎫⎛123⎫
=(a 1, a 2, a 3) 100⎪ 234⎪,
1-11⎪ 143⎪⎝⎭⎝⎭
由基a 1, a 2, a 3到基b 1, b 2, b 3的过渡矩阵为
-1
⎛111⎫⎛123⎫⎛234⎫
P = 100⎪ 234⎪= 0-10⎪. 1-11⎪ 143⎪ -10-1⎪⎝⎭⎝⎭⎝⎭
-1
第四章 向量组的线性相关性
1. 设v 1=(1, 1, 0)T , v 2=(0, 1, 1)T , v 3=(3, 4, 0)T , 求v 1-v 2及3v 1+2v 2-v 3. 解 v 1-v 2=(1, 1, 0)T -(0, 1, 1)T
T
=(1-0, 1-1, 0-1)
T
=(1, 0, -1) .
3v 1+2v 2-v 3=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).
2. 设3(a 1-a ) +2(a 2+a ) =5(a 3+a ) , 求a , 其中a 1=(2, 5, 1, 3)T , a 2=(10, 1, 5, 10), a 3=(4, 1, -1, 1). 解 由3(a 1-a ) +2(a 2+a ) =5(a 3+a ) 整理得 a =
T
T T
1
(3a 1+2a 2-5a 3) 6
1
3(2, 5, 1, 3) T +2(10, 1, 5, 10) T -5(4, 1, -1, 1) T ] 6
=
=(1, 2, 3, 4)T . 3. 已知向量组
A : a 1=(0, 1, 2, 3)T , a 2=(3, 0, 1, 2)T , a3=(2, 3, 0, 1)T ; B : b 1=(2, 1, 1, 2)T , b 2=(0, -2, 1, 1)T , b3=(4, 4, 1, 3)T , 证明B 组能由A 组线性表示, 但A 组不能由B 组线性表示. 证明 由
⎛0 1
(A , B ) =
2 3⎝⎛1r 0
~
0 0⎝
由
30122301
204⎫1-24⎪111⎪213⎪⎭⎛1
0~ 0 0⎝
r 031-24⎫32204⎪
1-6-15-7⎪2-8-17-9⎪⎭
031-24⎫
1-6-15-7⎪
041-35⎪00000⎪⎭
031-24⎫
1-6-15-7⎪0205-1525⎪041-35⎪⎭⎛1
r ~ 0
0 0⎝
知R (A ) =R (A , B ) =3, 所以B 组能由A 组线性表示.
⎛204⎫⎛102⎫⎛1 1-24⎪r 0-22⎪r 0
B = ~~111⎪ 01-1⎪ 0
213⎪ 01-1⎪ 0⎝⎭⎝⎭⎝
4. 已知向量组
A : a 1=(0, 1, 1), a 2=(1, 1, 0);
T
T
02⎫
1-1⎪
00⎪00⎪⎭
知R (B ) =2. 因为R (B ) ≠R (B , A ) , 所以A 组不能由B 组线性表示.
B : b 1=(-1, 0, 1), b 2=(1, 2, 1), b3=(3, 2, -1) , 证明A 组与B 组等价. 证明 由
T T T
⎛-11301⎫r ⎛-11301⎫r ⎛-11301⎫
(B , A ) = 02211⎪~ 02211⎪~ 02211⎪,
11-110⎪ 02211⎪ 00000⎪⎝⎭⎝⎭⎝⎭
知R (B ) =R (B , A ) =2. 显然在A 中有二阶非零子式, 故R (A ) ≥2, 又R (A ) ≤R (B , A ) =2, 所以R (A ) =2, 从而R (A ) =R (B ) =R (A , B ) . 因此A 组与B 组等价.
5. 已知R (a 1, a 2, a 3) =2, R (a 2, a 3, a 4) =3, 证明 (1) a1能由a 2, a 3线性表示; (2) a4不能由a 1, a 2, a 3线性表示.
证明 (1)由R (a 2, a 3, a 4) =3知a 2, a 3, a 4线性无关, 故a 2, a 3也线性无关. 又由R (a 1, a 2, a 3) =2知a 1, a 2, a 3线性相关, 故a 1能由a 2, a 3线性表示.
(2)假如a 4能由a 1, a 2, a 3线性表示, 则因为a 1能由a 2, a 3线性表示, 故a 4能由a 2, a 3线性表示, 从而a 2, a 3, a 4线性相关, 矛盾. 因此a 4不能由a 1, a 2, a 3线性表示.
6. 判定下列向量组是线性相关还是线性无关: (1) (-1, 3, 1)T , (2, 1, 0)T , (1, 4, 1)T ; (2) (2, 3, 0)T , (-1, 4, 0)T , (0, 0, 2)T .
解 (1)以所给向量为列向量的矩阵记为A . 因为
⎛-121⎫r ⎛-121⎫r ⎛-121⎫
A = 314⎪~ 077⎪~ 011⎪,
101⎪ 022⎪ 000⎪⎝⎭⎝⎭⎝⎭
所以R (A ) =2小于向量的个数, 从而所给向量组线性相关. (2)以所给向量为列向量的矩阵记为B . 因为
2-10
|B |=340=22≠0,
002
所以R (B ) =3等于向量的个数, 从而所给向量组线性相无关.
7. 问a 取什么值时下列向量组线性相关？ a 1=(a , 1, 1)T , a 2=(1, a , -1) T , a3=(1, -1, a ) T . 解 以所给向量为列向量的矩阵记为A . 由
a 11
|A |=1a -1=a (a -1)(a +1)
1-1a
知, 当a =-1、0、1时, R (A )
8. 设a 1, a 2线性无关, a 1+b , a 2+b 线性相关, 求向量b 用a 1, a 2线性表示的表示式. 解 因为a 1+b , a 2+b 线性相关, 故存在不全为零的数λ1, λ2使 λ1(a 1+b ) +λ2(a 2+b ) =0, 由此得 b =-
λλλλa 1-a 2=-a 1-(1-a 2,
λ1+λ2λ1+λ2λ1+λ2λ1+λ2
, 则
设c =-
λ1λ1+λ2
b =c a 1-(1+c ) a 2, c ∈R .
