Ma Tran Dinh Thuc Nghich Dao Va Hang Matran 6617

1C1. MA TRN - NH THC

1 Ma trn

2 nh thc

3 Ma trn nghc o

4 Hng ca ma trn

21. MA TRN1.1. CC NH NGHA1.1.1. nh ngha ma trn: Mt bng s ch nht c m hng v n ct gi l ma trn cp m x n

mn2m1m

n22221

n11211

a...aa............

a...aaa...aa

A

aij l phn t ca ma trn A hng i ct j. A = [aij]m x n = (aij)m x n

31. MA TRN1.1.2. Ma trn vung: Ma trn vung: Khi m = n , gi l ma trn vungcp n

nn2m1n

n22221

n11211

a...aa............

a...aaa...aa

A

a11,a22,ann c gi l cc phn t cho. ng thng xuyn qua cc phn t cho gi lng cho chnh.

41. MA TRN Ma trn tam gic trn: aij = 0 nu i > j

nn

n222

n11211

a...00............

a...a0a...aa

A

nn

n222

n11211

a......

a...aa...aa

A

Ma trn tam gic di: aij = 0 nu i < j

nn2m1n

2221

11

a...aa............0...aa0...0a

A

nn2m1n

2221

11

a...aa.........

aaa

A

51. MA TRN Ma trn cho: aij = 0 nu i j

nn

22

11

a...00............0...a00...0a

A

nn

22

11

a...

aa

A

Ma trn n v: I = [aij]n x n vi aii=1; aij = 0, ij

1...00............0...100...01

I

61. MA TRN1.1.3. Vect hng(ct): Ma trn ch c mt hng(ct)1.1.4. Ma trn khng:

0...00............0...000...00

1.1.4. Ma trn bng nhau: A=B

1) A=[aij]m x n; B=[bij]m x n2) aij = bij vi mi i,j

71. MA TRN1.1.5. Ma trn chuyn v: A=[aij]m x n => AT=[aji]n x m

312517181128192015132416181493027151210

A

81. MA TRN1.2. CC PHP TON TRN MA TRN:1.2.1. Php cng hai ma trn1. nh ngha: A=[aij]mxn; B=[bij]mxn => A+B =[aij+bij]mxn

31412231

23154132

2. Tnh cht:A + B = B + A (A + B) + C = A + (B + C) + A = A Nu gi -A = [-aij]m x n th ta c -A + A =

91. MA TRN1.2.2. Php nhn mt s vi ma trn:1. nh ngha: cho A=[aij]m x n, kR => kA=[kaij]m x n

401235021321

A

2. Tnh cht: cho k, h R: k(A + B) = kA + kB (k + h)A = kA + hA

Tnh 2A?

10

1. MA TRN1.2.3. Php nhn hai ma trn:1. nh ngha : A=[aik]m x p; B=[bkj]p x n => C=[cij]m x n:

p

1kkjikpjip2ji21ji1ij baba...babac

120301121321

023112

V d: Tnh tch 2 ma trn sau:

11

1. MA TRN2. Mt s tnh cht:

(A.B).C = A.(B.C) A(B+C) = AB + AC (B+C)A = BA + CA k(BC) = (kB)C = B(kC) Php nhn ni chung khng c tnh giao hon A=[aij]n x n => I.A = A.I = A

12

1. MA TRN1.3. V DV d 1: Tm lng hng bn trong hai thng.

Thng 1 A B C DCH1 10 2 40 15CH2 4 1 35 20

Thng 2 A B C DCH1 12 4 20 10CH2 10 3 15 15

13

1. MA TRNV d 2: Hy tnh nhu cu vt t cho tng phn xng theo k hoch sn xut cho bi 2 bng s liu sau:

Phn xng

Sn phmA B C

PX1 10 0 5PX2 0 8 4PX3 0 2 10

Sn phm

Vt liuVL1 VL2 VL3 VL4 VL5

A 1 2 0 2 0B 0 1 1 2 0C 0 0 2 1 3

14

2. NH THC2.1. CC NH NGHA:

A l ma trn vung cp 2:

A l ma trn vung cp 1:A= [a11] th det(A) = |A| = a11

2221

1211aaaa

A

th det(A) = a11a22 a12a21

15

2. NH THC

nn2m1n

n22221

n11211

a...aa............

a...aaa...aa

A

Aij l ma trn con cp n-1 nhn c t A bng cch xo hng i ct j. Cij = (-1)i+jdet(Aij) l phn b i s ca aij

A l ma trn vung cp n:

16

2. NH THC

V d: S dng nh ngha hy tnh nh thc:

987654321

A

nh thc cp n ca A l:det(A) = a11C11 + a12C12 + + a1nC1n

n

1jj1j1

j1n

1jj1j1 )Adet(a)1(Ca)Adet(

17

2. NH THC2.2. TNH CHT CA NH THC:

Tnh cht 1:AT=AH qu: Mt tnh cht ng khi pht biu v hng ca mt nh thc th n vn cn ng khi trong pht biu ta thay hng bng ct.

Tnh cht 2: i ch hai hng (hay hai ct) ca mt nh thc ta c mt nh thc mi bng nh thc c i du.

18

2. NH THC Tnh cht 3: Mt nh thc c hai hng (hay hai ct) nh nhau th bng khng. Tnh cht 4: Mt nh thc c mt hng (hay mt ct) ton l s khng th bng khng. Tnh cht 5: Khi nhn cc phn t ca mt hng (hay mt ct) vi cng mt s k th c mt nh thc mi bng nh thc c nhn vi k. H qu: Khi cc phn t ca mt hng (hay mt ct) c mt tha s chung, ta c th a tha s chung ra ngoi nh thc.

19

2. NH THC

Tnh cht 7: Dng th i no c aij = aij + aijth det(A) = det(A) + det(A)

nn2n1n

,in

,2i

,1i

n22221

n11211

,

a...aa...a

...

...

...

...

...a

...

...a

...a...aaa...aa

A

nn2n1n

"in

"2i

"1i

n22221

n11211

"

a...aa...a

...

...

...

...

...a

...

...a

...a...aaa...aa

A

Tnh cht 6: Mt nh thc c hai hng (hay hai ct) t l th bng khng.

20

2. NH THC Tnh cht 8: Nu nh thc c mt hng l t hp tuyn tnh ca cc hng khc th nh thc y bng khng. Tnh cht 9: Khi ta cng bi k ca mt hng vo mt hng khc th c mt nh thc mi bng nh thc c

516754312

)Adet(

21

2. NH THC Tnh cht 10: Cc nh thc ca ma trn tam gi bng tch cc phn t cho.

nn2211

nn

n222

n11211

a...aa

a...00............

a...a0a...aa

nn2211

nn2m1n

2221

11

a...aa

a...aa............0...aa0...0a

22

2. NH THC2.3. CC PHNG PHP TNH NH THC:

Phng php 1: Dng nh ngha.

Phng php 2: S dng cc bin i s cp.

Bin i s cp Tc dng L do

Nhn mt hng vi mt s k0 nh thc nhn vi k TC 5i ch hai hng nh thc i du TC 2Cng k ln hng r vo hng s nh thc khng i TC 9

23

2. NH THC

8432189043218765

)Adet(

V d: Tnh nh thc bng hai phng php:

24

3 MA TRN NGHCH O 3.1. Ma trn khng suy bin: Ta gi ma trn vung A cp n l mt ma trn khng suy bin nu det(A) 0.

3.2. Ma trn nghch o: Cho ma trn vung A cp n, nu tn ti ma trn vung B cp n tho mn: AB = BA = I th B c gi l ma trn nghch o ca A. Nu A c ma trn nghch o th A gi l ma trn kh nghch. K hiu: B = A-1, ngha l ta c AA-1 = A-1A = I

3.3. S duy nht ca ma trn nghch o:nh l: Nu A kh nghch th A-1 l duy nht.

25

3 MA TRN NGHCH O3.4. S tn ti v biu thc ma trn nghch o:nh l: Nu det(A)0 th ma trn A c nghch o A-1c tnh bi cng thc sau:

nnn2n1

2n2212

1n2111

T1

C...CC............

