高代课后习题详解第3章 线性方程组的进一步理论
发布时间：2024-11-02
丘维声编写的《高等代数》的课后习题详解
(
)
3
.1.P64,Ex3(1)
β
α1,α2,α3
c1α1+c2α2+c3α3=β 1c1+2c2 4c3=8 3c1+c3=3
7c2 2c3= 1 5c1 3c2+6c3= 25
c1,c2,c3
c1,c2,c3
丘维声编写的《高等代数》的课后习题详解
β=k1α1+k2α2+...+ksαs+l1α1+l2α2+...+lsαs=2)α2+...+(ks+ls)αs (k1+l1)α1+(k2+l k1+l1=k1l1=0 k2+l2=k2 l2=0
............ ...... k+l=k l=0
ssss
α1,α2,...,αs
β
α1,α2,...,αs
β
α1,α2,...,αs
α1,α2,...,αs
k1,k2,...,ks∈K
β=k1α1+k2α2+...+ksαsβ=l1α1+l2α2+...+lsαs
l1,l2,...,ls∈K
k1α1+k2α2+...+ksαs (l1α1+l2α2+...+lsαs)=0 (k1 l1)α1+(k2 l2)α2+...+(ks ls)αs=0 k1 l1 k l
22.. . k lss
α1,α2,...,αs
.3.P73,Ex8
=0l1 l=02 ..... .... l=0s
=k1=
...
k2
...
=ks
.
.
α1,...,αi 1,β,αi+1,...,αs0.
k1,...,ki 1,l,ki+1...,ks∈K
l=0.
k1α1+...+ki 1αi 1+lβ+ki+1αi+1+...+ksαs=0
l=0
k1,...,ki 1,l,ki+1...,ks
k1,...,ki 1,ki+1...,ks
α1,α2,...,αs
k1α1+...+ki 1αi 1+ki+1αi+1+...+ksαs=0
.
.
l=0.
β=b1α1+...+biαi+...+bsαsbi=0
k1α1+...+ki 1αi 1+l(b1α1+...+biαi+...+bsαs)+ki+1αi+1+...+ksαs=0 (k1+lb1)α1+...+(ki 1+lbi 1)αi 1+lbiαi+(ki+1+lbi+1)αi+1+...+(ks+
2
丘维声编写的《高等代数》的课后习题详解
lbs)αs=0
lbi=0
αi
α1,...,αi 1,αi+1,...,αs
.
..
αi(1<i≤s)
.4.P73,Ex9
α1,α2,...,αs
.
.
α1,α2,...,αs
.
α1,α2,...,αs
α1
.
α1,α2,...,αs
K(1<k≤s)
α1,...,αk 1
α1,...,αk 1,αk
(
k
α1,α2,...,αs
).αk
α1,...,αk 1
.
.
..
.
a11...
···
...
a1r
...
=0
.5.P73,Ex10
.
β1,β2,...,βr
ar1···arr
.
x1(a11α1+...+a1rαr)+...+xr(ar1α1+...+arrαr)=0 (a11x1+...+ar1xr)α1+...+(a1rx1+...+arrxr)αr=0
β1
... β
x1β1+x2β2+...+xrβr=0
=a11α1+...+a1rαr......
r
=ar1α1+...+arrαr
x1β1+x2β2+...+xrβr=0
α1,α2,...,αr
(a11x1+...+ar1xr)α1+...+(a1rx1+...+arrxr)αr=0
a11x1+···+ar1xr=0............ ax+···+ax=0
1r1
rrr
3
丘维声编写的《高等代数》的课后习题详解
x1,...,xr
a11...···
...a1r
...
T =0,
a11x+···
+a=0=
.1
..r1xr
. .... ...
..
.
ar1···arr
...
a1rx
1
+···+a=0 rrxr
x..1
.β1,β2,...,βr .
xr=0
.
.
β1,β2,...,βr
a11···
a .... .
.
.1r
..
a r1···arr
=0. β=
a
.1
.11α1+...+a .. ...1rαr
..