9. 设a 1, a 2线性相关, b 1, b 2也线性相关, 问a 1+b 1, a 2+b 2是否一定线性相关？试举例说明之.
解 不一定.
例如, 当a 1=(1, 2)T , a 2=(2, 4)T , b 1=(-1, -1) T , b 2=(0, 0)T 时, 有 a 1+b 1=(1, 2)T +b 1=(0, 1)T , a 2+b 2=(2, 4)T +(0, 0)T =(2, 4)T , 而a 1+b 1, a 2+b 2的对应分量不成比例, 是线性无关的.
10. 举例说明下列各命题是错误的:
(1)若向量组a 1, a 2, ⋅ ⋅ ⋅, a m 是线性相关的, 则a 1可由a 2, ⋅ ⋅ ⋅, a m 线性表示.
解 设a 1=e 1=(1, 0, 0, ⋅ ⋅ ⋅, 0), a 2=a 3= ⋅ ⋅ ⋅ =a m =0, 则a 1, a 2, ⋅ ⋅ ⋅, a m 线性相关, 但a 1不能由a 2, ⋅ ⋅ ⋅, a m 线性表示.
(2)若有不全为0的数λ1, λ2, ⋅ ⋅ ⋅, λm 使
λ1a 1+ ⋅ ⋅ ⋅ +λm a m +λ1b 1+ ⋅ ⋅ ⋅ +λm b m =0
成立, 则a 1, a 2, ⋅ ⋅ ⋅, a m 线性相关, b 1, b 2, ⋅ ⋅ ⋅, b m 亦线性相关. 解 有不全为零的数λ1, λ2, ⋅ ⋅ ⋅, λm 使
λ1a 1+ ⋅ ⋅ ⋅ +λm a m +λ1b 1+ ⋅ ⋅ ⋅ +λm b m =0,
原式可化为
λ1(a 1+b 1) + ⋅ ⋅ ⋅ +λm (a m +b m ) =0.
取a 1=e 1=-b 1, a 2=e 2=-b 2, ⋅ ⋅ ⋅, a m =e m =-b m , 其中e 1, e 2, ⋅ ⋅ ⋅, e m 为单位坐标向量, 则上式成立, 而a 1, a 2, ⋅ ⋅ ⋅, a m 和b 1, b 2, ⋅ ⋅ ⋅, b m 均线性无关.
(3)若只有当λ1, λ2, ⋅ ⋅ ⋅, λm 全为0时, 等式
λ1a 1+ ⋅ ⋅ ⋅ +λm a m +λ1b 1+ ⋅ ⋅ ⋅ +λm b m =0
才能成立, 则a 1, a 2, ⋅ ⋅ ⋅, a m 线性无关, b 1, b 2, ⋅ ⋅ ⋅, b m 亦线性无关. 解 由于只有当λ1, λ2, ⋅ ⋅ ⋅, λm 全为0时, 等式
由λ1a 1+ ⋅ ⋅ ⋅ +λm a m +λ1b 1+ ⋅ ⋅ ⋅ +λm b m =0
成立, 所以只有当λ1, λ2, ⋅ ⋅ ⋅, λm 全为0时, 等式
λ1(a 1+b 1) +λ2(a 2+b 2) + ⋅ ⋅ ⋅ +λm (a m +b m ) =0
成立. 因此a 1+b 1, a 2+b 2, ⋅ ⋅ ⋅, a m +b m 线性无关.
取a 1=a 2= ⋅ ⋅ ⋅ =a m =0, 取b 1, ⋅ ⋅ ⋅, b m 为线性无关组, 则它们满足以上条件, 但a 1, a 2, ⋅ ⋅ ⋅, a m 线性相关.
(4)若a 1, a 2, ⋅ ⋅ ⋅, a m 线性相关, b 1, b 2, ⋅ ⋅ ⋅, b m 亦线性相关, 则有不全为0的数, λ1, λ2, ⋅ ⋅ ⋅,
λm 使
λ1a 1+ ⋅ ⋅ ⋅ +λm a m =0, λ1b 1+ ⋅ ⋅ ⋅ +λm b m =0
同时成立.
解 a 1=(1, 0), a 2=(2, 0), b 1=(0, 3), b 2=(0, 4),
T
T
T
T
λ1a 1+λ2a 2 =0⇒λ1=-2λ2, λ1b 1+λ2b 2 =0⇒λ1=-(3/4)λ2,
⇒λ1=λ2=0, 与题设矛盾.
11. 设b 1=a 1+a 2, b 2=a 2+a 3, b3=a 3+a 4, b4=a 4+a 1, 证明向量组b 1, b 2, b 3, b 4线性相关. 证明 由已知条件得
a 1=b 1-a 2, a 2=b 2-a 3, a3=b 3-a 4, a4=b 4-a 1, 于是 a 1 =b 1-b 2+a 3
=b 1-b 2+b 3-a 4
=b 1-b 2+b 3-b 4+a 1, 从而 b 1-b 2+b 3-b 4=0,
这说明向量组b 1, b 2, b 3, b 4线性相关.
12. 设b 1=a 1, b 2=a 1+a 2, ⋅ ⋅ ⋅, b r =a 1+a 2+ ⋅ ⋅ ⋅ +a r , 且向量组a 1, a 2, ⋅ ⋅ ⋅ , a r 线性无关, 证明向量组b 1, b 2, ⋅ ⋅ ⋅ , b r 线性无关. 证明 已知的r 个等式可以写成
⎛1 0
(b 1, b 2, ⋅ ⋅ ⋅ , b r ) =(a 1, a 2, ⋅ ⋅ ⋅ , a r )
⋅⋅⋅ 0⎝
关.
13. 求下列向量组的秩, 并求一个最大无关组:
11⋅⋅⋅0⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅
1⎫1⎪, ⋅⋅⋅⎪1⎪⎭
上式记为B =AK . 因为|K |=1≠0, K 可逆, 所以R (B ) =R (A ) =r , 从而向量组b 1, b 2, ⋅ ⋅ ⋅ , b r 线性无
(1)a 1=(1, 2, -1, 4)T , a 2=(9, 100, 10, 4)T , a3=(-2, -4, 2, -8) T ; 解 由
9-2⎫⎛19-2⎫⎛1⎛1
r 2100-4⎪ 0820⎪r 0
(a 1, a 2, a 3) = ~~-1102⎪ 0190⎪ 0
4⎪ 0-320⎪ 04-8⎝⎭⎝⎭⎝
最大无关组.
