dstt cnttk6

8/7/2019 DSTT CNTTK6

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I S TUYN TNH

Bin son: Nguyn Minh Tr

Ngy 14 thng 2 nm 2011


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Nguyn Minh Tr 2


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Mc lc

1 Ma trn v nh thc 71.1 Mt s nh ngha Cc php ton trn ma trn . . . . . . . . . . . . 7

1.1.1 Mt s nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . 71.1.2 Cc php ton trn ma trn . . . . . . . . . . . . . . . . . . . . 81.1.3 Cc php bin i s cp trn ma trn . . . . . . . . . . . . . . 10

1.2 nh thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.2.1 nh thc cp 1, 2, 3 . . . . . . . . . . . . . . . . . . . . . . . . 101.2.2 Hon v . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.2.3 nh thc cp n . . . . . . . . . . . . . . . . . . . . . . . . . . 111.2.4 Cc tnh cht ca nh thc . . . . . . . . . . . . . . . . . . . . 121.2.5 nh l Laplace . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.3 Ma trn nghch o . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.3.1 Ma trn nghch o . . . . . . . . . . . . . . . . . . . . . . . . . 141.3.2 Cch tm ma trn nghch o . . . . . . . . . . . . . . . . . . . 15

1.4 Hng ca ma trn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.4.1 Khi nim v hng ca ma trn . . . . . . . . . . . . . . . . . . 161.4.2 Cch tnh hng ca ma trn . . . . . . . . . . . . . . . . . . . . 17

2 H phng trnh tuyn tnh 192.1 Khi nim tng qut . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.1.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.1.2 Tnh cht nghim . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.2 H Cramer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.2.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.2.2 Cng thc Cramer . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.3 H phng trnh tuyn tnh tng qut. Phng php Gauss . . . . . . . 20

2.3.1 H phng trnh tuyn tnh tng qut. Phng php Gauss . . 202.3.2 H phng trnh tuyn tnh thun nht . . . . . . . . . . . . . . 22

3 Khng gian vect 253.1 Cc khi nim c bn v khng gian vect . . . . . . . . . . . . . . . . 25

3.1.1 nh ngha khng gian vect . . . . . . . . . . . . . . . . . . . . 253.1.2 Cc v d khng gian vect . . . . . . . . . . . . . . . . . . . . 253.1.3 Cch tnh cht . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.2 S ph thuc tuyn tnh v c lp tuyn tnh . . . . . . . . . . . . . . 263.2.1 T hp tuyn tnh . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.2.2 S ph thuc tuyn tnh . . . . . . . . . . . . . . . . . . . . . . 273.2.3 S c lp tuyn tnh . . . . . . . . . . . . . . . . . . . . . . . . 273.3 Hng ca mt h vect . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.3.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3


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3.3.2 Cch tnh hng ca mt h vect . . . . . . . . . . . . . . . . . 283.4 C s, s chiu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.4.1 C s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.4.2 S chiu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.4.3 Ma trn chuyn c s . . . . . . . . . . . . . . . . . . . . . . . . 293.4.4 Quan h gia cc ta ca cng mt vect i vi cc c s

khc nhau . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.5 Khng gian vect con . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.5.1 nh ngha v v d . . . . . . . . . . . . . . . . . . . . . . . . . 303.5.2 Khng gian vect sinh bi mt tp . . . . . . . . . . . . . . . . 313.5.3 Khng gian con cc nghim ca h phng trnh tuyn tnh thun

nht . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4 nh x tuyn tnh 334.1 nh ngha v v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

4.1.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

4.1.2 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.1.3 Cc tnh cht c bn ca nh x tuyn tnh . . . . . . . . . . . 34

4.2 Ma trn ca nh x tuyn tnh . . . . . . . . . . . . . . . . . . . . . . 344.2.1 nh l c bn v s xc nh ca nh x tuyn tnh . . . . . . . 344.2.2 Ma trn ca nh x tuyn tnh . . . . . . . . . . . . . . . . . . 344.2.3 Biu thc ta ca nh x tuyn tnh . . . . . . . . . . . . . . 364.2.4 Ma trn ca nh x tuyn tnh trong cc c s khc nhau . . . 36

4.3 Ht nhn v nh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364.3.1 Cc khi nim c bn . . . . . . . . . . . . . . . . . . . . . . . . 364.3.2 Nhn xt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4.4 n cu, ton cu, ng cu . . . . . . . . . . . . . . . . . . . . . . . . 384.4.1 Cc nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . 384.4.2 Cc nh l v n cu, ton cu, ng cu . . . . . . . . . . . . 38

4.5 Gi tr ring, vect ring ca ma trn, nh x tuyn tnh . . . . . . . . 394.5.1 Vect ring v gi tr ring ca ma trn . . . . . . . . . . . . . . 394.5.2 Vect ring v gi tr ring ca php bin i tuyn tnh . . . . 41

4.6 Cho ha ma trn, cho ha nh x tuyn tnh . . . . . . . . . . . . . . 424.6.1 iu kin cho ha . . . . . . . . . . . . . . . . . . . . . . . . . 424.6.2 Cch cho ha ma trn v php bin i tuyn tnh . . . . . . . 42

5 Dng song tuyn tnh - Dng ton phng 475.1 Cc khi nim c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475.1.1 Dng song tuyn tnh . . . . . . . . . . . . . . . . . . . . . . . . 475.1.2 Dng ton phng . . . . . . . . . . . . . . . . . . . . . . . . . 475.1.3 Dng chnh tc . . . . . . . . . . . . . . . . . . . . . . . . . . . 485.1.4 Php i bin s . . . . . . . . . . . . . . . . . . . . . . . . . . 48

5.2 a dng ton phng v dng chnh tc . . . . . . . . . . . . . . . . . 485.2.1 Phng php Lagrange . . . . . . . . . . . . . . . . . . . . . . . 485.2.2 Phng php Jacobi . . . . . . . . . . . . . . . . . . . . . . . . 505.2.3 Lut qun tnh . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

5.3 Khng gian Euclide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525.3.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525.3.2 di vect , gc gia cc vect . . . . . . . . . . . . . . . . . 535.3.3 C s trc giao, c s trc chun . . . . . . . . . . . . . . . . . 53

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5.3.4 Cho ha ma trn i xng bng ma trn trc giao . . . . . . . 545.3.5 a dng ton phng v dng chnh tc bng php bin i

trc giao . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565.4 ng dng vo vic kho st ng v mt bc hai . . . . . . . . . . . . 57

5.4.1 nh ngha ng v mt bc hai . . . . . . . . . . . . . . . . . 57

5.4.2 Cc ng v mt bc hai c bn . . . . . . . . . . . . . . . . . 575.4.3 Cch nhn dng ng mt bc hai . . . . . . . . . . . . . . . . 58

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MC LC

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Chng 1Ma trn v nh thc

1.1 Mt s nh ngha Cc php ton trn matrn

1.1.1 Mt s nh ngha

Ma trn cp m n(m, n N) trn trng K (K l R hay C) l mt bng gm m ns c sp xp thnh m dng, n ct theo mt th t nht nh:

Ma trn A cp m n c vit di dng

A =

a11 a12 . . . a1na21 a22 . . . a2n

...... . . .

...

am1 am2 . . . amn

Ta cng k hiu ma trn A c m hng v n ct l A = (aij)mn; phn t nm hng th i, ct th j ca ma trn A c k hiu l aij

Tp hp cc ma trn cp m n trn K k hiu l Mmn(K)Ma trn cp n n c gi l ma trn vung cp n.Ma trn cp m 1 c gi l ma trn ct

A =

12

45

Ma trn cp 1 n c gi l ma trn dng.

B =

1 3 4 5

Ma trn khng nu mi phn t ca n u bng 0.Ma trn n v l ma trn vung m cc phn t aii = 1 cn cc phn t cn li

bng 0, k hiu In

In =

1 0 . . . 00 1 . . . 0...

... . . ....

0 0 . . . 1

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1.1. MT S NH NGHA CC PHP TON TRN MA TRN

Ma trn cho l ma trn vung m cc phn t aij = 0 vi i = j

A =

a1 0 . . . 00 a2 . . . 0...

... . . ....

0 0 . . . an

Cc ma trn vung cp n c dnga11 a12 . . . a1n0 a22 . . . a2n...

... . . ....

0 0 . . . ann

hay

a11 0 . . . 0a21 a22 . . . 0

...... . . .

...an1 an2 . . . ann

c gi l ma trn tam gic trn v ma trn tam gic di.

1.1.2 Cc php ton trn ma trn

1. Ma trn bng nhau Hai ma trn A = (aij)mn v B = (bij)mn c gi l

bng nhau nu chng c cng cp m n, v cc v tr tng ng cng bng nhauaij = bij, 1 i m, 1 j n

2. Ma trn chuyn v Cho ma trn A = (aij)mn, ma trn chuyn v ca A k hiul AT = (bij)nm vi

bij = aji

A =

1 2 5

9 7 4

AT =

1 92 75 4

3. Nhn mt s vi mt ma trn Cho A Mmn(K) v K

A = (aij)mn

Tnh cht:

1. 1.A = A

2. 0.A = 0

3. .0 = 0, K

4. (.A) = ()A, , K

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CHNG 1. MA TRN V NH THC

4. Cng hai ma trn A = (aij)mn v ma trn B = (bij)mn cng cp m n. Khi

A + B = (aij + bij)mn

V d:

A = 1 2 34 5 0 ; B = 1 3 52 4 6Khi

A + B =

2 5 86 9 6

Cc tnh cht

1. A + B = B + A

2. (A + B) + C = A + (B + C)

3. (u + v)A = uA + vA vi u, vK

4. u(A + B) = uA + uB vi u K5. (uA + vB)T = uAT + vBT vi u, v K

5. Php nhn hai ma trn Cho ma trn A = (aij)mn v B = (bij)np. Tch cahai ma trn A v B l mt ma trn cp m p, k hiu A.B = (cik)mp xc nh nhsau:

cik =n

j=1aijbjk

V d: Cho hai ma trn

A =

1 2 3 45 6 7 89 10 11 12

v B =

13 1415 1617 1819 20

Ta c ma trn C = A.B l ma trn cp 3 2 nh sau:

Phn t c11: ly cc phn t dng 1 ca ma trn A nhn tng ng vi cc phnt ca ct 1 ca ma trn B, sau cng tt c li vi nhau.

c11 = 1.13 + 2.15 + 3.17 + 4.19 = 170

Phn t c12: ly cc phn t dng 1 ca ma trn A nhn tng ng vi cc phnt ca ct 2 ca ma trn B, sau cng tt c li vi nhau.

c12 = 1.14 + 2.16 + 3.18 + 4.20 = 180

Tng t nh vy ta tnh c cc phn t cn li.

Vy C = A.B = 170 180426 452

682 724Tnh cht

1. (A.B).C = A.(B.C)

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1.2. NH THC

2. I.A = A

3. A(B + C) = AB + AC

4. (A + B)C = AC+ BC

5. (AB)T

= B

T

A

T

Ch : Php nhn ma trn khng c tnh giao hon.

1.1.3 Cc php bin i s cp trn ma trn

Ta gi php bin i s cp trn ma trn l php bin i c mt trong cc dng sau:

1. hi hj(ci cj): i ch hai hng (ct) cho nhau2. hi a.hi(ci a.ci), a = 0: nhn vo hng th i (ct i) vi s a = 0

3. hi hi + a.hj(ci ci + a.cj): nhn mt s a vo hng hj ri cng vo hng hi(nhn mt s a vo ct cj ri cng vo ci)Ta dng k hiu A B ch ma trn B nhn c t A sau hu hn cc php bini s cp trn A.

Ma trn bc thang l ma trn c dng

a11 a12 . . . . . . a1n0 a22 . . . . . . a2n...

... arr... arn

0 0 . . . . . . 00 0 . . . . . . 0

B Mi ma trn u c th a v dng bc thang nh cc php bin i s cp.

V d: Dng cc php bin i s cp a ma trn A v dng bc thang

A =

1 2 4 31 3 1 12 5 3 2

Gii:

A =

1 2 4 31 3 1 12 5 3 2

h2h1+h2h32h1+h3

1 2 4 30 1 5 40 1 5 4

h3h2+h3 1 2 4 30 1 5 4

0 0 0 0

1.2 nh thc

1.2.1 nh thc cp 1, 2, 3

nh thc cp 1, 2, 3 A l ma trn vung cp 1; A = [a11] th nh thc ca A l

mt s c k hiu l detA = a11A l ma trn vung cp 2:

A =

a11 a12a21 a22

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nh thc cp 2 ca A l mt s c k hiu

detA =

a11 a12a21 a22 = a11a22 a12a21

Nu A l mt ma trn cp 3

A =a11 a12 a13a21 a22 a23

a31 a32 a33

th nh thc cp 3 ca ma trn A l mt s, k hiu

detA =

a11 a12 a13a21 a22 a23a31 a32 a33

= a11a22a33+a13a21a32+a12a23a31a31a22a13a11a23a32a33a12a21

V d: Tnh nh thc ca ma trn

A =

1 2 34 5 67 8 9

1.2.2 Hon v

Hon v Cho tp N = {1, 2, 3, . . . , n}. Ta gi mt cch sp xp tp N theo mt tht no : {i1, i2, . . . , in} l mt hon v ca tp N

Nh vy c n! hon v ca tp N

V d: Nu N = {1, 2, 3, 4} th c 4! = 24 hon v1234 1243 1324 1342 1423 14322134 2143 2314 2341 2413 24313124 3142 3214 3241 3412 34214123 4132 4213 4231 4312 4321

Nghch th Trong hon v {i1, i2, . . . , ij, . . . , ik, . . . , in} ta ni cp s (ij, ik) lmthnh mt nghch th nu j < k m ij > ik

Trong hon v 3241 ta c cc nghch th (3,2), (3,1), (4,1), (2,1)Trong hon v 3124 ta c cc nghch th ...Hon v chn: hon v c s nghch th chn.Hon v l: hon v c s nghch th l.