2+...+xrβr=0
x β
r
=ar1α1+...+arrαr
x1β1+x2β1(a11α1+...+a1rαr)+...+xr(ar1α1+...+arrαr)=0 (a11x1+...+ar1xr)α1+...+(a1rx1+...+arrxr)αr=0
α1,α2,...,αr
(
a11x1+...+ar1xr)α1+...+(a1rx1+...+arrxr)αr=0 a11x1+···+ar1xr=0. . .......... a1rx1+···+arrxr
=0
a11x1+···+ar1xr=0β1,β2,...,βr ........0
.... a1rx1+···+arrxr
=0
a .11···a1r..... ... a r1···arr .6.P 79,Ex =0.
3
α1,α2,...,αs∈Kn
rαi1,...,αir
αj1,...,αjr
.
α1,α2,...,αsr
4
.
丘维声编写的《高等代数》的课后习题详解
.
αj1,...,αjr
αt∈{α1,α2,...,αs}α1,α2,...,αs
αj1,...,αjr,αt
.
αj1,...,αjr,αt
αj1,...,αjr,αt
α1,α2,...,αs
α1,α2,...,αs
αi1,...,αir
αj1,...,αjr,αt
αi1,...,αir
.
αj1,...,αjr,αt
..
.7.P79,Ex6
α1,α2,...,αn
α1,α2,...,αn
n
.
Kn
ε1,ε2,...,εn
n
ε1,ε2,...,εn
α1,α2,...,αn
ε1,ε2,...,εn
.
α1,α2,...,αn
α1,α2,...,αn
ε1,ε2,...,εn
.
rαi1,...,αir
r
.8.P79,Ex7
α1,α2,...,αs∈Kn
α1,α2,...,αs
αi1,...,αir
.
αi1,...,αir
α1,α2,...,αs
αi1,...,αir
αi1,...,αir
r
α1,α2,...,αs
αi1,...,αir
αi1,...,αir
α1,α2,...,αs
αi1,...,αir
α1,α2,...,αs
αi1,...,αir
r
β
.9.P79,Ex8
x1α1+x2α2+...+xnαn=β
α1,α2,...,αn
β
x1α1+x2α2+...+xnαn=β
β∈Kn
α1,α2,...,αn
Ex5,Ex6
β∈Kn
α1,α2,...,αn
α1,α2,...,αn
α1,α2,...,αn
|A|=0
.10.P79,Ex9
r
αi1,...,αir
α1,α2,...,αs
αi1,...,αir
α1,α2,...,αs,β
α1,α2,...,αs,β
r
αi1,...,αir
α1,α2,...,αs,β
α1,α2,...,αs,β
αi1,...,αir
αi1,...,αir
5
丘维声编写的《高等代数》的课后习题详解
α1,α2,...,αs
α1,α2,...,αs,β
α1,α2,...,αs
β
α1,α2,...,αs
βj1,βj2,...,βjm
.11.P79,Ex10
αi1,...,αir
α1,α2,...,αs
β1,β2,...,βt
α1,α2,...,αs,β1,β2,...,βt
αi1,...,αir,βj1,βj2,...,βjm
rank{α1,α2,...,αs,β1,β2,...,βt}≤rank{αi1,...,αir,βj1,βj2,...,βjm}≤r+m=rank{α1,α2,...,αs}+rank{β1,β2,...,βt}
β1,...,βr
.12.P83,Ex4
U
Kn
r
U
α1,...,αr
β1,...,βr
U
α1,...,αr
α1,...,αr
β1,...,βr
β1,...,βr
r
α1,...,αr
r
r
U
.13.P83,Ex5
U
rα1,...,αs
U
s
r=s
<α1,...,αs>=U
α1,...,αs(s=r)
U
r<s
<α1,...,αs>=U
r s
r s=1
<α1,...,αs>=U
αs+1∈Uαs+1
∈<α1,...,αs>
αs+1
α1,...,αs
α1,...,αs,αs+1
<α1,...,αs,αs+1>
s+1
r rank{α1,...,αs,αs+1}=k
αs+2,...,αr
:α1,...,αs,αs+1,αs+2,...,αr
U
.14.P89,Ex3
A
6
丘维声编写的《高等代数》的课后习题详解