(2)a 1T =(1, 2, 1, 3), a 2T =(4, -1, -5, -6) , a3T =(1, -3, -4, -7) . 解 由
9-2⎫
10⎪
,
00⎪00⎪⎭
知R (a 1, a 2, a3) =2. 因为向量a 1与a 2的分量不成比例, 故a 1, a 2线性无关, 所以a 1, a 2是一个
41⎫⎛141⎫⎛141⎫⎛1
2-1-3⎪r 0-9-5⎪r 0-9-5⎪
, (a 1, a 2, a 3) = ~~1-5-4⎪ 0-9-5⎪ 000⎪
3-6-7⎪ 0-18-10⎪ 000⎪⎝⎭⎝⎭⎝⎭
知R (a 1T , a 2T , a3T ) =R (a 1, a 2, a3) =2. 因为向量a 1T 与a 2T 的分量不成比例, 故a 1T , a 2T 线性无关, 所以a 1, a 2是一个最大无关组.
14. 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:
T
T
⎛25
75 (1)
75 25⎝
[***********]13213448
⎫⎪⎪; ⎪⎭
解 因为
⎛25 75 75 25⎝⎛1 0 (2)
2 1⎝
⎛1 0 2 1⎝
[***********]13213448
⎫⎪⎪⎪⎭
r 2-3r 1r 3-3r 1r 4-r 1
~
⎛25 0 0 0⎝
31111
1743⎫23⎪35⎪35⎪⎭
r 4-r 3r 3-r 2
~
⎛25
0 0 0⎝
31100
1743⎫23⎪
,
13⎪00⎪⎭
所以第1、2、3列构成一个最大无关组.
1
201213025-14
1⎫-1⎪
. 3⎪-1⎪⎭
⎛1
0 0 0⎝
12-20
21-1-2
25-52
1⎫-1⎪1⎪-2⎪⎭
⎛1
0 0 0⎝
120021-20
25201⎫-1⎪
, -2⎪0⎪⎭
解 因为
1
201213025-14
1⎫-1⎪3⎪-1⎪⎭
r 3-2r 1r 4-r 1
~
r 3+r 2r 3r 4
~
所以第1、2、3列构成一个最大无关组.
15. 设向量组
(a , 3, 1)T , (2, b , 3)T , (1, 2, 1)T , (2, 3, 1)T
的秩为2, 求a , b .
解 设a 1=(a , 3, 1), a 2=(2, b , 3), a3=(1, 2, 1), a 4=(2, 3, 1). 因为
T
T
T
T
13⎫r ⎛1113⎫⎛12a 2⎫r ⎛11
(a 3, a 4, a 1, a 2) = 233b ⎪~ 01a -1-1⎪~ 01a -1-1⎪,
1113⎪ 01⎪ 002-a b -5⎪1b -6⎝⎭⎝⎭⎝⎭
而R (a 1, a 2, a 3, a 4) =2, 所以a =2, b =5.
16. 设a 1, a 2, ⋅ ⋅ ⋅, a n 是一组n 维向量, 已知n 维单位坐标向量e 1, e 2,⋅ ⋅ ⋅, e n 能由它们线性表示, 证明a 1, a 2, ⋅ ⋅ ⋅, a n 线性无关.
证法一 记A =(a 1, a 2, ⋅ ⋅ ⋅, a n ) , E =(e 1, e 2,⋅ ⋅ ⋅, e n ) . 由已知条件知, 存在矩阵K , 使
E =AK .
两边取行列式, 得
|E |=|A ||K |.
可见|A |≠0, 所以R (A ) =n , 从而a 1, a 2, ⋅ ⋅ ⋅, a n 线性无关.
证法二 因为e 1, e 2,⋅ ⋅ ⋅, e n 能由a 1, a 2, ⋅ ⋅ ⋅, a n 线性表示, 所以
R (e 1, e 2,⋅ ⋅ ⋅, e n ) ≤R (a 1, a 2, ⋅ ⋅ ⋅, a n ) ,
而R (e 1, e 2,⋅ ⋅ ⋅, e n ) =n , R (a 1, a 2, ⋅ ⋅ ⋅, a n ) ≤n , 所以R (a 1, a 2, ⋅ ⋅ ⋅, a n ) =n , 从而a 1, a 2, ⋅ ⋅ ⋅, a n 线性无关.
17. 设a 1, a 2, ⋅ ⋅ ⋅, a n 是一组n 维向量, 证明它们线性无关的充分必要条件是: 任一n 维向量都可由它们线性表示.
证明 必要性: 设a 为任一n 维向量. 因为a 1, a 2, ⋅ ⋅ ⋅, a n 线性无关, 而a 1, a 2, ⋅ ⋅ ⋅, a n , a 是n +1个n 维向量, 是线性相关的, 所以a 能由a 1, a 2, ⋅ ⋅ ⋅, a n 线性表示, 且表示式是唯一的.
充分性: 已知任一n 维向量都可由a 1, a 2, ⋅ ⋅ ⋅, a n 线性表示, 故单位坐标向量组e 1, e 2, ⋅ ⋅ ⋅, e n 能由a 1, a 2, ⋅ ⋅ ⋅, a n 线性表示, 于是有
n =R (e 1, e 2, ⋅ ⋅ ⋅, e n ) ≤R (a 1, a 2, ⋅ ⋅ ⋅, a n ) ≤n ,
即R (a 1, a 2, ⋅ ⋅ ⋅, a n ) =n , 所以a 1, a 2, ⋅ ⋅ ⋅, a n 线性无关.
18. 设向量组a 1, a 2, ⋅ ⋅ ⋅, a m 线性相关, 且a 1≠0, 证明存在某个向量a k (2≤k ≤m ) , 使a k 能由a 1, a 2, ⋅ ⋅ ⋅, a k -1线性表示.