1.2.3 nh thc cp n

Cho A l ma trn vung cp nXt tch a1i1 , a2i2, . . . , anin (1)trong i1, i2, . . . , in l mt hon v bc n.

Ta c n! tch c dng (1).Mi tch dng (1) c 1 du: Nu hon v i1, i2, . . . , in l chn th tch ly du

+, nu l hon v l th tch l du nh thc cp n ca ma trn A l tng ca n! hng t c dng (1).K hiu nh thc ca ma trn A l |A| hay detAnh thc ca ma trn vung cp n k hiu l Dn

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1.2. NH THC

1.2.4 Cc tnh cht ca nh thc

|A| = |AT|V d: Cho ma trn

A = 1 2 34 5 67 8 0

v AT = 1 4 72 5 83 6 0

Khi detA = v detAT =

Nu ta i ch hai dng (hai ct) ca nh thc cho nhau th nh thc i du.V d: Ta c

1 2 34 5 67 8 0

= 4 5 61 2 37 8 0

Nu trong mt nh thc c hai dng (hoc 2 ct) t l th nh thc bng 0.

V d:

1 2 32 4 67 8 0

= 0 nh thc khng i nu ta ly cc phn t ca mt dng nhn vi 1 s ri cng

vi cc phn t tng ng ca mt dng khc

a11 a12 . . . a

1k + a

1k . . . a1n

a21 a22 . . . a

2k + a

2k . . . a2n...

......

... . . ....

an1 an2 . . . a

nk + a

nk . . . ann

=a11 a12 . . . a

1k . . . a1n

a21 a22 . . . a

2k . . . a2n...

......

... . . ....

an1 an2 . . . a

nk . . . ann

+a11 a12 . . . a

1k . . . a1n

a21 a22 . . . a

2k . . . a2n...

......

... . . ....

an1 an2 . . . a

nk . . . ann

a11 a12 . . . ra1k . . . a1na21 a22 . . . ra2k . . . a2n

......

...... . . .

...an1 an2 . . . rank . . . ann

= r.a11 a12 . . . a1k . . . a1na21 a22 . . . a2k . . . a2n

......

...... . . .

...an1 an2 . . . ank . . . ann

V d: 1 2 34 5 67 8 0h2 4h1 + h2h37h1+h3 1 2 30 3 60 6 21

1.2.5 nh l Laplace

nh thc con Cho nh thc D cp n. nh thc con cp k (1 k n) ca nhthc D l nh thc ca ma trn vung cp k gm cc phn t nm giao ca k dngv k ct ty ca nh thc D.

V d:

D3 = a11

a12

a13a21 a22 a23

a31 a32 a33

c 9 nh thc con cp 1, 9 nh thc con cp 2 v 1 nh thc con cp 3.

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nh thc con b Cho D l nh thc cp n. Ta gi Mij l nh thc con b caphn t aij nu Mij l nh thc con cp n 1 nhn c t D bng cch xa i hngth i v ct th j.

D = a11 a12 a13 a14 a15a21 a22 a23 a24 a25a31 a32 a33 a34 a35a41 a42 a43 a44 a45a51 a52 a53 a54 a55

Ta c nh thc con b ca a24 l M24 =

a11 a12 a13 a15a31 a32 a33 a35a41 a42 a43 a35a51 a52 a53 a55

nh thc con b ca nh thc con M cp k trong D l nh thc con M thu

c bng cch xa i k dng v k ct xc nh nn nh thc con M.

V d: Cho D l nh thc cp 5. nh thc con b ca nh thc con

M =

a12 a14 a15a22 a24 a25a52 a54 a55

l nh thc

M

=

a31 a33a41 a43

Phn b i s ca phn t aij k hiu l Aij = (1)i+j

Mij trong Mij l nhthc con b ca phn t aijPhn b i s ca a24 l

(1)2+4M24Gi s nh thc con M c lp nn t cc phn t nm giao im ca k dng

i1, i2, . . . , ik v k ct j1, j2, . . . , jk v gi s M l nh thc con b ca nh thc con

M trong nh thc D. Ta nh ngha phn b i s ca M trong D l:

(1)i1+i2++ik+j1+j2++jkM

V d: Cho D l nh thc cp 5. Phn b i s ca ca nh thc con

M =

a12 a14 a15a22 a24 a25a52 a54 a55

l

(1)1+2+5+2+4+5a31 a33a41 a43

B : Cho nh thc D cp n. Vi mi i, j = 1, 2, . . . , n ta c:

Khai trin nh thc theo dng i nh sau

D = ai1Ai1 + ai2Ai2 + + ainAin

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1.2. NH THC

Khai trin nh thc theo ct j nh sau

D = a1jA1j + a2jA2j + + anjAnj

V d: Tnh cc nh thc

D =

0 4 0 01 3 2 1

2 5 3 13 7 2 2

; D

=

1 4 3 20 3 2 10 0 3 10 0 0 2

Tnh nh thc D ta khai trin theo dng th 1:

D = 4.(1)1+2

1 2 12 3 13 2 2

=

Tnh nh thc D

ta khai trin theo ct th 1:D

= 1.(1)1+13 2 10 3 10 0 2

= Cch tnh nh thc

1. S dng cc tnh cht ca nh thc l trit tiu cc phn t nm di ngcho chnh ca nh thc.

2. nh thc bng tch cc phn t trn ng cho chnh.V d: Tnh nh thc

D =

1 7 2 11 3 2 1

2 5 3 13 7 2 2

By gi ta s tng qut ha cng thc khai trin nh thc theo dng hoc ct.

nh l Laplace: Nu trong mt nh thc D ta ly ra k dng (hoc k ct), 1k n 1 th tng tt c cc nh thc con cp k cha trong cc dng (hoc cc ct)

y nhn vi phn b i s ca chng bng nh thc D.V d: Tnh nh thc

D =

2 3 0 04 5 0 06 7 1 22 3 2 1

Gii: Theo nh l Laplace, ta s khai trin nh thc theo 2 dng u tin. C tt

c 6 nh thc con cp 2 nhn c t 2 dng u tin nhng ch c 1 nh thc con

khc 0. Do ta c

D =

2 34 5 .(1)1+2+1+2 1 22 1

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1.3 Ma trn nghch o

1.3.1 Ma trn nghch o

nh l 1 Cho 2 ma trn vung A, B cp n. Khi det(AB) = det(A).det(B)

Ma trn khng suy bin Ma trn vung A cp n l ma trn khng suy bin nudet(A) = 0

Ma trn nghch o A l ma trn vung cp n, nu c ma trn vung B cp n saocho: A.B = B.A = I( I l ma trn n v) th B l ma trn nghch o ca ma trnA, k hiu B = A1

Khi ta ni A l ma trn kh nghch.

Ma trn ph hp Cho ma trn vung A cp n khng suy bin. Aij l phn b i

s ca phn t aij khi

PA =

A11 A21 . . . An1A12 A22 . . . An2

...... . . .

...A1n A2n . . . Ann

l ma trn ph hp ca A.

nh l 2 A khng suy bin

A l ma trn kh nghch v A1 =1

detA

PA.

1.3.2 Cch tm ma trn nghch o

Phng php dng nh thc Cho ma trn vung A cp n khng suy bin. Aijl phn b i s ca phn t aij khi

A1 =1

|A|

A11 A21 . . . An1A12 A22 . . . An2

...... . . .

...

A1n A2n . . . Ann

V d 1: Tm ma trn nghch o ca ma trn

A =

3 1 22 1 10 2 1

Gii: Ta c |A| = 3

A11 = 3, A12 = 2, A13 = 4A21 = 3, A22 = 3, A23 =

6

A31 = 3, A32 = 1, A33 = 5

Vy A1 =1

3

3 3 32 3 14 6 5

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1.4. HNG CA MA TRN

V d 2: Tm ma trn nghch o ca ma trn

B =

1 2 32 5 31 0 8

Phng php dng cc php bin i s cp tm ma trn nghch o cama trn vung A cp n ta lm nh sau:

Vit k A ma trn n v In Dng cc php bin i s cp trn dng a ma trn A thnh ma trn n

v

Khi ta c ma trn bn cnh ma trn In l ma trn nghch o ca AV d 3: Dng cc php bin i s cp tm ma trn nghch o ca ma trn

A = 3 1 22 1 1

0 2 1

Gii: 3 1 2 1 0 02 1 1 0 1 0

0 2 1 0 0 1

h1h2+h1 1 2 1 1 1 02 1 1 0 1 0

0 2 1 0 0 1

h22h1+h2 1 2 1 1 1 00 5 1 2 3 0

0 2 1 0 0 1

h22h3+h2

1 2 1 1 1 00 1 1 2 3 20 2 1 0 0 1

h32h2+h3

1 2 1 1 1 00 1 1 2 3 20 0 3

4

6 5

h31

3h3 1 2 1 1 1 00 1 1 2 3 2

0 0 1 4/3 2 5/3

h2h3+h2h1h3+h1

1 2 0 7/3 3 5/30 1 0 2/3 1 1/30 0 1 4/3 2 5/3

h12h2+h1

1 0 0 1 1 10 1 0 2/3 1 1/30 0 1 4/3 2 5/3

Vy A1 =

1 1 12/3 1 1/34/3 2 5/3

V d 4: Tm ma trn nghch o ca ma trn B = 2 7 33 9 4

1 5 3

1.4 Hng ca ma trn

1.4.1 Khi nim v hng ca ma trn

Cho ma trn A. Ta gi hng ca ma trn A l s nguyn dng r sao cho:

Trong A c mt nh thc con cp r khc 0. Mi nh thc con cp ln hn r trong ma trn A u bng 0.K hiu: rankANu A = 0 th ta qui c rankA = 0

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nh lrankA = rankAT

1.4.2 Cch tnh hng ca ma trnDng cc php bin i s cp trn dng (hay ct) a ma trn v dng

a11 a12 a13 . . . a1r . . . a1n

0 a22 a23 . . . a2r . . . a2n

0 0 a33 . . . a3r . . . a3n.

.

.

.

.

.

.

.

.

.

.

.

.

.

. . . .

.

.

.

0 0 0 . . . arr . . . arn

0 0 0 . . . 0 . . . 0

.

.

.

.

.

.

.

.

.

.

.

.

.

.

. . . .

.

.

.

0 0 0 . . . 0 . . . 0

Khi rankA = rV d 1: Dng cc php bin i s cp tnh hng ca ma trn A

A =

1 2 4 31 3 1 12 5 3 2

Gii: Trc tin, ta dng cc php bin i s cp a A v dng ma trn bcthang.

A = 1 2 4 31 3 1 12 5 3 2 h2h1+h2h32h1+h3

1 2 4 30 1 5 40 1 5 4 h3h2+h3

1 2 4 30 1 5 40 0 0 0

Vy rankA = 2

V d 2: Dng cc php bin i s cp tnh hng ca ma trn B

B =

1 3 50 2 21 5 73 7 13

Bi tp

Bi 1. Cho ma trn

A =

1 2 63 4 52 3 4

a. Tm ma trn X sao cho 2A + 3X = Ib. Tm ma trn X sao cho 2A 3X = I

Bi 2. Cho cc ma trn

A = 1 23 4 , B = 2 2 33 0 7 , C = 1 2 35 4 1Tnh A2 3A; BTA BT; A(B + C); BCT + ABi 3. Tnh cc nh thc

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1.4. HNG CA MA TRN

a.a + b a ba b a + b

b. x 1 1x3 x2 + x + 1 c. sin x cos x cos x sin x

d.

2 2 13 4 25 1 3

e.

1 2 34 5 67 8 9

f.

1 a bc1 b ac1 c ab

Bi 4. Tnh cc nh thc

a.

1 1 1 11 1 2 31 1 1 31 1 1 1

b.2 1 1 11 1 2 31 1 1 31 1 1 1

c.0 1 1 11 1 2 31 1 1 31 1 1 1

Bi 5. Tnh hng ma trn

a.

1 1 1 11 1 2 31 1 1 31 1 1 1

b.

0 1 1 11 1 2 31 1 1 31 1 1 1

c.

1 1 5 11 1 2 33 1 8 11 3 9 7

d.

3

1

1 2 11 1 2 4 51 1 3 6 9

12 2 1 2 10

e.4 1

1 3 2

2 2 3 0 12 3 2 3 34 1 3 1 1

Bi 6. Tm hng ca ma trn theo tham s m

3 1 1 4m 4 10 11 7 17 32 2 4 3

Bi 7. Xc nh a hng ca ma trn sau bng 2:

A =

1 a 1 22 1 a 51 10 6 1

Bi 8. Tm ma trn nghch o (nu c) ca cc ma trn sau:

a.

2 34 5

b.

2 34 6

c.

1 3 72 1 2

7 1 4

d.

1 0 10 0 2

1 3 1

e.

1 1 1 11 1 1 11 1 0 00 0 1 1

Bi 9. Tm ma trn X sao cho

a. X.