2 7A=
13 0 1 34
5201 → 13 2 7
1 0 1 34
0 131036
11013
2 7
→
14
13 0 1 34
4
00
49
16 0000 r(A)=3A
A
3A
.15.P89,Ex4
12
A= 1λ 12 2 1λ5 → 1λ
λ+21
5 12
1
λ
→ 61 12 1λ 0 1 2λ
1100 01λ+2 1 2λ 10 λ1
1λ+2 1 2λ
0 1 510 λ0λ 39 3λ
λ=3
r(A)=2
→λ= 0
3
r(A)=3
.16.P89,Ex6
A
4
1im
i
2m
i
3m
1111
A1234
1im+1i2(m+1)i3(m+1) 1i
i2i3
1234 =
1im+2i2(m+2)i3(m+2) 1im+3i2(m+3)i3(m+3)
=i6m
1i2i4i6
1i3i6i9
4
4
.17.P90,Ex8
A1
αk1,αk2,...,αkl
A
A
r(A)=r
r l
A1
s
7
丘维声编写的《高等代数》的课后习题详解
m s
r l≤m s l≥r+s m
A1
r+s m
n2 (n 1)
.18.P90,Ex9,10
n0
n 1
00
.19.P90,Ex11
(1)A
|A|=0
A
β1,β2,...,βn
k1,k2,...,kn
k1β1+k2β2+...+knβn=0
|ki |
|k1|,|k2|,...,|kn|
|ki |=0
k1β1+k2β2+...+knβn=0
k1ai 1+k2ai 2+...+ki ai i +...+knai n=0
ai n=ai i ai n|≤
n
k1ai 1+...+ki 1ai i 1+ki +1ai i +1+...+knai n= ki ai i
k1
ki ki
k1
ai i 1+
ki +1
ki +1 ki ki +1
|
ki
|,||
1
|
k1
ki
ki
ai i 1+
ki
|
kl
||ai l|≤
l=i
(2)
(1 )|A(t)|(2)
t∈[ 1,1]
|A|0 =
a11
at 21
|A(t)|= .
. .
an1t
n
l=i
|ai l|<|ai i |
t
akk>
a12t···a1nt
a22···a2nt
.... .···.
an2t···ann
l=k
(3 )
n
akk>0
t∈[ 1,1]
n
|akl|≥
l=k
n
|aklt|
|A(t)|=0
|A(0)|=
|A|=|A(1)|>0
k=1
.20.P92,Ex1,2,3
(1)
8
丘维声编写的《高等代数》的课后习题详解
|A|=1im
i
2m
i3m
1im+1i2(m+1)i3(m+1)
=0
(2) +2
1i
mi
2(m+2)i
3(m+2)
1im+3i2(m+3)i3(m+3)
a···as 1
|A|= 1 1a2···
a2(s 1)
.
.... s.···..
as···as(s 1)
=0 .s
r(
1
A=
1111
abcd
3
a2b2c2d2a3b3c3d3111
r(A)=3
abc a2b2c2 =0
A
B
A
A
r(B)
A)=r(B)
r(A)=r(
s
r(A)=
3
r(A)=r(
r(
丘维声编写的《高等代数》的课后习题详解
γ1,γ2,...,γs
γ1,γ2,...,γs
ηη
η1,η2,...,ηtη
η1,η2,...,ηt
γ1,γ2,...,γs
γ1,γ2,...,γs
γ1,γ2,...,γs
.23.P99,Ex3
n
r
n r
n r
.24.P99,Ex4
n
r
n r
n r
rank{δ1,δ2,...,δm}≤n r
.25.P99,Ex5
A
Akl=0
A
n 1
|A|=0
r(A)=n 1|A|(=0)i=kai1Ak1+ai2Ak2+...+ainAkn=
i=k=0,
i=1,2,...,n
η1
η1=0
η1
.26.P99,Ex6
B
A
A
B
A
Ex5
Dj
A
(n,j)
Ex5
.27.P99,Ex7
A1
A
10
0
丘维声编写的《高等代数》的课后习题详解
n r(A1)=n r(A)r(A1)=r(A)
A1
A
s
A1
A
A
s
A1
.28.P103,Ex1,(1)
7k1
+
17k1
1
72
1
1
丘维声编写的《高等代数》的课后习题详解
.29.P104,Ex2
n
n
n
12
高代课后习题详解第3章 线性方程组的进一步理论.doc
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