证明 因为a 1, a 2, ⋅ ⋅ ⋅, a m 线性相关, 所以存在不全为零的数λ1, λ2, ⋅ ⋅ ⋅, λm , 使
λ1a 1+λ2a 2+ ⋅ ⋅ ⋅ +λm a m =0,
而且λ2, λ3,⋅ ⋅ ⋅, λm 不全为零. 这是因为, 如若不然, 则λ1a 1=0, 由a 1≠0知λ1=0, 矛盾. 因此存在k (2≤k ≤m ) , 使
λk ≠0, λk +1=λk +2= ⋅ ⋅ ⋅ =λm =0,
于是
λ1a 1+λ2a 2+ ⋅ ⋅ ⋅ +λk a k =0,
a k =-(1/λk )(λ1a 1+λ2a 2+ ⋅ ⋅ ⋅ +λk -1a k -1) ,
即a k 能由a 1, a 2, ⋅ ⋅ ⋅, a k -1线性表示.
19. 设向量组B : b 1, ⋅ ⋅ ⋅, b r 能由向量组A : a 1, ⋅ ⋅ ⋅, a s 线性表示为
(b 1, ⋅ ⋅ ⋅, b r ) =(a 1, ⋅ ⋅ ⋅, a s ) K , 其中K 为s ⨯r 矩阵, 且A 组线性无关. 证明B 组线性无关的充分必要条件是矩阵K 的秩R (K ) =r .
证明 令B =(b 1, ⋅ ⋅ ⋅, b r ) , A =(a 1, ⋅ ⋅ ⋅, a s ) , 则有B =AK . 必要性: 设向量组B 线性无关.
由向量组B 线性无关及矩阵秩的性质, 有 r =R (B ) =R (AK ) ≤min{R (A ) , R (K )}≤R (K ) , 及 R (K ) ≤min{r , s }≤r . 因此R (K ) =r .
充分性: 因为R (K ) =r , 所以存在可逆矩阵C , 使KC (b 1, ⋅ ⋅ ⋅, b r ) C =( a1, ⋅ ⋅ ⋅, a s ) KC =(a 1, ⋅ ⋅ ⋅, a r ) .
因为C 可逆, 所以R (b 1, ⋅ ⋅ ⋅, b r ) =R (a 1, ⋅ ⋅ ⋅, a r ) =r , 从而b 1, ⋅ ⋅ ⋅, b r 线性无关.
20. 设
⎛E ⎫
= r ⎪为K 的标准形. 于是 ⎝O ⎭
⎧β1= α2+α3+ ⋅ ⋅ ⋅ +αn ⎪β2=α1 +α3+ ⋅ ⋅ ⋅ +αn
⎨ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ , ⎪
⎩βn =α1+α2+α3+ ⋅ ⋅ ⋅ +αn -1
证明向量组α1, α2, ⋅ ⋅ ⋅, αn 与向量组β1, β2, ⋅ ⋅ ⋅, βn 等价. 证明 将已知关系写成
⎛0 1
(β1, β2, ⋅ ⋅ ⋅ , βn ) =(α1, α2, ⋅ ⋅ ⋅ , αn ) 1
⋅⋅⋅ ⎝1
将上式记为B =AK . 因为
101⋅⋅⋅1110⋅⋅⋅1⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅
1⎫1⎪1⎪, ⋅⋅⋅⎪⎪0⎭
01|K |=1
⋅⋅⋅1101⋅⋅⋅1110⋅⋅⋅1⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅11
1=(-1) n -1(n -1) ≠0, ⋅⋅⋅0
所以K 可逆, 故有A =BK -1. 由B =AK 和A =BK -1可知向量组α1, α2, ⋅ ⋅ ⋅, αn 与向量组β1, β2, ⋅ ⋅ ⋅, βn 可相互线性表示. 因此向量组α1, α2, ⋅ ⋅ ⋅, αn 与向量组β1, β2, ⋅ ⋅ ⋅, βn 等价.
21. 已知3阶矩阵A 与3维列向量x 满足A 3x =3A x -A 2x , 且向量组x , A x , A 2x 线性无关.
(1)记P =(x , A x , A 2x ) , 求3阶矩阵B , 使AP =PB ; 解 因为
AP =A (x , A x , A 2x ) =(A x , A 2x , A 3x ) =(A x , A 2x , 3A x -A 2x )
⎛000⎫
=(x , A x , A x ) 103⎪,
01-1⎪⎝⎭
2
⎛000⎫
所以B = 103⎪.
01-1⎪⎝⎭
(2)求|A |.
解 由A 3x =3A x -A 2x , 得A (3x -A x -A 2x ) =0. 因为x , A x , A 2x 线性无关, 故3x -A x -A 2x ≠0, 即方程A x =0有非零解, 所以R (A )
x -8x 2+10x 3+2x 4=0⎧⎪1
(1)⎨2x 1+4x 2+5x 3-x 4=0;
⎪⎩3x 1+8x 2+6x 3-2x 4=0
解 对系数矩阵进行初等行变换, 有
40⎫⎛1-8102⎫r ⎛10
A = 245-1⎪ ~ 01-3/4-1/4⎪,
386-2⎪ 00⎪00⎝⎭⎝⎭
于是得
⎧x 1=-4x 3
⎨x =(3/4) x +(1/4) x . ⎩234
T
T
T
T
取(x 3, x 4) =(4, 0), 得(x 1, x 2) =(-16, 3); 取(x 3, x 4) T =(0, 4)T , 得(x 1, x 2) T =(0, 1)T . 因此方程组的基础解系为
ξ1=(-16, 3, 4, 0), ξ2=(0, 1, 0, 4).
T
T
2x -3x 2-2x 3+x 4=0⎧⎪1
(2)⎨3x 1+5x 2+4x 3-2x 4=0.