3 4 60 1 12 3 4

=1 10 1

2 2

b.3 4 60 1 1

2 3 4

.X = 1 1 20 1 2

Bi 10. Tm ma trn nghch o ca ma trn cp n sau:

a.

1 1 1 10 1 1 10 0 1 1...

.

.....

.. .

.

..0 0 0 1

b.

1 1 0 00 1 1 00 0 1 0...

.

.....

.. .

.

..0 0 0 1

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Chng 2H phng trnh tuyn tnh

2.1 Khi nim tng qut

2.1.1 nh nghaH phng trnh tuyn tnh gm m phng trnh n n l h thng c dng

a11x1 + a12x2 + + a1nxn = b1a21x1 + a22x2 + + a2nxn = b2.....................................

am1x1 + am2x2 + + amnxn = bm

(1)

Trong x1, x2, . . . , xn l cc n, cc aij, biK (K = R hay K = C)

H trn c vit di dng: AX = B

Trong A = (aij)mn , X =

x1x2...

xn

v B =

b1b2...

bn

A c gi l ma trn h s ca h phng trnh (1).

A =

a11 a12 . . . a1n b1a21 a22 . . . a2n b2

...... . . .

......

am1 am2 . . . amn bm

gi l ma trn b sung

2.1.2 Tnh cht nghim

Ta gi mt nghim ca h phng trnh (1) l mt b n s (c1, c2, . . . , cn) sao chokhi thay vo cc n x1, x2, . . . , xn tng ng th hai v ca h phng trnh (1)tr thnh cc ng nht thc.

Mt h phng trnh tuyn tnh c th c mt nghim duy nht, c th c nhiuhn 1 nghim hoc cng c th khng c nghim.

Hai h phng trnh tuyn tnh ca n n gi l tng ng nu tp hp ccnghim ca chng trng nhau.

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2.2. H CRAMER

2.2 H Cramer

2.2.1 nh ngha

H phng trnh Cramer l mt h n phng trnh tuyn tnh n n m nh thc cama trn cc h s khc khng.

Tc l h phng trnh AX = B, trong A l ma trn vung v detA = 0

2.2.2 Cng thc Cramer

H phng trnh Cramer c duy nht nghim xc nh bi cng thc sau:

xi =|Ai||A|

trong Ai l ma trn thu c t ma trn A bng cch thay ct th i bng ct B

V d 1: Gii h phng trnh sau bng phng php Cramerx1 + x2 x3 = 22x1 + 3x2 + x3 = 15x1 + 8x2 + 2x3 = 3

Gii: |A| =

1 1 12 3 15 8 2

= 2 = 0, |A1| =

2 1 11 3 13 8 2

= 18, |A2| =

1 2 12 1 15 3 2

=

14,

|A3| = 1 1 22 3 15 8 3 = 8Vy ta c nghim ca h phng trnh

x1 =|A1||A| = 9, x2 =

|A2||A| = 7, x3 =

|A3||A| = 4

V d 2: Gii h phng trnh sau bng phng php Cramer

x1 + 2x2 + x3 = 1

2x1 + 5x2 + x3 = 6

x1 4x2 + 2x3 = 2

2.3 H phng trnh tuyn tnh tng qut. Phngphp Gauss

2.3.1 H phng trnh tuyn tnh tng qut. Phng phpGauss

i vi nhng h phng trnh m s phng trnh khc s n hay s phng trnh

bng s n nhng |A| = 0 th ta khng p dng c phng php Cramer.Ta dng phng php kh dn n s hay cn gi l phng php GaussTa gi cc php bin i s cp trn h phng trnh tuyn tnh l cc php bin

i di y:

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CHNG 2. H PHNG TRNH TUYN TNH

Loi 1: i ch 2 phng trnh cho nhau, cn cc phng trnh khc gi nguyn.

Loi 2: Nhn 1 s khc 0 v mt phng trnh, cn cc phng trnh cn li ginguyn.

Loi 3: Cng vo mt phng trnh mt phng trnh khc nhn vi mt s khc

0, cn cc phng trnh cn li gi nguyn.Cc php bin i s cp a h phng trnh tuyn tnh thnh mt h mi tngng.

Ta c th gii h phng trnh tuyn tnh bng phng php sau: Thc hin ccphp bin i s cp a h phng trnh v dng bc thang (tc l ma trn h s bsung A c dng bc thang - cc phn t nm di ng cho chnh u bng 0). T ta tm cc nghim.

V d 1: Gii h phng trnh sau:

x1 + x2

2x3 = 6

2x1 + 3x2 7x3 = 165x1 + 2x2 + x3 = 16

3x1 x2 + 8x3 = 0

Gii: Xt ma trn A

A =

1 1 2 62 3 7 165 2 1 163

1 8 0

h22h1+h2h35h1+h3h43h1+h4

1 1 2 60 1 3 40 3 7 140

4 14

18

h33h2+h3h44h2+h4

1 1 2 60 1 3 40 0 2 20 0 2 2

h4h3+h4

1 1 2 60 1 3 40 0 2 20 0 0 0

T suy ra nghim ca h phng trnh l:

x3 = 1x2 = 4 + 3x3 = 1

x1 = 6 + 2x3 x2 = 3V d 2: Gii h phng trnh sau:

x1 2x2 + 3x3 4x4 = 23x1 + 3x2 5x3 + x4 = 32x1 + x2 + 2x3 3x4 = 53x1 + 3x3 10x4 = 8

nh l 3 (nh l Kronecker - Capelli) H phng trnh tuyn tnh (1) c nghimkhi v ch khi hng ca ma trn A bng hng ca ma trn b sung A.

Tc rankA = rankA

Ch :

1. Nu rankA = rankA = n (s n ca h phng trnh) th h phng trnh cnghim duy nht

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2.3. H PHNG TRNH TUYN TNH TNG QUT. PHNG PHP GAUSS

2. Nu rankA < rankA th h v nghim

3. Nu rankA = rankA = k < n th h phng trnh c v s nghim ph thucvo n k tham s.

V d 3: Gii h phng trnh sau:x1 + 2x2 3x3 + 5x4 = 1x1 + 3x2 x3 + x4 = 22x1 + 5x2 4x3 + 6x4 = 1

Gii: Ta xt ma trn b sung

1 2 3 5 11 3 1 1 22 5 4 6 1

h2h1+h2h32h1+h3

1 2 3 5 10 1 2 4 30 1 2 4 3

h3h2+h3

1 2 3 5 10 1 2 4 30 0 0 0 0

Ta thy rankA = rankA = 2 < 4 nn h phng trnh c v s nghim v ph thucvo 2 tham s.

Nu ta t x3 = t1; x4 = t2 th nghim ca h phng trnh l:x1 = 1 5x4 + 3x3 2x2 = 7 13t2 + 7t1x2 = 3 + 4x4 2x3 = 3 + 4t2 2t1x3 = t1

x4 = t2

(t1, t2 ty )

V du 4: Cho h phng trnh sau v nghimx1 + 2x2 3x3 + 5x4 = 1x1 + 3x2 x3 + x4 = 22x1 + 5x2 4x3 + 6x4 = m

a. Tm m h phng trnh c nghim.b. Tm m h phng trnh v nghim.

2.3.2 H phng trnh tuyn tnh thun nht

H phng trnh tuyn tnh thun nht l h phng trnh dng AX = 0(2) viA = (aij)mn

H ny lun c 1 nghim x1 = x2 = . . . = xn = 0 (nghim tm thng)

1. Nu rankA = n th h phng trnh (2) c nghim duy nht (nghim tm thng)

2. Nu rankA = k < n th h phng trnh (2) c v s nghim,ph thuc n ktham s.

Gi s n k tham s ny ta k hiu l t1, t2, . . . , tnk.Ln lt cho:

t1 = 1, t2 = 0, . . . tnk = 0 tm c 1 nghim k hiu X1t1 = 0, t2 = 1, . . . tnk = 0 tm c 1 nghim k hiu X2. . .t1 = 0, t2 = 0, . . . tnk = 1 tm c 1 nghim k hiu Xnk

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CHNG 2. H PHNG TRNH TUYN TNH

Khi X1, X2, . . . , X nk gi l h nghim c bn ca h thun nht.V d 4: Tm h nghim c bn v nghim tng qut ca h phng trnh

x1 + 2x2 3x3 + 5x4 = 0x1 + 3x2 x3 + x4 = 0

2x1 + 5x2 4x3 + 6x4 = 0Gii: Ta xt ma trn b sung 1 2 3 5 01 3 1 1 0

2 5 4 6 0

h2h1+h2h32h1+h3

1 2 3 5 00 1 2 4 00 1 2 4 0

h3h2+h3 1 2 3 5 00 1 2 4 0

0 0 0 0 0

Ta thy rankA = rankA = 2 < 4 nn h phng trnh c v s nghim v ph thucvo 2 tham s.

Nu ta t x3 = t1; x4 = t2 th nghim ca h phng trnh l:

x1 =

5x4 + 3x3

2x2 =

13t2 + 7t1

x2 = 4x4 2x3 = 4t2 2t1x3 = t1

x4 = t2

(t1, t2 ty )

Cho x3 = 1, x4 = 0 ta c 1 nghim c bn l X1 = (7, 2, 1, 0).Cho x3 = 0, x4 = 1 ta c 1 nghim c bn l X2 = (13, 4, 0, 1).

V d 5: Tm h nghim c bn v nghim tng qut ca h phng trnh

2x1 + 2x2 + 4x3 3x4 = 0x1 + 3x2

x3 + x4 = 0

3x1 + 5x2 + 3x3 2x4 = 0Bi tp

Bi 11. Gii cc h phng trnh sau y theo qui tc Crame

a.

x 3y = 12x + y = 8

b.

7x + 2y + 3z = 15

5x 3y + 2z = 1510x 11y + 5z = 36

c.

x1 + x2 5x3 = 102x1 + 3x3 x4 = 104x1 + 4x2 + 5x3 + 5x4 = 0

3x2 + 2x3 = 1

Bi 12. Gii cc h phng trnh sau bng phng php Gauss

a.

x1 x 2 + x3 = 22x1 + x2 2x3 = 6x1 + 2x2 + 3x3 = 2

b.

2x1 x2 + 3x3 = 1x1 + 2x2 x3 = 3x1 + x2 + 2x3 = 4

c.

3x1 + 2x2 + x3 = 52x1 + 3x2 + x3 = 1

2x1 + x2 + 3x3 = 11

d.

x1 + x2 + x3 = 1

2x1 x2 + x3 = 3x1 x2 + 2x3 = 53x1 6x2 + 5x3 = 6

e.

x1 + x2 + 2x3 + 3x4 = 1

2x1 + 3x2 x3 x4 = 63x1 x2 x3 2x4 = 4x1 + 2x2 + 3x3 x4 = 4

f.2x1 x2 + 5x4 = 3

4x

1 x2

+ 4x3

12x4

= 16

2x1 5x2 + 7x3 4x4 = 366x1 3x3 + 20x4 = 4

Bi 13. Tm nghim tng qut ca cc h phng trnh sau bng phng php Gauss

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2.3. H PHNG TRNH TUYN TNH TNG QUT. PHNG PHP GAUSS

a.

x1 + 2x2 3x3 = 1x1 + 3x2 x3 = 1

b.

x1 + 2x2 3x3 = 12x1 + 5x2 8x3 = 43x1 + 8x2 13x3 = 7

c.

x1 3x2 + 2x3 x4 = 24x1 + x2 + 3x3 2x4 = 12x1 + x2 x3 + x4 = 1

d.x1 + 2x2 3x3 + 5x4 = 1x1

+ 3x2

13x3

+ 22x4

=

1

3x1 + 5x2 + x3 2x4 = 52x1 + 3x2 + 4x3 7x4 = 1

e.x1 2x2 + x3 x4 + x5 = 03x

1 2x

2 x3

+ x4

2x5

=

1

2x1 + x2 x3 + 2x4 3x5 = 12x1 5x2 + x3 2x4 + 2x5 = 2

Bi 14. Gii v bin lun cc h phng trnh sau

a.

(m + 1)x1 + x2 + x3 = 1

x1 + (m + 1)x2 + x3 = 1

x1 + x2 + (m + 1)x3 = 1

b.

(m + 1)x1 + x2 + x3 = 1

x1 + (m + 1)x2 + x3 = m

x1 + x2 + (m + 1)x3 = m2

c.x1 + x2 + (1 m)x3 = m + 2(1 + m)x1

x2 + 2x3 = 0

2x1 mx2 + x3 = m + 2d.

x1 + x2 + x3 + mx4 = 1

x1 + x2 + mx3 + x4 = 1

x1 + mx2 + x3 + x4 = 1mx1 + x2 + x3 + x4 = 1

e.

2x1 + x2 + x3 + x4 = 1

x1 + 2x2 x3 + 4x4 = 2x1 + 7x2 4x3 + 11x4 = m4x1 + 8x2 4x3 + 16x4 = m + 1

f.

