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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.