⎪⎩8x 1+7x 2+6x 3-3x 4=0
解 对系数矩阵进行初等行变换, 有
⎛2-3-21⎫r ⎛102/19-1/19⎫
4-2⎪ ~ 0114/19-7/19⎪, A = 35
876-3⎪ 0000⎪⎝⎭⎝⎭
于是得
⎧x 1=-(2/19) x 3+(1/19) x 4
⎨x =-(14/19) x +(7/19) x . ⎩234
取(x 3, x 4) T =(19, 0)T , 得(x 1, x 2) T =(-2, 14)T ; 取(x 3, x 4) T =(0, 19)T , 得(x 1, x 2) T =(1, 7)T . 因此方程组的基础解系为
ξ1=(-2, 14, 19, 0)T , ξ2=(1, 7, 0, 19)T .
(3)nx 1 +(n -1) x 2+ ⋅ ⋅ ⋅ +2x n -1+x n =0. 解 原方程组即为
x n =-nx 1-(n -1) x 2- ⋅ ⋅ ⋅ -2x n -1.
取x 1=1, x 2=x 3= ⋅ ⋅ ⋅ =x n -1=0, 得x n =-n ;
取x 2=1, x 1=x 3=x 4= ⋅ ⋅ ⋅ =x n -1=0, 得x n =-(n -1) =-n +1; ⋅ ⋅ ⋅ ;
取x n -1=1, x 1=x 2= ⋅ ⋅ ⋅ =x n -2=0, 得x n =-2. 因此方程组的基础解系为 ξ1=(1, 0, 0, ⋅ ⋅ ⋅, 0, -n ) T , ξ2=(0, 1, 0, ⋅ ⋅ ⋅, 0, -n +1) T , ⋅ ⋅ ⋅,
ξn -1=(0, 0, 0, ⋅ ⋅ ⋅, 1, -2) T .
23. 设A = R (B ) =2.
解 显然B 的两个列向量应是方程组AB =0的两个线性无关的解. 因为 A =
r
⎛2-213⎫, 求一个4⨯2矩阵B , 使AB =0, 且
⎪
9-528⎝⎭
1/8⎫⎛2-213⎫ ~ ⎛10-1/8
⎪ 01-5/8-11/8⎪,
9-528⎝⎭⎝⎭
所以与方程组AB =0同解方程组为
⎧x 1=(1/8) x 3-(1/8) x 4
⎨x =(5/8) x +(11/8) x . ⎩234
取(x 3, x 4) T =(8, 0)T , 得(x 1, x 2) T =(1, 5)T ; 取(x 3, x 4) T =(0, 8)T , 得(x 1, x 2) T =(-1, 11)T . 方程组AB =0的基础解系为
ξ1=(1, 5, 8, 0)T , ξ2=(-1, 11, 0, 8)T .
⎛1
5
因此所求矩阵为B =
8 0⎝
-1⎫11⎪
. 0⎪8⎪⎭
24. 求一个齐次线性方程组, 使它的基础解系为
ξ1=(0, 1, 2, 3)T , ξ2=(3, 2, 1, 0)T .
解 显然原方程组的通解为
⎛x 1⎫⎧x 1=3k 2⎛0⎫⎛3⎫ x ⎪ 1⎪ 2⎪⎪x 2=k 1+2k 2
2
, 即=k +k ⎪1 ⎪2 ⎪⎨x =2k +k , (k 1, k 2∈R ) ,
21x 312
⎪⎪3 3⎪ 0⎪
⎝⎭⎝⎭⎩x 4=3k 1⎝x 4⎭
消去k 1, k 2得
⎧2x 1-3x 2+x 4=0
⎨x -3x +2x =0, ⎩134
此即所求的齐次线性方程组.
25. 设四元齐次线性方程组 I :
⎧x 1+x 2=0
⎨x -x =0 , II: ⎩24⎧x 1-x 2+x 3=0
⎨x -x +x =0. ⎩234
求: (1)方程I 与II 的基础解系; (2) I与II 的公共解. 解 (1)由方程I 得⎨
⎧x 1=-x 4
.
x =x ⎩24
取(x 3, x 4) T =(1, 0)T , 得(x 1, x 2) T =(0, 0)T ; 取(x 3, x 4) T =(0, 1)T , 得(x 1, x 2) T =(-1, 1)T . 因此方程I 的基础解系为
ξ1=(0, 0, 1, 0)T , ξ2=(-1, 1, 0, 1)T . 由方程II 得⎨
⎧x 1=-x 4
.
x =x -x ⎩234
T
T
T
取(x 3, x 4) T =(1, 0)T , 得(x 1, x 2) T =(0, 1)T ; 取(x 3, x 4) =(0, 1), 得(x 1, x 2) =(-1, -1) . 因此方程II 的基础解系为
ξ1=(0, 1, 1, 0)T , ξ2=(-1, -1, 0, 1)T . (2) I与II 的公共解就是方程
T
⎧x 1+x 2=0⎪x -x 4=0
III : ⎨2
x 1-x 2+x 3=0⎪
⎩x 2-x 3+x 4=0
的解. 因为方程组III 的系数矩阵
⎛1100⎫⎛1 010-1⎪r 0
A = ~
1-110⎪ 0 01-11⎪ 0⎝⎭⎝
所以与方程组III 同解的方程组为
100
01⎫0-1⎪
,
1-2⎪00⎪⎭
x =-x 4⎧⎪1
⎨x 2=x 4.
⎪⎩x 3=2x 4
取x 4=1, 得(x 1, x 2, x 3) =(-1, 1, 2), 方程组III 的基础解系为 ξ=(-1, 1, 2, 1)T .
因此I 与II 的公共解为x =c (-1, 1, 2, 1)T , c ∈R .
26. 设n 阶矩阵A 满足A =A , E 为n 阶单位矩阵, 证明
R (A ) +R (A -E ) =n .
证明 因为A (A -E ) =A 2-A =A -A =0, 所以R (A ) +R (A -E ) ≤n . 又R (A -E ) =R (E -A ) , 可知
R (A ) +R (A -E ) =R (A ) +R (E -A ) ≥R (A +E -A ) =R (E ) =n ,
由此R (A ) +R (A -E ) =n .
27. 设A 为n 阶矩阵(n ≥2) , A *为A 的伴随阵, 证明
2T
T
⎧n 当R (A ) =n
⎪
R (A *)=⎨1 当R (A ) =n -1.