2x1 x2 + x3 2x4 + 3x5 = 3x1 + x2 x3 x4 + x5 = 13x1 + 2x2 + x3 3x4 + 4x5 = 65x1 + 2x2 5x4 + 7x5 = 9 m

Bi 15. Tm nghim tng qut v h nghim c bn ca cc h phng trnh sau

a.2x1 x2 2x3 = 0x1 + x2 + x3 = 0

b.

x1 + 2x2 3x3 = 02x1 + 5x2

8x3 = 0

3x1 + 8x2 13x3 = 0c.

x1 3x2 + 2x3 x4 = 04x1 + x2 + 3x3

2x4 = 0

2x1 + x2 x3 + x4 = 0

d.

x1 + 2x2 3x3 + 5x4 = 0x1 + 3x2 13x3 + 22x4 = 03x1 + 5x2 + x3 2x4 = 02x1 + 3x2 + 4x3 7x4 = 0

e.

x1 2x2 + x3 x4 + x5 = 03x1 2x2 x3 + x4 2x5 = 02x1 + x2 x3 + 2x4 3x5 = 02x1 5x2 + x3 2x4 + 2x5 = 0

Nguyn Minh Tr 24


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Chng 3Khng gian vect

3.1 Cc khi nim c bn v khng gian vect

3.1.1 nh ngha khng gian vectCho tp E= v trng K (K = R hay K = C) vi hai php ton sau:

+ : E E E(x, y) x + y

: K E E(k, x) k.x

sao cho tha 8 tnh cht sau:

1. a + b = b + a,

a, b

E

2. (a + b) + c = a + (b + c), a,b,c E3. Tn ti phn t 0 E sao cho a + 0 = 0 + a = a, a E4. a E, tn ti a E sao cho a + (a) = (a) + a = 05. k(a + b) = ka + kb

6. (k + m)a = ka + ma

7. (km)a = k(ma)

8. 1a = a

th E c gi l mt K-khng gian vectCc phn t ca E gi l vect , cc phn t ca K gi l v hng.

3.1.2 Cc v d khng gian vect

V d 1: E = R3 = {(a1, a2, a3)|ai R} v K = RTa nh ngha hai php ton: a = (a1, a2, a3), b = (b1, b2, b3) v k R

a + b = (a1 + b1, a2 + b2, a3 + b3)

k.a = (ka1, ka2, ka3)

Khi R3 cng vi hai php ton trn l mt R-khng gian vect

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3.2. S PH THUC TUYN TNH V C LP TUYN TNH

Vect khng 0 = (0, 0, 0)Vect i ca vect a = (a1, a2, a3) l a = (a1, a2, a3)

V d 2: Cho Mn(R) l tp tt c cc ma trn vung cp n vi cc phn t trong matrn l cc s thc. t V = Mn v K = R.

Trn Mn ta xt php cng l php cng hai ma trn v php nhn l php nhn

ca mt s vi mt ma trn.Khi Mn cng vi hai php ton trn lp thnh mt R-khng gian vect .

3.1.3 Cch tnh cht

Cho E l mt K-khng gian vect , m, n K, x, y E1. Vect 0 l duy nht

2. mx = 0 m = 0 hoc x = 0

3. (m n)x = mx nx4. m(x y) = mx my

3.2 S ph thuc tuyn tnh v c lp tuyn tnh

3.2.1 T hp tuyn tnh

Cho cc vect x1, x2, . . . , xn E.Vect y c gi l mt t hp tuyn tnh ca cc vect x1, x2, . . . , xn nu tn ti cc

s k1, k2, . . . , kn K sao cho:y = k1x1 + k2x2 + + knxn

Hay ta ni y biu th tuyn tnh c qua h cc vect x1, x2, . . . , xnV d 1: Cho h cc vect e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) R3

Khi vect y = (3, 2, 4) l mt t hp tuyn tnh ca h v:

y = 3e1 + 2e2 + 4e3

V d 2: Vect a = (1, 2, 5) c l t hp tuyn tnh ca cc vect x = (1, 1, 1), y =(1, 2, 3), z = (2, 1, 1) trong R3 khng?

Gii: Gi s

a = k1x + k2y + k3z(1, 2, 5) = k1(1, 1, 1) + k2(1, 2, 3) + k3(2, 1, 1)(1, 2, 5) = (k1 + k2 + 2k3, k1 + 2k2 + k3, k1 + 3k2 + k3)

Ta c h phng trnh

k1 + k2 + 2k3 = 1

k1 + 2k2 + k3 = 2k1 + 3k2 + k3 = 5

k1 + k2 + 2k3 = 1

k2 k3 = 12k2 k3 = 4

k1 + k2 + 2k3 = 1

k2 k3 = 1k2 = 3

k1 = 6k2 = 3k3 = 2

Vy a = 6x + 3y + 2z

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CHNG 3. KHNG GIAN VECT

3.2.2 S ph thuc tuyn tnh

H cc vect x1, x2, . . . , xn c gi l ph thuc tuyn tnh nu c n s k1, k2, . . . , knkhng ng thi bng 0, sao cho:

k1x1 + k2x2 + + knxn = 0

H vect x1 = (2, 1, 0), x2 = (1, 3, 2), x3 = (0, 7, 4) l ph thuc tuyn tnh v:1x1 + 2x2 + 1x3 = (2, 1, 0) + (2, 6, 4) + (0, 7, 4) = (0, 0, 0)

3.2.3 S c lp tuyn tnh

H cc vect x1, x2, . . . , xn c gi l c lp tuyn tnh nu c

k1x1 + k2x2 + + knxn = 0khi v ch khi k1 = k2 = . . . = kn = 0

Cho h cc vect e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) c lp tuyn tnh.Tht vy, xt k1e1 + k2e2 + k3e3 = 0 th k1 = k2 = k3 = 0

V d 1: Xt s c lp tuyn tnh v ph thuc tuyn tnh ca cc vect trong R3

x = (1, 1, 0), y = (0, 1, 1), z = (1, 0, 1)

Gii: Gi sk1x + k2y + k3z = 0

k1(1, 1, 0) + k2(0, 1, 1) + k3(1, 0, 1) = (0, 0, 0)

k1 + k3 = 0

k1 + k2 = 0

k2 + k3 = 0

k1 = 0

k2 = 0

k3 = 0

Vy ba vect x,y,z c lp tuyn tnh.V d 2: Trong khng gian R2, xt s c lp tuyn tnh ca hai vect a =

(1, 2), b = (3, 6)

3.3 Hng ca mt h vect

3.3.1 nh ngha

Cho h gm m vect a1, a2, . . . , am (2)S r l hng ca h vect (2) nu:

1. C mt h con gm r vect ca h (2) c lp tuyn tnh.

2. Mi h con gm ln hn r vect u ph thuc tuyn tnh.

K hiu: rank(a1, a2, . . . , am) = rNhn xt: Gi s rank(a1, a2, . . . , am) = r

Nu r < m th h cc vect (2) ph thuc tuyn tnh. Nu r = m th h cc vect (2) c lp tuyn tnh.Nu h (2) gm cc vect khng th hng ca h bng 0.

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3.4. C S, S CHIU

3.3.2 Cch tnh hng ca mt h vect

Tnh hng cc vect

a1 = (a11, a12,...,a1n)a2 = (a21, a22,...,a2n)

...am = (am1, am2,...,amn)

1. Lp ma trn

A =

a11 a12 . . . a1na21 a22 . . . a2n

...... . . .

...am1 am2 . . . amn

2. Tnh rank(A)

3. rank(a1, a2, . . . , am) = rank(A)

V d: Tnh hng ca h cc vect sau a = (1, 1, 0), b = (0, 1, 1), c = (1, 0, 1), d =(2, 1, 3).T kt lun v s c lp tuyn tnh ca h cc vect trn.

3.4 C s, s chiu

3.4.1 C s

Mt tp cc vect B = {u1, u2, . . . , un} E c gi l mt c s ca khng gianvect E nu:

1. H vect B c lp tuyn tnh.

2. Mi vect x E u biu th tuyn tnh c qua cc vect ca B.Tc l x = t1u1 + t2u2 + + tnun

Cc s t1, t2, . . . , tn gi l ta ca vect x i vi c s B. Ta vit [x]/B =

t1t2...

tn

hay x(B) = (t1, t2, . . . , tn)V d 1: H cc vect 3 chiu e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) l c s ca

R-khng gian R3. C s ny c gi l c s chnh tc ca R3.Tng qut hn, R-khng gian vect Rn c c s chnh tc l

e1 = (1, 0, . . . , 0), e2 = (0, 1, 0, . . . , 0), . . . , en = (0, 0, . . . , 1)

V d 2: Chng minh tp cc vect B = {(1, 1, 1), (0, 1, 1), (0, 0, 1)} l mt c sca R-khng gian R3

Nhn xt: Mt khng gian vect c th c nhiu c s.

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3.4.2 S chiu

nh l 4 Trong khng gian vect E, s vect trong hai c s bt k u bng nhau

S vect trong mt c s bt k ca K-khng gian vect E gi l s chiu ca E v khiu l dim(E)

V d: Chng minh dimR3 = 3

nh l 5 Gi s E l mt khng gian vect c dim(E) = n, n N. Khi mi hn vect c lp tuyn tnh ca E u l mt c s ca E

V d: Chng minh tp cc vect B = {(1, 1, 1), (1, 0, 1), (0, 0, 1)} l mt c s caR-khng gian R3

Gii:Ta c dimR3 = 3Ta cn chng minh h 3 vect trn c lp tuyn tnh.

3.4.3 Ma trn chuyn c s

Mt khng gian vect c nhiu c s. Ta ca mt vect i vi mt c s l duynht. Vy khi thay c c ny bng mt c s khc th ta vect thay i nh thno?

Cho K-khng gian vect E c 2 c s:U = {u1, u2, . . . , un} v V = {v1, v2, . . . , vn}

v1 = t11u1 + t12u2 + + t1nunv2 = t21u1 + t22u2 +

+ t2nun

..........................................vn = tn1u1 + tn2u2 + + tnnun

TUV =

t11 t21 . . . tn1t12 t22 . . . tn2...

... . . ....

t1n t2n . . . tnn

gi l ma trn chuyn c s t U sang V.

Nu SV U l ma trn chuyn c s t V sang U th SV U chnh l ma trn nghch

o ca ma trn chuyn c s t U sang VV d: Trong R-khng gian vect R2 cho 2 c s:

U = {u1 = (1, 0), u2 = (0, 1)} v V = {v1 = (1, 2), v2 = (3, 4)}

Tm ma trn chuyn c s t U sang V v ma trn chuyn c s t V sang U.Gii: Tm ma trn chuyn c s t U sang V

v1 = u1 + 2u2

v2 = 3u1 + 4u2

Nh vy TUV = 1 32 4Tng t ta tm c ma trn chuyn c s t V sang U l TV U =

2 3/21 1/2

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3.5. KHNG GIAN VECT CON

3.4.4 Quan h gia cc ta ca cng mt vect i vi ccc s khc nhau

Cho vect a E. Ta ca a i vi c s U l x1, x2, . . . , xn, ta ca a i vic s V l y1, y2, . . . , yn. Khi :

[a]/U = TUV [a]/V

vi [a]/U =

x1x2...

xn

v [a]/V =

y1y2...

yn

, TUV l ma trn chuyn c s t U sang V.V d: Trong R-khng gian vect R2 cho 2 c s:

U = {u1 = (1, 0), u2 = (0, 1)} v V = {v1 = (2, 1), v2 = (3, 1)}

Ta c ma trn chuyn c s l

TUV =

2 31 1

v TV U =

1 31 2

Tm ta ca vect x = (5, 2) i vi 2 c s trn.

3.5 Khng gian vect con

3.5.1 nh ngha v v dnh ngha Gi s E l mt K-khng gian vect , F E. Ta ni F l mt khnggian vect con ca E khi n tha cc iu kin sau:

1. F =

2. x, y F : x + y F

3. x F, k K : k.x F

Hay ta c mt iu kin tng ng nh sau:F l khng gian vect con ca E x, y F, a, b K : a.x + b.y F()

V d

1. Tp {0} v E l hai khng gian vect con ca E

2. Trong R-khng gian vect R3, cho tp

A ={

(x1, x2, x3)R3

|x1 + x2 + x3 = 0

}l khng gian vect con ca R3

(Dng tiu chun (*) chng minh.)