C...CCC...CC

A1C

A1A

Trong Cij l phn b i s ca phn t aij. CT: ma trn chuyn v ca ma trn phn b i s

26

3 MA TRN NGHCH O3.5. Phng php tm ma trn nghch o:3.5.1. Phng php dng nh thc:

nnn2n1

2n2212

1n2111

T1

C...CC............

C...CCC...CC

A1C

A1A

120112213

A

V d: tm ma trn nghch o ca ma trn:

27

3 MA TRN NGHCH O3.5.1. Phng php dng php bin i s cp ca Gauss - Jordan:1. Nhn mt dng no ca ma trn vi mt s thc khc khng2. Cng vo mt dng ca ma trn mt dng khc nhn vi mt s thc

3. i ch hai dng ca ma trn cho nhau

tm ma trn nghch o dng cc php bin i s cp sau cho: [AI] = [IA-1]

28

3 MA TRN NGHCH OV d: tm ma trn nghch o:

342221211

A

29

4 HNG CA MA TRN4.1. Ma trn con: ma trn A cp m x n, gi p l mt s nguyn dng, p

30

4 HNG CA MA TRN4.2. Hng ca ma trn: nh ngha: Hng ca ma trn A l cp cao nht ca nh thc con khc khng ca A.

Nu r l hng ca ma trn nu: Trong A tn ti mt nh con cp r khc 0. Mi nh thc con cp ln hn r trong ma trn A u bng 0. K hiu: rankA = r

V d: Tm hng A

212141122431

A

31

4 HNG CA MA TRN4.3. Ma trn bc thang:4.3.1. nh ngha:

Mt dng ca ma trn c gi l dng 0 nu n ch gm nhng phn t 0.

Ngc li, nu mt dng ca ma trn c t nht mt phn t khc 0 th c gi l dng khc 0.

Phn t khc 0 u tin ca mt dng c gi l phn t chnh ca dng .

32

4 HNG CA MA TRNMa trn A c gi l ma trn bc thang khi tho cc iu kin sau:

A khng c dng 0 hoc dng 0 lun di cc dng khc 0. Nu A c t nht 2 dng khc 0 th i vi 2 dng khc 0 tu ca A, phn t chnh ca dng di lun nm bn phi ct cha phn t chnh ca dng trn.

0000100002104321

A

100042

B

000012432

C

310000021

D

33

4 HNG CA MA TRN4.3.2. nh l v hng ca ma trn:

Cho A, B l hai ma trn cng cp. Nu B l ma trn nhn c t A sau mt s hu hn cc php bin i s cp th rankA = rankB.

H qu: Hng ca ma trn A bng s dng khc khng ca ma trn dng bc thang thu c t A sau mt s hu hn cc php bin i s cp.

34

4 HNG CA MA TRN4.3.3. Thut ton a mt ma trn v ma trn dng bc thang

Bin i sao cho phn t chnh dng mt v v tr ct u tin so vi p phn t chnh cc dng khc.

Bin i sao cho cc phn t nm pha di phn t chnh ca dng u tin u bng 0.

Lm tng t i vi hng 3, 4.

35

4 HNG CA MA TRN4.4. Cc phng php tm hng ma trn.4.4.1. Phng php 1: S dng nh ngha.Bc 1: Tnh cc nh thc con cp p cao nht c trong A:- Nu gp mt nh thc khc 0 th kt lun ngay rankA bng cp ca nh thc .- Nu tt c cc nh thc u bng 0 th tip tc bc 2.Bc 2: Tnh cc nh thc con cp p-1 c trong A:- Nu gp mt nh thc khc 0 th ta kt lun ngay rankA bng cp ca nh thc .- Nu tt c cc nh thc u bng 0 th tip tc bc 3.Bc 3, 4, cho n khi tm c rankA

36

4 HNG CA MA TRNV d: Tm hng ca ma trn

1221671113152

A

37

4 HNG CA MA TRN4.4.2. Phng php 2: S dng cc php bin i s cp. tm hng ca ma trn A ta bin i ma trn A v dng bc thang, s dng khc dng 0 l hng ca ma trn A.

212141122431

A

V d: Tm hng ca ma trn.