⎪⎩0 当R (A ) ≤n -2
证明 当R (A ) =n 时, |A |≠0, 故有 |AA *|=||A |E |=|A |≠0, |A *|≠0, 所以R (A *)=n .
当R (A ) =n -1时, |A |=0, 故有 AA *=|A |E =0,
即A *的列向量都是方程组A x =0的解. 因为R (A ) =n -1, 所以方程组A x =0的基础解系中只含一个解向量, 即基础解系的秩为1. 因此R (A *)=1.
当R (A ) ≤n -2时, A 中每个元素的代数余子式都为0, 故A *=O , 从而R (A *)=0. 28. 求下列非齐次方程组的一个解及对应的齐次线性方程组的基础解系:
x +x 2=5⎧⎪1
(1)⎨2x 1+x 2+x 3+2x 4=1;
⎪⎩5x 1+3x 2+2x 3+2x 4=3
解 对增广矩阵进行初等行变换, 有
⎛11005⎫r ⎛1010-8⎫
B = 21121⎪ ~ 01-1013⎪.
53223⎪ 00012⎪⎝⎭⎝⎭
与所给方程组同解的方程为
x =-x 3-8⎧⎪1
⎨x 2= x 3+13. ⎪⎩x 4= 2
当x 3=0时, 得所给方程组的一个解η=(-8, 13, 0, 2). 与对应的齐次方程组同解的方程为
T
x =-x 3⎧⎪1
⎨x 2= x 3. ⎪⎩x 4=0
当x 3=1时, 得对应的齐次方程组的基础解系ξ=(-1, 1, 1, 0).
T
x -5x 2+2x 3-3x 4=11⎧⎪1
(2)⎨5x 1+3x 2+6x 3-x 4=-1.
⎪⎩2x 1+4x 2+2x 3+x 4=-6
解 对增广矩阵进行初等行变换, 有
⎛1-52-311⎫r ⎛109/7-1/21⎫ B = 536-1-1⎪ ~ 01-1/71/2-2⎪.
2421-6⎪ 000⎪00⎝⎭⎝⎭
与所给方程组同解的方程为
⎧x 1=-(9/7) x 3+(1/2) x 4+1⎨x =(1/7)x -(1/2) x -2. ⎩234
当x 3=x 4=0时, 得所给方程组的一个解
η=(1, -2, 0, 0)T .
与对应的齐次方程组同解的方程为
⎧x 1=-(9/7) x 3+(1/2) x 4⎨x =(1/7)x -(1/2) x . ⎩234
分别取(x 3, x 4) =(1, 0), (0, 1), 得对应的齐次方程组的基础解系
T
T
T
ξ1=(-9, 1, 7, 0)T . ξ2=(1, -1, 0, 2)T .
29. 设四元非齐次线性方程组的系数矩阵的秩为3, 已知η1, η2, η3是它的三个解向量. 且
η1=(2, 3, 4, 5)T , η2+η3=(1, 2, 3, 4)T ,
求该方程组的通解.
解 由于方程组中未知数的个数是4, 系数矩阵的秩为3, 所以对应的齐次线性方程组的基础解系含有一个向量, 且由于η1, η2, η3均为方程组的解, 由非齐次线性方程组解的结构性质得
2η1-(η2+η3) =(η1-η2) +(η1-η3) = (3, 4, 5, 6)
为其基础解系向量, 故此方程组的通解:
x =k (3, 4, 5, 6)T +(2, 3, 4, 5)T , (k ∈R ) .
30. 设有向量组A : a 1=(α, 2, 10), a 2=(-2, 1, 5), a3=(-1, 1, 4), 及b =(1, β, -1) , 问α, β为何值时
(1)向量b 不能由向量组A 线性表示;
(2)向量b 能由向量组A 线性表示, 且表示式唯一;
(3)向量b 能由向量组A 线性表示, 且表示式不唯一, 并求一般表示式.
T
T
T
T
T
⎛-1-2α1⎫
12β⎪ 解 (a 3, a 2, a 1, b ) = 1
4510-1⎪⎝⎭1⎫⎛-1-2α
~ 0-11+αβ+1⎪. 004+α-3β⎪⎝⎭
r
(1)当α=-4, β≠0时, R (A ) ≠R (A , b ) , 此时向量b 不能由向量组A 线性表示.
(2)当α≠-4时, R (A ) =R (A , b ) =3, 此时向量组a 1, a 2, a 3线性无关, 而向量组a 1, a 2, a 3, b 线
性相关, 故向量b 能由向量组A 线性表示, 且表示式唯一.
(3)当α=-4, β=0时, R (A ) =R (A , b ) =2, 此时向量b 能由向量组A 线性表示, 且表示式不唯一.
当α=-4, β=0时,
⎛-1-2-41⎫
(a 3, a 2, a 1, b ) = 1120⎪
4510-1⎪⎝⎭
方程组(a 3, a 2, a 1) x =b 的解为
⎛10-21⎫
~ 013-1⎪, 0000⎪⎝⎭
r
⎛x 1⎫⎛2⎫⎛1⎫⎛2c +1⎫
x 2⎪=c -3⎪+ -1⎪= -3c -1⎪, c ∈R .
⎪ 1⎪ 0⎪ c ⎪
⎭⎝⎭⎝⎭⎝x 3⎭⎝
因此 b =(2c +1) a 3+(-3c -1) a 2+c a 1, 即 b = ca 1+(-3c -1) a 2+(2c +1) a 3, c∈R .
31. 设a =(a 1, a 2, a 3) T , b =(b 1, b 2, b 3) T , c=(c 1, c 2, c 3) T , 证明三直线 l 1: a 1x +b 1y +c 1=0,
l 2: a 2x +b 2y +c 2=0, (a i +b i ≠0, i =1, 2, 3) l 3: a 3x +b 3y +c 3=0,
相交于一点的充分必要条件为: 向量组a , b 线性无关, 且向量组a , b , c 线性相关. 证明 三直线相交于一点的充分必要条件为方程组
2
2
a x +b 1y +c 1=0a x +b 1y =-c 1⎧⎧⎪1⎪1
⎨a 2x +b 2y +c 2=0, 即⎨a 2x +b 2y =-c 2 ⎪⎪⎩a 3x +b 3y +c 3=0⎩a 3x +b 3y =-c 3
有唯一解. 上述方程组可写为x a +y b =-c . 因此三直线相交于一点的充分必要条件为c 能由a , b 唯一线性表示, 而c 能由a , b 唯一线性表示的充分必要条件为向量组a , b 线性无关, 且向量组a , b , c 线性相关.