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3.5.2 Khng gian vect sinh bi mt tp

Cho E l mt khng gian vect , a1, a2, . . . , an l h cc vect ca E. Ta nh ngha

a1, a2, . . . , an := {x = k1a1 + k2a2 + + knan|ki K} E

Dng tiu chun khng gian vect con ta chng minh c a1, a2, . . . , an l khnggian vect con ca E.a1, a2, . . . , an c gi l khng gian vect sinh bi h cc vect a1, a2, . . . , anKhi ta ni a1, a2, . . . , an l h sinh ca a1, a2, . . . , anMi h con c lp tuyn tnh ti i ca cc vect a1, a2, . . . , an l mt c s ca

a1, a2, . . . , an

3.5.3 Khng gian con cc nghim ca h phng trnh tuyntnh thun nht

Cho h phng trnh tuyn tnh thun nht gm m phng trnh v n n sa11x1 + a12x2 + + a1nxn = 0a21x1 + a22x2 + + a2nxn = 0. . .

am1x1 + am2x2 + + amnxn = 0

()

hay A.X = 0t N l tp tt c cc nghim ca h phng trnh tuyn tnh thun nht (**),

khi N l mt khng gian vect con ca Rn

Nu rankA = r < n th s chiu ca khng gian nghim N chnh l n r. C sca N chnh l h nghim c bn ca (**)V d: Tm c s v s chiu ca khng gian nghim

x1 + 2x2 + 4x3 3x4 = 0x1 + 3x2 x3 + x4 = 02x1 + 5x2 + 3x3 2x4 = 0

Gii: Ta gii h phng trnh choBin i ma trn cc h s b sung:

A =

1 2 4 3 01 3 1 1 02 5 3 2 0

h2h1+h2h32h1+h3

1 2 4 3 00 1 5 4 00 1 5 4 0

h3h2+h3 1 2 4 3 00 1 5 4 0

0 0 0 0 0

rankA = 2 < 4 h phng trnh c v s nghim ph thuc 2 tham s l x3, x4. Ta c

x2 = 5x3 4x4x1 = 2x2 4x3 + 3x4 = 14x3 + 11x4Vy nghim tng qut ca h l

x1

=

14t1

+ 11t2

x2 = 5t1 4t2x3 = t1

x4 = t2

, t1, t2 R

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Cho t1 = 1, t2 = 0 ta c x1 = 14, x2 = 5, x3 = 1, x4 = 0, ta c vecta1 = (14, 5, 1, 0)

Cho t1 = 0, t2 = 1 ta c x1 = 11, x2 = 4, x3 = 0, x4 = 1, ta c vecta2 = (11, 4, 0, 1)

Vy c s ca khng gian nghim l {a1, a2} v s chiu ca khng gian nghim l

2.Bi tp

Bi 16. Cho A l mt tp ty , k hiu M(A) l tp tt c cc nh x t A n R.Trong M(A) ta nh ngha 2 php ton nh sau:

(f + g)(x) = f(x) + g(x)

(kf)(x) = k.f(x)

vi mi f, g M(A), k R, x A Chng minh rng M(A) l mt R-khng gian vect

.Bi 17. Chng minh rng tp hp Pn(x) cc a thc bc nh hn hoc bng n ca xvi h s thc lp thnh mt R-khng gian vect cng vi php cng a thc v phpnhn 1 s thc vi mt a thc thng thng.Bi 18. Trong R3, xt xem vect u c phi l t hp tuyn tnh ca u1, u2, u3 haykhng?

a. u = (1, 2, 1); u1 = (0, 1, 1); u2 = (1, 0, 1); u3 = (1, 1, 0)b. u = (2, 2, 1); u1 = (2, 1, 3); u2 = (4, 1, 2); u3 = (6, 0, 5)c. u = (7, 2, 15); u1 = (2, 3, 5); u2 = (3, 7, 8); u3 = (1, 6, 1)

Bi 19. Tm m sao cho x l t hp tuyn tnh ca u1, u2, u3a. x = (5, 9, m); u1 = (4, 4, 3); u2 = (7, 2, 1); u3 = (4, 6, 1)b. x = (7, 2, m); u1 = (2, 3, 5); u2 = (3, 7, 8); u3 = (1, 6, 1)c. x = (1, 3, 5); u1 = (3, 2, 5); u2 = (2, 4, 7); u3 = (5, 6, m)

Bi 20. Trong R-khng gian vect R3, xt s c lp tuyn tnh ca cc h vect sau:a. x1 = (2, 1, 1), x2 = (1, 3, 1), x3 = (1, 2, 0)b. x1 = (2, 3, 0), x2 = (0, 1, 2), x3 = (2, 4, 1)

Bi 21. Trong R-khng gian vect R4, vi gi tr no ca m th h vect sau c lptuyn tnh:

x1 = (0, 1, 1, 1), x2 = (1, 0, 1, 1), x3 = (1, 1, 0, 1), x4 = (1, 1, 1, m)

Bi 22. Trong R-khng gian vect R3, hy tm hng ca cc h vect sau:a. {(1, 2, 3), (0, 1, 1), (1, 3, 4)}b. {(1, 2, 1), (0, 1, 1), (2, 3, 4)}c. {(1, 2, 1), (0, 3, 3), (2, 3, 3), (1, 1, 2)}d. {(1, 3, 1, 0), (2, 0, 1, 1), (0, 1, 4, 3)}e. {(1, 0, 2, 3, 1), (2, 1, 3, 2, 0), (3, 2, 5, 1, 4), (0, 1, 4, 6, 5)}

Bi 23. Trong R-khng gian vect R3, chng minh h vect {x1 = (1, 1, 1), x2 =(1, 1, , 2), x3 = (1, 2, 3)} l mt c s ca R3. Tm ta ca vect x = (2, 1, 9) trongc s .Bi 24. Trong R-khng gian vect R3, cho hai h vect

B = {(1, 1, 1), (1, 1, 2), (1, 2, 3)}

B

= {(2, 1, 1), (3, 2, 5), (1, 1, m)}

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a. Tm m B l mt c s ca R3

b. Tm ma trn chuyn c s t B sang B

c. Tm ta ca vect a = (1, 0, 0) i vi 2 c s trn.Bi 25. Chng minh A = {(x1, x2, x3) R3|x1 + x2 + x3 = 0} l R-khng gian vectcon ca R3

Bi 26. TrongR3

cho khng gian vect con F = (1, 1, 1), (2, 3, 1), (5, 1, 2). Tmmt c s ca F v dim(F)Bi 27. Cho h phng trnh tuyn tnh thun nht

x1 + x2 + x3 + x4 = 0

2x1 + 3x2 x3 + x4 = 0

Tm c s v s chiu ca khng gian nghim

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Chng 4

nh x tuyn tnh

4.1 nh ngha v v d

4.1.1 nh nghaCho E, F l hai K-khng gian vect , nh x f : E F l nh x tuyn tnh nu ftha 2 iu kin:

1. f(a + b) = f(a) + f(b) a, b E2. f(ka) = kf(a) a E, k KMt nh x tuyn tnh f : E E c gi l mt php bin i tuyn tnh ca ENh vy mun chng minh f l mt nh x tuyn tnh th ta cn kim tra 2 iu

kin nh trn.

4.1.2 Cc v d

1. nh x khng:0 : E F

a 0(a) = 0l nh x tuyn tnh

2. nh x ng nhtid : E

E

a id(a) = al nh x tuyn tnh

3. nh xp : R3 R2

(x1, x2, x3) p(x1, x2, x3) = (x1, x2)l nh x tuyn tnh.

4. nh xf : R3

R3

(x1, x2, x3) f(x1, x2, x3) = (x1 + x2, x1 + x3, x3)l nh x tuyn tnh.

Chng minh:

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4.2. MA TRN CA NH X TUYN TNH

x, y R3, ta c x = (x1, x2, x3), y = (y1, y2, y3):f(x + y) = f(x1 + y1, x2 + y2, x3 + y3)

= ((x1 + y1) + (x2 + y2), (x1 + y1) + (x3 + y3), x3 + y3)= ((x1 + x2) + (y1 + y2), (x1 + x3) + (y1 + y3), x3 + y3)= (x1 + x2, x1 + x3, x3) + (y1 + y2, y1 + y3, y3)

= f(x) + f(y) x R3, k R ta c x = (x1, x2, x3)

f(kx) = f(kx1, kx2, kx3)= (kx1 + kx2, kx1 + kx3, kx3)= k.(x1 + x2, x1 + x3, x3)= k.f(x)

4.1.3 Cc tnh cht c bn ca nh x tuyn tnh

Cho E, F l hai K-khng gian vect , f : E

F l nh x tuyn tnh, khi

1. f(0E) = 0F, f(a) = f(a)

2. nh x f : E F l nh x tuyn tnh khi v ch khi

f(ax + by) = a.f(x) + b.f(y) , a, b K; x, y E

3. Vi mi x1, x2, . . . , xn E v k1, k2, . . . , kn K ta c

f(k1x1 + k2x2 + + knxn) = k1f(x1) + k2f(x2) + + knf(xn)

4. nh x tuyn tnh bin mt h ph thuc tuyn tnh thnh mt h ph thuctuyn tnh

5. nh x tuyn tnh khng lm tng hng ca mt h vect

4.2 Ma trn ca nh x tuyn tnh

4.2.1 nh l c bn v s xc nh ca nh x tuyn tnh

nh l Cho E l khng gian vect n chiu (dimE = n), B ={

e1, e2, . . . , en}

l mtc s ca E, F l khng gian vect ty v b1, b2, . . . , bn l h cc vect ty trongF. Khi tn ti duy nht mt nh x tuyn tnh f : E F tha mn f(ei) = bi vimi i = 1, 2, . . . , n

T nh l ny ta thy mt nh x tuyn tnh hon ton c xc nh nu nh tabit c nh ca mt c s ca n. V cho 1 nh x tuyn tnh ta ch cn cho nhca mt c s l .

4.2.2 Ma trn ca nh x tuyn tnh

Gi s E, F l hai K-khng gian vect , dimE = n, dimF = m v nh x tuyn tnh

f : E F.Gi s B = {e1, e2, . . . , en} l mt c s ca E; C = {f1, f2, . . . , f m} l mt c s

ca F;V f(ei) F nn f(ei) biu th tuyn tnh c qua h cc vect ca C. Ta c

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CHNG 4. NH X TUYN TNH

f(e1) = a11f1 + a12f2 + + a1mfmf(e2) = a21f1 + a22f2 + + a2mfm

f(en) = an1f1 + an2f2 + + anmfm

Ma trn

A =

a11 a21 . . . an1a12 a22 . . . an2

...... . . .

...a1m a2m . . . anm

gi l ma trn ca nh x tuyn tnh f trong cp c s B, C. Ta k hiu A = Af/B,C

Trng hp c bit khi f l php bin i tuyn tnh ca E, f : E E v B Cth ma trn ca nh x tuyn tnh f trong cp c c B, B c gi l ma trn ca ftrong c s B v k hiu l Af/B

V d 1: Cho nh x tuyn tnh f : R2 R3

f(x1, x2) = (x1 + 2x2, x1 x2, x2)

Tm ma trn ca nh x tuyn tnh f trong cp c s B, C vi cc c s B, C cho nhsau:

B = {e1 = (1, 1), e2 = (1, 0)}C = {f1 = (1, 1, 1), f2 = (1, 2, 1), f3 = (1, 3, 2)}

Gii: Ta c

f(e1) = a1f1 + a2f2 + a3f3 = (3, 0, 1) (4.1)f(e2) = b1f1 + b2f2 + b3f3 = (1, 1, 0) (4.2)

Theo nh ngha th ma trn ca nh x tuyn tnh f i vi cp c s B, C l Af/B,C

Af/B,C =

a1 b1a2 b2a3 b3

Gii cc phng trnh (1) v (2) tm a1, a2, a3 v b1, b2, b3. Cc phng trnh (1), (2)tng ng vi h phng trnh tuyn tnh m ma trn cc h s b sung nh sau: 1 1 1 3 11 2 3 0 1

1 1 2 1 0

h2h1+h2h3h1+h3

1 1 1 3 10 3 2 3 00 2 1 4 1

h2h3+h2 1 1 1 3 10 1 1 1 1

0 2 1 4 1h32h2+h3

1 1 1 3 10 1 1 1 10 0 1 6 3

H (1): a3 = 6; a2 = 1

a3 =

5; a1 = 3

a3 + a2 =

8

H (2): b3 = 3; b2 = 1 b3 = 2; b1 = 1 + b2 b3 = 4

Vy Af/B,C =

a1 b1a2 b2a3 b3

= 8 45 2

6 3

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4.3. HT NHN V NH

V d 2: Cho nh x tuyn tnh f : R3 R3

f(x1, x2, x3) = (x1 + 2x2 x3, x2 + x3, x1 + x2 2x3)Tm ma trn ca f i vi c s B = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}

4.2.3 Biu thc ta ca nh x tuyn tnhCho E, F l cc K-khng gian vect , B = {e1, e2, . . . , en} l mt c s ca E; C ={f1, f2, . . . , f m} l mt c s ca F. Cho f : E F l nh x tuyn tnh. tA = Af/B,C l ma trn ca f trong cp c s B, C

Vi mi vect x E, gi s

[x]/B =

x1x2...

xn

v [f(x)]/C =

y1y2...

ym

Khi cng thc sau gi l biu thc ta ca nh x tuyn tnh fy1y2...

ym

= A.

x1x2...

xn

4.2.4 Ma trn ca nh x tuyn tnh trong cc c s khc

nhauCho E, F l cc K-khng gian vect , B = {e1, e2, . . . , en}, B = {e1, e2, . . . , en} l haic s ca E; C = {f1, f2, . . . , f m}; C = {f1, f2, . . . , f m} l hai c s ca F. Cho nhx tuyn tnh f : E F, khi ta c cng thc lin h gia ma trn ca f trong cpma trn B, C vi ma trn ca f trong cp c s B, C nh sau:

Af/B,C

= T1CC

.Af/B,C .TBB

trong TBB l ma trn chuyn c s t B sang B

Nu f : E

E l php bin i tuyn tnh v B ={

e1, e2, . . . , en

}, B

=

{e1, e2, . . . , en} l hai c s ca E, ta cAf/

B

= T1BB

.Af/B .TBB

4.3 Ht nhn v nh

4.3.1 Cc khi nim c bn

Cho E, F l cc K-khng gian vect , f : E F l nh x tuyn tnh

K hiu Kerf = {x E|f(x) = 0} gi l ht nhn ca nh x tuyn tnh f K hiu Imf = f(E) = {f(x)|x E} gi l nh ca nh x tuyn tnh f

Ta c Kerf v Imf u l cc K-khng gian vect

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4.3.2 Nhn xt

Cch tm ht nhn ca mt nh x tuyn tnh f : E F.Chn B = {e1, e2, . . . , en} l mt c s ca E; C = {f1, f2, . . . , f m} l mt c sca F. Ta c [f(x)]/C = A.[x]/B

x Kerf f(x) = 0

[f(x)]/C =

00...0

A.[x]/B =

00...0

()

Nh vy x Kerf khi v ch khi ta ca x trong c s B l nghim ca hphng trnh tuyn tnh thun nht (*)

T tm ht nhn ca nh x tuyn tnh f : E F ta lm nh sau:

1. Tm ma trn ca nh x tuyn tnh f i vi cp c s B, C (tm Af/B,C)

2. Gii h phng trnh tuyn tnh thun nht

A.

x1x2...

xn

= 00...0

()Tp hp tt c cc vect c ta i vi c s B l nghim c bn ca hphng trnh (*) l c s ca Kerf

3. Tm nghim ca (*)

4. Kerf l tp tt c cc nghim ca (*). (H nghim c bn ca (*) chnh l

c s ca Kerf) Cch tm nh ca nh x f

V e1, e2, . . . , en l h sinh ca E nn f(e1), f(e2), . . . , f (en) l h sinh ca Imf.

