高代课后习题详解第3章 线性方程组的进一步理论

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