32. 设矩阵A =(a 1, a 2, a 3, a 4) , 其中a 2, a 3, a 4线性无关, a 1=2a 2- a 3. 向量b =a 1+a 2+a 3+a 4, 求方程A x =b 的通解.
解 由b =a 1+a 2+a 3+a 4知η=(1, 1, 1, 1)T 是方程A x =b 的一个解. 由a 1=2a 2- a 3得a 1-2a 2+a 3=0, 知ξ=(1, -2, 1, 0)T 是A x =0的一个解.
由a 2, a 3, a 4线性无关知R (A ) =3, 故方程A x =b 所对应的齐次方程A x =0的基础解系中含
一个解向量. 因此ξ=(1, -2, 1, 0)T 是方程A x =0的基础解系. 方程A x =b 的通解为
x =c (1, -2, 1, 0)T +(1, 1, 1, 1)T , c ∈R .
33. 设η*是非齐次线性方程组A x =b 的一个解, ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r ,是对应的齐次线性方程组的一个基础解系, 证明:
(1)η*, ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性无关; (2)η*, η*+ξ1, η*+ξ2, ⋅ ⋅ ⋅, η*+ξn -r 线性无关.
证明 (1)反证法, 假设η*, ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性相关. 因为ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性无关, 而
η*, ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性相关, 所以η*可由ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性表示, 且表示式是唯一的, 这说
明η*也是齐次线性方程组的解, 矛盾.
(2)显然向量组η*, η*+ξ1, η*+ξ2, ⋅ ⋅ ⋅, η*+ξn -r 与向量组η*, ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 可以相互表示, 故这两个向量组等价, 而由(1)知向量组η*, ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性无关, 所以向量组η*, η*+ξ1,
η*+ξ2, ⋅ ⋅ ⋅, η*+ξn -r 也线性无关.
34. 设η1, η2, ⋅ ⋅ ⋅, ηs 是非齐次线性方程组A x =b 的s 个解, k 1, k 2, ⋅ ⋅ ⋅, k s 为实数, 满足k 1+k 2+ ⋅ ⋅ ⋅ +k s =1. 证明
x =k 1η1+k 2η2+ ⋅ ⋅ ⋅ +k s ηs
也是它的解.
证明 因为η1, η2, ⋅ ⋅ ⋅, ηs 都是方程组A x =b 的解, 所以 A ηi =b (i =1, 2, ⋅ ⋅ ⋅, s ) ,
从而 A (k 1η1+k 2η2+ ⋅ ⋅ ⋅ +k s ηs ) =k 1A η1+k 2A η2+ ⋅ ⋅ ⋅ +k s A ηs =(k 1+k 2+ ⋅ ⋅ ⋅ +k s ) b =b . 因此x =k 1η1+k 2η2+ ⋅ ⋅ ⋅ +k s ηs 也是方程的解.
35. 设非齐次线性方程组A x =b 的系数矩阵的秩为r , η1, η2, ⋅ ⋅ ⋅, ηn -r +1是它的n -r +1个线性无关的解. 试证它的任一解可表示为
x =k 1η1+k 2η2+ ⋅ ⋅ ⋅ +k n -r +1ηn -r +1, (其中k 1+k 2+ ⋅ ⋅ ⋅ +k n -r +1=1).
证明 因为η1, η2, ⋅ ⋅ ⋅, ηn -r +1均为A x =b 的解, 所以ξ1=η2-η1, ξ2=η3-η1, ⋅ ⋅ ⋅, ξn -r =η
n -r +1-1均为
η
A x =b 的解.
用反证法证: ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性无关.
设它们线性相关, 则存在不全为零的数λ1, λ2, ⋅ ⋅ ⋅, λn -r , 使得 λ1ξ1+ λ2ξ2+ ⋅ ⋅ ⋅ + λ n-r ξ n-r =0,
即 λ1(η2-η1) + λ2(η3-η1) + ⋅ ⋅ ⋅ + λ n-r (ηn -r +1-η1) =0, 亦即 -(λ1+λ2+ ⋅ ⋅ ⋅ +λn -r ) η1+λ1η2+λ2η3+ ⋅ ⋅ ⋅ +λ n-r ηn -r +1=0, 由η1, η2, ⋅ ⋅ ⋅, ηn -r +1线性无关知
-(λ1+λ2+ ⋅ ⋅ ⋅ +λn -r ) =λ1=λ2= ⋅ ⋅ ⋅ =λn -r =0,
矛盾. 因此ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性无关. ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 为A x =b 的一个基础解系.
设x 为A x =b 的任意解, 则x -η1为A x =0的解, 故x -η1可由ξ1, ξ2, ⋅ ⋅ ⋅, ξn -r 线性表出, 设 x -η1=k 2ξ1+k 3ξ2+ ⋅ ⋅ ⋅ +k n -r +1ξn -r
=k 2(η2-η1) +k 3(η3-η1) + ⋅ ⋅ ⋅ +k n -r +1(ηn -r +1-η1) , x =η1(1-k 2-k 3 ⋅ ⋅ ⋅ -k n -r +1) +k 2η2+k 3η3+ ⋅ ⋅ ⋅ +k n-r +1ηn -r +1. 令k 1=1-k 2-k 3 ⋅ ⋅ ⋅ -k n -r +1, 则k 1+k 2+k 3 ⋅ ⋅ ⋅ -k n -r +1=1, 于是 x =k 1η1+k 2η2+ ⋅ ⋅ ⋅ +k n -r +1ηn -r +1.