Hay Imf = f(e1), f(e2), . . . , f (en). Ta tm mt h con c lp tuyn tnh tii ca f(e1), f(e2), . . . , f (en) l c s ca Imf (S vect c lp tuyn tnhti i bng hng ca h cc vect f(e1), f(e2), . . . , f (en))

V d Cho nh x tuyn tnh f : R3

R3

f(x1, x2, x3) = (x1 + 2x2 x3, x2 + x3, x1 + x2 2x3)

Tm ht nhn v nh ca f.

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4.4. N CU, TON CU, NG CU

Gii:

(x1, x2, x3) Kerf f(x1, x2, x3) = 0 (x1, x2, x3) l nghim h phng trnh:

x1 + 2x2 x3 = 0x2 + x3 = 0x1 + x2 2x3 = 0

Ta bin i ma trn h s b sung 1 2 1 00 1 1 01 1 2 0

h3h1+h3 1 2 1 00 1 1 0

0 1 1 0

h3h2+h3 1 2 1 00 1 1 0

0 1 1 0

H c v s nghim ph thuc 1 tham s l x3. Ta c:

x3 = tx2 = x3 = tx1 = 2x3 + x3 = 3x3 = 3t

, t R

Nghim c bn ca h l (3, 1, 1)Vy {(3, 1, 1)} l c s ca Kerf v dim Kerf = 1

Tm ImfTa tm nh ca f i vi c s chnh tc B =

{e1 = (1, 0, 0), e2 = (0, 1, 0), e3 =

(0, 0, 1)}. Ta cf(e1) = (1, 0, 1), f(e2) = (2, 1, 1), f(e3) = (1, 1, 2)Imf = f(e1), f(e2), f(e3)Tm h con c lp tuyn tnh ti i ca h f(e1), f(e2), f(e3).

Tm hng ca h cc vect f(e1), f(e2), f(e3)

1 0 12 1 1

1 1

2

f(e1)f(e2)f(e3)

h22h1+h2h3h1+h3

1 0 10 1 10 1

1

f(e1)f(e2)f(e3)

h3h2+h3

1 0 10 1 10 0 0

f(e1)f(e2)f(e3)

Vy c s ca Imf l {f(e1), f(e2)}

4.4 n cu, ton cu, ng cu

4.4.1 Cc nh ngha

Cho E, F l cc K-khng gian vect v nh x tuyn tnh f : E F. Khi :

f gi l n cu nu f n nh

f gi l ton cu nu f ton nh

f gi l ng cu nu f song nh

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4.4.2 Cc nh l v n cu, ton cu, ng cu

nh l 6 Cho E, F l cc K-khng gian vect v nh x tuyn tnh f : E F. Khi:

1. f l n cu Kerf = {0}2. f l ton cu Imf = F

Chng minh

nh l 7 Cho E, F l cc K-khng gian vect hu hn chiu v nh x tuyn tnhf : E F. Khi f l ng cu khi v ch khi dim f(E) =dim F

4.5 Gi tr ring, vect ring ca ma trn, nh xtuyn tnh

4.5.1 Vect ring v gi tr ring ca ma trn

nh ngha: Cho A l ma trn vung cp n, K gi l gi tr ring ca A nunh tn ti vect x = (x1, x2, . . . , xn) = 0 ca Kn sao cho

A

x1x2...

xn

=

x1x2...

xn

()

Khi vect x = (x1, x2, . . . , xn) gi l vect ring ca ma trn A ng vi gi tr ring

T () ta c:

[A I]

x1x2...

xn

= 0

Cch tm vect ring v gi tr ring

1. Tnh det(A I)Ch : t PA() =det(A I) gi l a thc c trng ca ma trn A

2. Gii phng trnh PA() = 0

Tt c cc nghim ca phng trnh u l gi tr ring ca ma trn A

3. ng vi gi tr ring 0 (0 l mt nghim ca phng trnh PA() = 0)

Ta gii h phng trnh

[A 0I]

x1x2...

xn

=

00...0

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4.5. GI TR RING, VECT RING CA MA TRN, NH X TUYN TNH

Tt c cc nghim khc 0 ca h () u l vect ring ca A ng vi gi trring 0

V d: Tm vect ring v gi tr ring ca ma trn A trn tp s thc R

A =0 1 11 0 1

1 1 0

Gii:

1. Tm a thc c trng PA()

PA() = det(A I)

= 1 1

1

11 1

= ()3 + 3 + 2

2. Tm gi tr ring

3 + 3 + 2 = 0 ( + 1)(2 + + 2) = 0 = 1 hoc = 2Ta c 2 vect ring =

1, = 2

3. Tm vect ring

= 1Ta gii h phng trnh

[A (1)I]

x1x2...

xn

=

00...

0

1 1 1 01 1 1 0

1 1 1 0

h2h1+h2h3h1+h3

1 1 1 00 0 0 00 0 0 0

H c v s nghim ph thuc 2 tham s

x1 = a bx2 = a

x3 = b

ng vi gi tr ring = 1 ta c cc vect ring (a b,a,b) vi a, b Rv a, b khng ng thi bng 0.

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= 2Ta gii h phng trnh

[A

2.I]

x1x2...

xn

=

00...0

2 1 1 01 2 1 01 1 2 0

h1h3 1 1 2 01 2 1 0

2 1 1 0

h2h1+h2h32h1+h3

1 1 2 00 3 3 00 3 3 0

h3h2+h3 1 1 2 00 3 3 0

0 0 0 0

H phng trnh c v s nghim ph thuc 1 tham s

x1 = x2 + 2x3 = cx2 = x3 = c

x3 = c

ng vi gi tr ring = 2 ta c cc vect ring (c,c,c) vi c R v c = 0

4.5.2 Vect ring v gi tr ring ca php bin i tuyn tnh

nh x tuyn tnh f : V

V gi l php bin i tuyn tnh trn V

nh ngha: Cho f : V V l php bin i tuyn tnh. K gi l gi tr ringca f nu tn ti vect x V(x = 0) sao cho f(x) = x.

Khi ta ni x l mt vect ring ca f i vi gi tr ring

Cch tm gi tr ring v vect ring ca php bin i tuyn tnh Chof : V V l php bin i tuyn tnh.

E = {e1, e2, . . . , en} l mt c s bt k ca V.A = Af/E l ma trn ca php bin i tuyn tnh f trong c s EGi s u l vect ring ca f ng vi gi tr ring , khi ta c

f(u) = u

vi [u]/E =

x1x2...

xn

(ta ca u i vi c s E). Do [f(u)]/E =

x1x2

...xn

Hay

[f(x)]/E = A.[u]/E

x1x2...

xn

= A.x1x2...

xn

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4.6. CHO HA MA TRN, CHO HA NH X TUYN TNH

Ta c (x1, x2, . . . , xn) l vect ring ca ma trn A ng vi gi tr ring ca A l gi tr ring ca f l gi tr ring ca A (x1, x2, . . . , xn) l vect ring ca ma trn A ng vi gi tr ring u =

x1e1 + x2e2 +

+ xnen l vect ring ca f ng vi gi tr ring

4.6 Cho ha ma trn, cho ha nh x tuyn tnh

4.6.1 iu kin cho ha

nh ngha 1: Cho php bin i tuyn tnh f : V V, f gi l cho ha c nutn ti mt c s B sao cho Af/B l ma trn cho.

Cho ha php bin i tuyn tnh f tc l tm c s B sao cho ma trn Af/B lma trn cho.

nh ngha 2: Cho A, B l cc ma trn vung cp n. Ma trn A ng dng vi matrn B nu tn ti ma trn C vung cp n khng suy bin sao cho B = C1.A.C

Hai ma trn ng dng th c cng a thc c trng.

nh ngha 3: Cho ma trn vung A cp n. Ma trn A gi l cho ha c nu Ang dng vi ma trn cho. Tc l tn ti mt ma trn T sao cho T1AT l ma trncho

Nhn xt: Php bin i tuyn tnh f cho ha c khi v ch khi Af/B cho hac.

nh l 8 Php bin i tuyn tnh f : V V cho ha c khi v ch khi trong Vc mt c s l cc vect ring ca f

Hay ta c th ni cch khc nh sau: Php bin i tuyn tnh f cho ha c khi vch khi f c n vect ring c lp tuyn tnh (vi n=dimV)

nh l 9 Ma trn vungA cp n cho ha c khi v ch khi A c n vect ringc lp tuyn tnh.

nh l 10 Nu ma trnA c n gi tr ring khc nhau tng i mt th n c n vectring c lp tuyn tnh

4.6.2 Cch cho ha ma trn v php bin i tuyn tnh

Cho ha ma trn: cho ha ma trn A cp n ta lm nh sau:

1. Tm PA(), gii phng trnh PA() = 0 tm cc nghim l tt c cc gi trring ca A.

Gi s ta c cc gi tr ring 1, 2, . . . , k. Ta k hiu mi l s bi ca i (tc l

nu 1 l nghim kp th m1 = 2, nu j l nghim bi 3 th mj = 3) Nu m1 + m2 + + mk < n th ma trn A khng cho ha c Nu m1 + m2 + + mk = n th ma trn A cho ha c

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2. Tm cc vect ring c lp tuyn tnh.

Vi mi i ta gii h phng trnh

[A

iI]

x1x2...

xn

= 0 ()

Vi mi gi tr 1 i k, h phng trnh c v s nghim v ph thuc mitham s th ma trn A cho ha c. Ngc li th khng cho ha c

Vi mi gi tr i, chn ra mi vect l h nghim c bn ca () l cc vectring ca A

Gp li ta c n vect ring c lp tuyn tnh ca A l a1, a2, . . . , an

3. Lp ma trn T sao cho cc ct ca T ln lt l cc vect ring c lp tuyntnh a1, a2, . . . , an. Khi T1AT l ma trn cho c cc phn t trn ng chochnh l cc gi tr ring. (Nu ta xt cc gi tr ring theo th t 1, 2, . . . , kth cc phn t trn ng cho chnh cng theo th t 1, 2, . . . , k)

V d: Cho ha ma trn

A =

15 18 169 12 84 4 6

Gii:

1. Tm a thc c trng PA()

PA() = det(A .I) =

15 18 169 12 84 4 6

= 3 32 + 4 + 12Gii phng trnh PA() = 0

3 32 + 4 + 12 = 0

( + 3)( + 2)(

2) = 0

Vy ma trn A c cc vect ring l 1 = 3; 2 = 2; = 2(theo nh l 10 th ma trn A c 3 gi tr ring khc nhau nn A cho ha c)

2. Tm cc vect ring c lp tuyn tnh

Vi 1 = 3 ta gii h phng trnh A 1I = 0 18 18 16 09 9 8 04 4 3 0

h12h2+h1h3

4

9h2+h3

0 0 0 09 9 8 00 0 5

90

H phng trnh c v s nghim ph thuc 1 tham s: x3 = 0; x2 = a; x1 = aNghim tng qut ca h phng trnh (a,a, 0) vi a RNghim c bn ca h phng trnh l (1, 1, 0)Chn 1 vect ring ca A ng vi gi tr ring 1 = 3 l a1 = (1, 1, 0)

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Vi 2 = 2 ta gii h phng trnh A 2I = 0 17 18 16 09 10 8 04 4 4 0

h12h2+h1 1 2 0 09 10 8 0

4 4 4 0

h29h1+h2h34h1+h3

1 2 0 00 8 8 00 4 4 0

h22h3+h2

1 2 0 00 0 0 00 4 4 0

H phng trnh c v s nghim ph thuc 1 tham s: x3 = b; x2 = b; x1 =2x2 = 2b

Nghim tng qut ca h phng trnh (2b,b,b) vi b RNghim c bn ca h phng trnh l (2, 1, 1)Chn 1 vect ring ca A ng vi gi tr ring 2 = 2 l a2 = (2, 1, 1)

Vi 3 = 2 ta gii h phng trnh A 3I = 0

13 18 16 09 14 8 04 4 8 0 h13h3+h1

1

6 8 0

9 14 8 04 4 8 0 h29h1+h2h34h1+h3

1

6 8 0

0 40 80 00 20 40 0

h22h3+h2 1 6 8 00 0 0 0

0 20 40 0

H phng trnh c v s nghim ph thuc 1 tham s: x3 = c; x2 = 2x3 =2c; x1 = 6x2 8x3 = 4cNghim tng qut ca h phng trnh (4c, 2c, c) vi c RNghim c bn ca h phng trnh l (4, 2, 1)

Chn 1 vect ring ca A ng vi gi tr ring 3 = 2 l a3 = (4, 2, 1)

3. Lp ma trn T (c cc ct l cc vect ring a1, a2, a3)

T =

1 2 41 1 20 1 1

Khi ta c ma trn cho T1AT =

3 0 0

0 2 00 0 2

Ch : Nu ta xt cc gi tr ring theo cc th t khc nhau th ta s c cc matrn T khc nhau, khi ma trn cho cng khc nhau. i vi v d pha trn ta xtcc gi tr ring theo th t 3; 2; 2 th ta c ma trn cho c cc phn t trnng cho cng theo th t 3; 2; 2 ny. Nu ta xt theo th t i; j; k th matrn cho thu c c cc phn t nm trn ng cho cng theo th t i; j; k

Cho ha php bin i tuyn tnh

Cho php bin i tuyn tnh F : V

V. Ly B l mt c s bt k ca V v

dimV = n, khi ta c ma trn ca nh x tuyn tnh l A = Af/BTa tin hnh cho ha ma trn Af/B . Nu Af/B l ma trn cho ha c th ta c

n vect ring c lp tuyn tnh. Chn n vect ny lp thnh mt c s ca V, khi ma trn ca php bin i tuyn tnh trong c s va lp c chnh l ma trn cho.