36. 设
V 1={x =(x 1, x 2, ⋅ ⋅ ⋅, x n ) T | x 1, ⋅ ⋅ ⋅, x n ∈R 满足x 1+x 2+ ⋅ ⋅ ⋅ +x n =0}, V 2={x =(x 1, x 2, ⋅ ⋅ ⋅, x n ) | x 1, ⋅ ⋅ ⋅, x n ∈R 满足x 1+x 2+ ⋅ ⋅ ⋅ +x n =1},
问V 1, V 2是不是向量空间？为什么？ 解 V 1是向量空间, 因为任取
α=(a 1, a 2, ⋅ ⋅ ⋅, a n ) T ∈V 1, β=(b 1, b 2, ⋅ ⋅ ⋅, b n ) T ∈V 1, λ∈∈R , 有 a 1+a 2+ ⋅ ⋅ ⋅ +a n =0, b 1+b 2+ ⋅ ⋅ ⋅ +b n =0,
从而 (a 1+b 1) +(a 2+b 2) + ⋅ ⋅ ⋅ +(a n +b n ) =(a 1+a 2+ ⋅ ⋅ ⋅ +a n ) +(b 1+b 2+ ⋅ ⋅ ⋅ +b n ) =0, λa 1+λa 2+ ⋅ ⋅ ⋅ +λa n =λ(a 1+a 2+ ⋅ ⋅ ⋅ +a n ) =0, 所以 α+β=(a 1+b 1, a 2+b 2, ⋅ ⋅ ⋅, a n +b n ) T ∈V 1, λα=(λa 1, λa 2, ⋅ ⋅ ⋅, λa n ) T ∈V 1. V 2不是向量空间, 因为任取
α=(a 1, a 2, ⋅ ⋅ ⋅, a n ) T ∈V 1, β=(b 1, b 2, ⋅ ⋅ ⋅, b n ) T ∈V 1, 有 a 1+a 2+ ⋅ ⋅ ⋅ +a n =1, b 1+b 2+ ⋅ ⋅ ⋅ +b n =1,
从而 (a 1+b 1) +(a 2+b 2) + ⋅ ⋅ ⋅ +(a n +b n ) =(a 1+a 2+ ⋅ ⋅ ⋅ +a n ) +(b 1+b 2+ ⋅ ⋅ ⋅ +b n ) =2,
T
所以 α+β=(a 1+b 1, a 2+b 2, ⋅ ⋅ ⋅, a n +b n ) T ∉V 1.
37. 试证: 由a 1=(0, 1, 1), a 2=(1, 0, 1), a3=(1, 1, 0)所生成的向量空间就是R . 证明 设A =(a 1, a2, a 3) , 由
T
T
T
3
011
|A |=101=-2≠0,
110
知R (A ) =3, 故a 1, a2, a 3线性无关, 所以a 1, a2, a 3是三维空间R 3的一组基, 因此由a 1, a2, a 3所生成的向量空间就是R .
38. 由a 1=(1, 1, 0, 0) T , a 2=(1, 0, 1, 1) T 所生成的向量空间记作V 1, 由b 1=(2, -1, 3, 3) T , b 2=(0, 1, -1, -1) 所生成的向量空间记作V 2, 试证V 1=V 2. 证明 设A =(a 1, a2) , B =(b 1, b2) . 显然R (A ) =R (B ) =2, 又由
T
3
⎛1 1
(A , B ) =
0 0⎝
10112-133
0⎫⎛11⎪r 0 ~ -1⎪ 0
0-1⎪⎭⎝
1
-1002-300
0⎫1⎪, 0⎪0⎪⎭
知R (A , B ) =2, 所以R (A ) =R (B ) =R (A , B ) , 从而向量组a 1, a2与向量组b 1, b2等价. 因为向量组a 1, a2与向量组b 1, b2等价, 所以这两个向量组所生成的向量空间相同, 即V 1=V 2.
39. 验证a 1=(1, -1, 0) T , a 2=(2, 1, 3) T , a 3=(3, 1, 2) T 为R 3的一个基, 并把v 1=(5, 0, 7) T , v 2=(-9, -8, -13) T 用这个基线性表示. 解 设A =(a 1, a2, a3) . 由
123
|(a 1, a 2, a 3) |=-111=-6≠0,
032
知R (A ) =3, 故a 1, a2, a3线性无关, 所以a 1, a2, a3为R 的一个基. 设x 1a 1+x 2a 2+x 3a 3=v 1, 则
3
x +2x 2+3x 3=5⎧⎪1
⎨-x 1+x 2+x 3=0, ⎪⎩3x 2+2x 3=7
解之得x 1=2, x 2=3, x 3=-1, 故线性表示为v 1=2a 1+3a 2-a 3. 设x 1a 1+x 2a 2+x 3a 3=v 2, 则
x +2x 2+3x 3=-9⎧⎪1
⎨-x 1+x 2+x 3=-8, ⎪⎩3x 2+2x 3=-13
解之得x 1=3, x 2=-3, x 3=-2, 故线性表示为v 2=3a 1-3a 2-2a 3.
40. 已知R 的两个基为
a 1=(1, 1, 1), a 2=(1, 0, -1) , a 3=(1, 0, 1), b 1=(1, 2, 1)T , b 2=(2, 3, 4)T , b 3=(3, 4, 3)T . 求由基a 1, a 2, a 3到基b 1, b 2, b 3的过渡矩阵P . 解 设e 1, e 2, e 3是三维单位坐标向量组, 则
T
T
T
3
⎛111⎫
(a 1, a 2, a 3) =(e 1, e 2, e 3) 100⎪,
1-11⎪⎝⎭
⎛111⎫
(e 1, e 2, e 3) =(a 1, a 2, a 3) 100⎪,
1-11⎪⎝⎭⎛123⎫
于是 (b 1, b 2, b 3) =(e 1, e 2, e 3) 234⎪
143⎪⎝⎭
-1
⎛111⎫⎛123⎫
=(a 1, a 2, a 3) 100⎪ 234⎪,
1-11⎪ 143⎪⎝⎭⎝⎭
由基a 1, a 2, a 3到基b 1, b 2, b 3的过渡矩阵为
-1
⎛111⎫⎛123⎫⎛234⎫
P = 100⎪ 234⎪= 0-10⎪. 1-11⎪ 143⎪ -10-1⎪⎝⎭⎝⎭⎝⎭
-1