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Bi tp

Bi 28. Trong cc nh x sau, nh x no l nh x tuyn tnh?a. f : R3 R3, f(x1, x2, x3) = (2x2 x3, x1 + x3, 3x1 + 2x2)b. f : R3 R3, f(x1, x2, x3) = (2x1 + x3, x1, x2 + 2)

Bi 29. Cho nh x tuyn tnh f : R2

R3 xc nh bi

f(1, 2) = (3, 1, 5); f(0, 1) = (2, 1, 1)Hy xc nh f(x, y)Bi 30. Cho f : R3 R3 l nh x tuyn tnh xc nh nh sau:

f(x1, x2, x3) = (x1, x2, 0)

Tm ma trn ca nh x trong c s chnh tc ca R3

Bi 31. Cho f : R2 R2 l nh x tuyn tnh c tnh chtf(1, 1) = (2, 0); f(0, 1) = (3, 1)

a. Tnh f(1, 0)b. Tm ma trn ca f trong c s chnh tc ca R2

Bi 32. Cho f : R3 R3 l nh x tuyn tnh c ma trn trong c s

U = {u1 = (1, 1, 1), u2 = (1, 1, 0), u3 = (1, 0, 0)} l A = 8 4 24 2 1

0 0 0

Tm f(x1, x2, x3)

Bi 33. Cho f : R3 R3 l nh x tuyn tnh xc nh bif(1, 0, 0) = (1, 1, 1); f(

1, 1, 0) = (

2,

1, 0); f(0,

1, 1) = (2, 1, 3)

a. Hy xc nh f(x1, x2, x3)b. Tm c s v s chiu ca kerf

Bi 34. Tm gi tr ring v vect ring ca ma trn

a. A =

5 11 3

b. A =

1 1 11 1 01 0 1

c. A =3 1 12 4 2

1 1 3

Bi 35. Cho ma trn A =

1 42 3

a. Tm tr ring v cc vect ring ca A

b. Tm ma trn kh nghch S sao cho A cho ha c, tc l ma trn S

1AS cdng cho.Bi 36. Tm ma trn kh nghch S lm cho ha c cc ma trn sau:

a. A =

1 22 3

b. A =

3 1 12 4 21 1 3

c. A =0 1 01 0 1

0 1 0

Bi 37. Tnh A10 vi

a. A =

1 22 3

b. A =

3 1 12 4 21 1 3

Bi 38. Cho f :

R3

R3

l nh x tuyn tnh xc nh nh sau:f(x1, x2, x3) = (z1 + x2 + x3, 2x2 + x3, 2x2 + 3x3)

Tm cc tr ring v vect ring ca f

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Chng 5Dng song tuyn tnh - Dng tonphng

5.1 Cc khi nim c bn5.1.1 Dng song tuyn tnh

Dng song tuyn tnh ca 2n bin x1, x2, . . . , xn v y1, y2, . . . , yn l biu thc dng:

B = a11x1y1 + a12x1y2 + + a1nx1yn++a21x2y1 + a22x2y2 + + a2nx2yn++ ++an1xny1 + an2xny2 + + annxnyn

Nh vy

B =ni=1

nj=1

aijxiyj

V d: Dng song tuyn tnh ca 4 bin x1, x2, y1, y2 nh sau:

f(x1, x2, y1, y2) = 2x1y1 3x1y2 2x2y1 + x2y2

5.1.2 Dng ton phng

Dng ton phng ca n bin x1, x2, . . . , xn l biu thc dng:

f(x1, x2, . . . , xn) = a11x21 + a12x1x2 + + a1nx1xn+

+a21x2x1 + a22x22 + + a2nx2xn+

+ +an1xnx1 + an2xnx2 + + annx2n

f(x1, x2, . . . , xn) =ni=1

nj=1

aijxixj v aij = aji

T cc h s trong dng ton phng ta lp c ma trn

A =

a11 a12 . . . a1na21 a22 . . . a2n

...... . . .

...an1 an2 . . . ann

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5.2. A DNG TON PHNG V DNG CHNH TC

Ma trn A c gi l ma trn ca dng ton phng f(x1, x2, . . . , xn). Ma trnca dng ton phng l ma trn i xng (aij = aji).

V d: f(x1, x2, x3) = x21+2x22+3x

238x1x24x1x36x2x3 l mt dng ton phng.

Ta c th vit li nh sau: f(x1, x2, x3) = x214x1x22x1x34x2x1+2x223x2x32x3x13x3x2 + 3x

23

A = 1 4 24 2 32 3 3

Ta t X =

x1x2...

xn

Khi dng ton phng c th c vit li nh sau f(X) = XTAX

5.1.3 Dng chnh tcDng ton phng c gi l dng chnh tc nu n c dng

f(x1, x2, . . . , xn) = a1x21 + a2x

22 + + anx2n

(tc l f ch gm cc bnh phng)

Nhn xt Ma trn ca dng chnh tc l ma trn cho

5.1.4 Php i bin s

Cho cc bin x1, x2, . . . , xn v cc bin y1, y2, . . . , yn. Php bin i cc bin s x1, x2, . . . , xnti cc bin s y1, y2, . . . , yn nu ta cx1 = s11y1 + s12y2 + + s1nynx2 = s21y1 + s22y2 + + s2nyn.....................................

xn = sn1y1 + sn2y2 + + snnynMa trn cc h s

P = s11 s12 . . . s1ns21 s22 . . . s2n... ... . . . ...

sn1 sn2 . . . snn

gi l ma trn ca php bin i bin s cho.

Ta c X = P Y

5.2 a dng ton phng v dng chnh tc

5.2.1 Phng php Lagrange

Cho dng ton phng

f(x1, x2, . . . , xn) =ni=1

nj=1

aijxixj ()

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CHNG 5. DNG SONG TUYN TNH - DNG TON PHNG

vi ma trn ca dng ton phng l

A =

a11 a12 . . . a1na21 a22 . . . a2n

...... . . .

...

an1 an2 . . . ann

1. Nu dng ton phng () c cc tt c cc aii = 0, i = 1, 2, . . . , n v c mtaij = 0(i = j), ta c th gi s a12 = 0 th ta bin i nh sau:

x1 = y1 y2x2 = y1 + y2

xi = yi , i = 2, 3, . . . , n

Khi ta a () v dng

f(y1, y2, . . . , yn) = a12(y1y2)(y1+y2)+a21(y1+y2)(y1y2)+ = 2a12y212a12y22+

Nh vy dng ton phng (*) c h s bnh phng khc 0.

2. Nh vy ta c th gi s dng ton phng () c t nht 1 h s ca bnhphng khc 0, chng hn a11 = 0

nh l 11 Tn ti mt php bin i bin s a dng ton phng v dngchnh tc

V d : a dng ton phng v dng chnh tc

Q(x1, x2, x3) = 2x1x2 + 2x1x3 6x2x3

y khng c bnh phng no. Ta lm xut hin bnh phng bng cch i bins nh sau

x1 = y1 y2x2 = y1 + y2

x3 = y3

v nh vy ta c ma trn i bin s P1 =1 1 01 1 0

0 0 1

v X = P1.YKhi

Q(y1, y2, y3) = 2(y1 y2)(y1 + y2) + 2(y1 y2)y3 6(y1 + y2)y3= 2y21 2y22 4y1y3 8y2y3

Tip theo ta bin i nh sau

Q(y1, y2, y3) = 2y21

2y22

4y1y3

8y2y3

= 2y21 4y1y3+2y23 2y22 8y2y32y23= 2(y1 y3)2 2y22 8y2y38y23 + 8y23 2y23= 2(y1 y3)2 2(y22 + 4y2y3 + 4y23) + 6y23= 2(y1 y3)2 2(y2 + 2y3)2 + 6y23

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5.2. A DNG TON PHNG V DNG CHNH TC

Ta thc hin php i bin s nh sau

t1 = y1 y3t2 = y2 + 2y3

t3 = y3

y1 = t1 + t3

y2 = t2 2t3y3 = t3

v ta c ma trn ca php i bin s l P2 =

1 0 10 1 20 0 1

Y = P2.T

Ta c dng chnh tc ca dng ton phng l

Q(t1, t2, t3) = 2t21 2t22 + 6t23

vi ma trn i bin s l P = P1.P2 = 1 1 31 1 10 0 1

Vy php i bin s t x1, x2, x3 n t1, t2, t3 l:

x1 = t1 t2 + 3t3x2 = t1 + t2 t3x3 = t3

Ch : Dng chnh tc c th khc nhau do ta dng cc php i bin s khc nhauNu ta dng php i bin s

u1 =

2t1

u2 =

2t2

u3 = t3

hay

t1 =

u12

t2 =u2

2t3 = u3

th ta c dng chnh tc l

Q(u1, u2, u3) = u2

1 u2

2

+ 6u2

3

5.2.2 Phng php Jacobi

p dng cho cc dng ton phng m ma trn A = (aij) ca n tha:

1 = a11 = 0, 2 =a11 a12a21 a22

= 0, ,

i = a11 a12 . . . a1i

a21 a22 . . . a2i...... . . .

...ai1 ai2 . . . aii

= 0, , n = |A| = 0()

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nh l 12 Nu dng ton phng bc hai tha cc iu kin(**) th c php i bins x1, x2, . . . , xn ti cc bin s y1, y2, . . . , yn a Q v dng chnh tc

Q(y1, y2, . . . , yn) = 1y21 +

21

y22 + +n

n1y2n

Vi php bin i bin s

x1 = y1 + 21y2 + 31y3 + + n1ynx2 = y2 + 32y3 + + n2yn................................

xi = yi + i+1iyi+1 + i+2iyi+1 + + niyn.................................

xn = yn

( )

trong cc cc h s ij c tnh nh sau:

ji = (1)i+j Dj1,ij1

( )

Dj1,i l nh thc con ca |A| c lp trn giao ca cc dng 1, 2, . . . , j 1 v ccct th 1, 2, . . . , i 1, i + 1, . . . , j

V d: a dng ton phng

Q(x1, x2, x3) = 2x21 2x22 4x1x3 8x2x3

v dng chnh tc bng phng php Jacobi

Gii:Ta c ma trn ca dng ton phng l:

A =

2 0 20 2 42 4 0

Trc ht ta c:

1 = 2 = 0, 2 =2 00 2

= 4 = 0, 3 = detA = 24 = 0

Ta thc hin php i bin sx1 = y1 + 21y2 + 31y3

x2 = y2 + 32y3

x3 = y3

trong

21 = (1)2+1D21,121

= D1,11

=0

2= 0

(D1,1 l giao ca hng 1 v ct 2; giao ca 1 hng v 1 ct)

31 = (1)3+1D31,131

=D2,12

=

0 22 4

4 = 1

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5.3. KHNG GIAN EUCLIDE

(D2,1 l giao ca dng 1,2 v ct 2,3; giao ca 2 hng v 2 ct)

32 = (1)3+2D31,231

= D2,22

=

2 20 4

4 = 2

(D2,2 l giao ca dng 1,2 v ct 1,3; giao ca 2 hng v 2 ct)Nh vy php i bin s lx1 = y1 + 0y2 + 1y3

x2 = y2 2y3x3 = y3

Ta a dng ton phng Q(x1, x2, x3) = 2x21 2x22 4x1x3 8x2x3 v dng chnhtc nh sau:

Q(y1, y2, y3) = 2y21

2y22 + 6y

23

5.2.3 Lut qun tnh

Lut qun tnh Mt dng ton phng c th a v dng chnh tc bng nhiucch khc nhau. Tuy nhin ngi ta chng minh c nh l sau y gi l lut quntnh ca dng ton phng

nh l 13 Nu mt dng ton phng c a v dng chnh tc bng nhiu cchkhc nhau th cc h s dng v cc h s m trong dng chnh tc ca chng l nhnhau ch khc nhau v cch sp xp

Dng ton phng xc nh dng Dng ton phng trn tp s thc f(x1, x2, . . . , xn)l xc nh dng khi v ch khi dng chnh tc ca n l:

f = a1y21 + a2y

22 + + any2n

vi ai > 0 vi i = 1, 2, . . . , n

nh l Sylvester Dng ton phng trn tp s thc f(x1, x2, . . . , xn) l xc nhdng khi v ch khi tt c cc

i > 0, i = 1, 2, . . . , n

5.3 Khng gian Euclide

5.3.1 nh ngha

Cho X l mt R-khng gian vect . Ta gi mt tch v hng trn X l mt qui tct hai vect bt k x, y X tng ng vi mt s thc x, y tha cc iu kin diy vi mi x,y,z X, R

1. x, y = y, x2. x,y = x, y3. x + y, z = x, z + y, z

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4. x, x 0, x, x = 0 x = 0Khng gian vect thc X cng vi tch v hng trn X gi l mt mt gian Euclide.

Ch rng: Nu t f(x, y) = x, y th f l mt dng song tuyn tnh i xngtrn X. Do ta c th nh ngha tch v hng l mt dng song tuyn tnh ixng xc nh dng, tc l tha mn iu kin 4.

V d: Vi mi x = (x1, x2, . . . , xn), y = (y1, y2, . . . , yn) Rn, tx, y = x1y1 + x2y2 + + xnyn

Ta c mt tch v hng trn Rn. Rn l mt khng gian Euclide cng vi tch vhng ny.

5.3.2 di vect , gc gia cc vect

Cho khng gian Euclide X v vect x X. Ta gi di hay chun ca vect x l:

x

= x, xV d: Trong Rn th x = x21 + x22 + + x2n

Mt s tnh cht

x 0, x = 0 x = 0 x = ||. x |x, y| x . y x + y x + y

Gc gia hai vect x, y, k hiu (x, y) xc nh nh sau:cos(x, y) = x, y x . y

Nu (x, y) = 2

ta ni vect x trc giao vi vect y. K hiu xy

xy x, y = 0

5.3.3 C s trc giao, c s trc chun

Cho mt khng gian Euclide n chiu. Xt mt h gm n vect khc 0 ca X c gil mt c s trc giao ca X nu chng i mt trc giao vi nhau.Mt c s trc giao bao gm cc vect c di bng 1 c gi l mt c s trc

chun. Nh vy, {e1, e2, . . . , en} l mt c s trc chun nu

ei, ej =

1 nu i = j0 nu i = j

Nu {e1, e2, . . . , en} l mt c s trc giao th e1 e1 ,

e2 e2 , ,

en en

l mt

c s trc chun, gi l trc chun ha c s trc giao cho.

Mi c s trc giao ca khng gian Euclide X u l c s ca khng gian vectX.

T mt c s bt k ta c th xy dng nn mt c s trc giao, qu trnh cgi l trc giao ha mt h vect (hay cn gi l phng php Gram-Schmidt)

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nh l 14 Cho {f1, f2, . . . , f n} l mt c s ca khng gian Euclide X. Khi ta xydng mt c s trc giao {e1, e2, . . . , en} nh sau:

e1 = f1

e2 = f2

f2, e1

e1, e1e1

e3 = f3 f3, e1e1, e1e1 f3, e2e2, e2e2

ei = fi i1k=1

fi, ekek, ekek

V d: Trong R3, trc giao ha h

f1 = (1, 1, 1, ), f2 = (1, 1, 0), f3 = (1, 0, 0)

Gii: Ta c

e1 = f1 = (1, 1, 1)e2 = f2 f2, e1e1, e1e1 = (1, 1, 0)

2

3(1, 1, 1) =

13

,1

3,23

e3 = f3 f3, e1e1, e1e1

f3, e2e2, e2e2 = (1, 0, 0)

1

3(1, 1, 1)

1323

13

,1

3,23

=1

2,12

, 0

5.3.4 Cho ha ma trn i xng bng ma trn trc giao

Ma trn A c gi l ma trn i xng nu A = AT

V d:

A = 1 2 32 4 03 0 1

Ma trn vung P gi l ma trn trc giao nu P l ma trn vung khng suy bintha mn

PT = P1

tc ma trn nghch o ca P bng ma trn chuyn v ca P.T iu kin ny ta c PTP = IMt s tnh cht ca ma trn trc giao P:

1.|P

|= 1

2. Tng bnh phng cc phn t trn 1 dng hay 1 ct ca P bng 1.

3. Hai dng khc nhau th trc giao vi nhau

nh l 15 Vi mi ma trn i xngA tn ti ma trn trc giao P sao cho P1APl ma trn cho.

Hn na, nu 1, 2, . . . , n l n vect ring ca A (c th trng nhau) th ta c thchn ma trn trc giao P

P1AP = 1 0 00 2 0... ... . . . ...0 0 n

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nh l 16 Cho A l ma trn i xng cp n. Khi trongRn tn ti mt c s trcchun gm nhng vect ring ca A.

By gi ta ch ra phng php xy dng c s ni trong nh l 16 v cng l phngphp tm ma trn P ni trong nh l 15

Cho ma trn i xng A. V A cho ha c nn mi gi tr ring bi m ca A

c ng m vect ring c lp ng vi 0 v sau trc chun ha h m vect nyta c mt h trc chun gm m vect . Hin nhin m vect ny cng l vect ringng vi gi tr ring 0. Tin hnh nh vy i vi tt c cc gi tr ring ta c mth gm n vect E = {e1, e2, . . . , en}. Gi P l ma trn c cc vect ca E l cc ctth P l trc giao v

P1AP =

1 0 00 2 0...

... . . ....

0 0 n

trong i l gi tr ring tng ng vi vect ring ei.V d: Cho ha trc giao ma trn i xng

A =

1 2 22 1 22 2 1

Gi: Phng trnh c trng

|A

I

|=

1 2 22 1

2

2 2 1 = 0 ( 5)( + 1)2 = 0

Vy ta c cc gi tr ring: 1 = 5 (n), 2 = 1 (kp) Vi = 5, ta gii h phng trnh tm vect ring l 4 2 2 02 4 2 0

2 2 4 0

2 1 1 00 1 1 0

0 0 0 0

Vy h phng trnh c nghim:

x1 = c

x2 = c

x3 = c

, c R

Ta c 1 vect ring c lp l v1 = (1, 1, 1) (cho c = 1), trc chun ha tac:

e1 = 1

3,

13

,1

3

Vi =

1, ta gii h phng trnh tm vect ring l 2 2 2 02 2 2 0

2 2 2 0

1 1 1 00 0 0 0

0 0 0 0

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Vy h phng trnh c nghim:

x1 = c1 c2x2 = c1

x3 = c2

, c1, c2 R

Ta c 2 vect ring c lp l v2 = (1, 1, 0) (cho c1 = 1, c2 = 0) v v2 =(1, 0, 1) (cho c1 = 0, c2 = 1), trc giao ha ta c:

f2 = v2 = (1, 1, 0)

f3 = v3 v3, f2f2, f2f2 = (1, 0, 1) 1

2(1, 1, 0) =

1

2, 1

2, 1

Trc chun ha f2, f3 ta c

e2 =f2

f2=

1

2,

1

2, 0

e3 =f3

f3 =

16

, 16

,2

6

T , vi

P =

13

12

16

13

12

16

1

30

2

6

Do ta c

P1AP =

5 0 00 1 00 0 1

5.3.5 a dng ton phng v dng chnh tc bng php

bin i trc giao

Cho dng ton phng f(x) trn Rn c ma trn trong c s chnh tc l ma trn i

xng cp n, A = (aij)Gi E = {e1, e2, . . . , en} l mt c s trc chun ca Rn gm cc vect ring caA. P l ma trn c cc ct l e1, e2, . . . , en. Khi P l ma trn trc giao v l matrn chuyn t c s chnh tc sang c s E

Trong c s E, ma trn ca dng ton phng f(x) l

P1AP =

1 0 00 2 0...

... . . ....

0 0 n

trong i l gi tr ring tng ng vi vect ring ei.

V vy trong c s ny f(x) c dng chnh tc

f(x) = 1x21 + 2x

22 + + nx2n

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V d: a dng ton phng

f(x) = x21 + x22 + x

23 + 4x1x2 + 4x1x3 + 4x2x3

v dng chnh tc bng php bin i trc giao.Gii: Ma trn ca dng ton phng l

A =

1 2 22 1 22 2 1

Tin hnh cc bc nh trong v d bn trn ta c c s E = {e1, e2, e3} v ma trnca php bin i trc giao l

P =

13

12

16

1

31

2 1

613

02

6

v dng chnh tc ca f(x) l:

f(x) = 5x21 x22 x23

5.4 ng dng vo vic kho st ng v mt bc

hai5.4.1 nh ngha ng v mt bc hai

nh ngha 1: ng bc hai l ng c phng trnh dng

ax2 + 2bxy + cy2 + dx + ey + f = 0

Vi a,b,c,d,e,f R

nh ngha 2: Mt bc hai l mt m phng trnh c dng

ax2 + by2 + cz2 + 2dxy + 2exz + 2f yz + gx + hy + kz + m = 0

vi cc h s a,b,c,d,e,f,g,h,k,m R

5.4.2 Cc ng v mt bc hai c bn

1.x2

a2+

y2

b2= 1 (Elip)

2.x2

a2 y

2

b2= 1 (Hypebol)

3. y2 = 2px (Parabol)

4.x2

a2+

y2

b2+

z2

c2= 1 (Mt Elipsoide)

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5.4. NG DNG VO VIC KHO ST NG V MT BC HAI

5.x2

a2+

y2

b2 z

2

c2= 1 (Mt Hypeboloide 1 tng)

6.x2

a2+

y2

b2 z

2

c2= 1 (Mt Hypeboloide 2 tng)

7.x2

a2 y2

b2 z2

c2 = 0 (Mt nn)

8.x2

a+

y2

b= 2z(p,q R+) (Mt paraboloide - eliptic)

9.x2

a y

2

b= 2z(p,q R+) (Mt paraboloide - hypebolic)

5.4.3 Cch nhn dng ng mt bc hai

nhn dng cc ng, mt bc hai ta tin hnh a v phng trnh ca chng v

dng chnh tc nh cc php bin i bin s, sau dng php tnh tin v xem nthuc vo dng no.V d: Nhn dng ng cong bc hai sau:

3x2 + 2xy + 3y2 + 8

2y 4 = 0()Gii: Xt f = 3x2 + 2xy + 3y2

Ma trn ca f l A =

3 11 3

.

a thc c trng |A I| = 0 1 = 2, 2 = 4Khi ta c cc vect trc chun tng ng:

e1 = 1

2, 1

2

, e2 = 12

, 12

Ma trn ca php i bin s P =

1

2

12

12

12

Vy vi php i bin s

x =

12

x1 +1

2y1

y = 12

x1 + 12

y1

Thay vo () ta cf(x1, y1) = x

21 + 2y

21 4x1 + 4y1 2 = 0

Ta s dng php tnh tin. Vit li phng trnh trn dng

(x1 2)2 + 2(y1 + 1)2 = 8

t x2 = x1 2y2 = y1 + 1

Vy ta c phng trnh x22 + 2y22 = 8 hayx228

+y224

= 1

y l phng trnh ca mt elip.

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Bi tp

Bi 39. Tm ma trn ca cc dng ton phng saua. f(x1, x2) = 3x21 + 4x1x2 5x22b. f(x1, x2, x3) = x21 2x1x3 + 4x2x3 + 3x22 9x23

Bi 40. Dng phng php Lagrange, tm php bin i tuyn tnh a dng tonphng v dng chnh tc:

a. f(x1, x2, x3) = 2x21 + 3x22 + 4x23 2x1x2 + 4x1x3 3x2x3b. f(x1, x2, x3) = x1x2 + x2x3 + x1x3

Bi 41. Dng phng php Jacobi a cc dng ton phng sau v dng chnh tca. f(x1, x2, x3) = 3x21 + 4x1x2 2x1x3 + 2x22 2x2x3 + 6x23b. f(x1, x2, x3) = x21 + 5x

22 + 2x

23 + 4x1x2 + 2x1x3 + 4x2x3

Bi 42. Cho dng ton phng:

f(x1, x2, x3) = 5x21 + x

22 + mx

23 + 4x1x2 2x1x3 2x2x3

Vi gi tr no ca m th dng ton phng trn xc nh dng?Bi 43. Trc giao ha theo phng php Gram-Schmidta. H {(1, 2.3), (0, 2, 0), (0, 0, 3)}b. H {(1, 1, 0), (1, 0, 1), (0, 1, 1)}

Bi 44. Cho ha trc giao cc ma trn i xng sau:

a.

5 66 0

b.

3 2 02 4 20 2 5

Bi 45. Tm php bin i trc giao a dng ton phng v dng chnh tc.

a. 6x21 + 5x22 + 7x23 4x1x2 + 4x1x3b. x21 + x

22 + 5x

23 6x1x2 2x1x3 + 2x2x3

Bi 46. Nhn dng cc ng v cc mt bc hai sau:a. 5x21 4x1x2 + 8x22 36 = 0b. x21 + 2x1x2 + 2x1x3 + 3x22 2x2x3 + 3x23 4x1 + 5x2 + 5x3 + 13 = 0

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5.4. NG DNG VO VIC KHO ST NG V MT BC HAI

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Ti liu tham kho

[1] Ng Thu Lng, Nguyn Minh Hng. Bi tp ton cao cp 2 (i s tuyn tnh).NXBHQG TP. H Ch Minh, 2000

[2] Nguyn nh Tr (cb), T Vn nh, Nguyn H Qunh. Ton hc cao cp tp 1(i s v hnh hc gii tch), NXB Gio dc, 1998

[3] Trn Lu Cng (cb), Nguyn nh Huy, Hunh B Ln, Nguyn B Thi, Nguyn

Quc Ln, ng Vn Vinh. Ton cao cp 2 - i s tuyn tnh, NXB Gio Dc,2007

[4] Trn Vn Ho. i s tuyn tnh dng trong kinh t. NXB Khoa hc v k thut,1997

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