Giáo trình đại số tuyến tính ths nguyễn cao đạt, ths nguyễn quốc huy

De bieu dien ma tran don vi, ngodi ta con dung ky hieu Kronecker : I = 5ii = v Ma tran tam giac tren diidi : la ma tran vuong ma cac phan td d phia dutfi d phia tren diidng cheo chinh d

Nguydn Cao Dat - Nguyen Quoc Huy

Trifdng Dai hoc Dan lap Cifu Long muon khang dinh la trung tam dao tao nguon nhan lire co trinh do cao cho khu vifc dong b^ng song Cii’u Long de phuc vu cho cong cube cong nghiep hoa, hien dai hoa dat nude thi phai nang cao chat lifpng dao tao toan dien Mot trong nhCrng yeu to trong yeu la dpi ngu thay co, he thong cac giao trinh va trang thiet bi day hoc.Cung vdi cac nganh cac cap trong toan quoc dang day nhanh tien do xay dirng va cung cd vi the trong xu the hoi nhap, trirdng Dai hoc Dan lap Cdu Long dan tting butfc hoan thien de diTOng dau vdi nhiing thach thud do, gop phan dufa nen giao due dai hoc Viet Nam diing vCTng, ngang tam khu vifc va the gio’i.

Ke hoach xay du’ng ve mot Bo giao trinh sd dung cho cac giang vien lam tai lieu co ban de giang day, cho sinh vien nghien cdu tham khao hoc tap la nhu cau cap bach cua nha trufing

Hu’dng ting ke hoach tren, la Can bo, Giang vien co hdu cua nha trudng, chung tbi bien soan cudn giao trinh nay sau nhieu.nam giang day va chinh sti'a de hoan thien va gibi thieu vdi cac dong nghiep sinh vien trudng Dai hoc Dan lap Ctiu Long Mong rbng, qua sti dung de giang day va hoc tap, quy thay co dong nghiep va cac anh chi sinh vien dong gop de lan tai ban sau, cudn giao trinh nay tot hon, dap u’ng dope yeu cau day va hoc cua Tru’bng

ThS Nguyen Cao Dat

trong nha trifdng

Tran trong gidi thieu giao trinh nay den Quy thay co va

cac em sinh vien triTdng Dai hoc Dan lap Ciiu Long

Vinh Long, thang 09 nam 2007

Q HIEU TRUdNG

a12 a22

hay

m x nchi so hang nam d dong thd i, cot

m dong va n cot

aln a2n

am2 amn 7

Ma tran la mot khai niem co ban trong dai so tuyen tinh

Nd dac trOng cho he phoong trinh tuyen tinh, bieu dien vecto bang ma tran toa do cung nhif bieu dien phep ddi toa do trong khdng gian vecto Hon ntfa, ma tran cung dung de bieu dien anh xa tuyen tinh, dang song tuyen tinh cung nhif dang toan phoong tren cac khdng gian vecto Trong hng dung, ma tran con dung de bieu dien nhieu ddi tOpng khac nhau Do do, trong choong dau tien nay, ta nghien cdu ma tran nham phuc vu cho cac chaong sau cung nho cho nhieu dng dung sau nay

1 MA TRAN

1.1 Dinh nghia ma tran.

Mot bang so hinh chu' nhat gom

ma tran cap m x n dope ky hieu la Mmx„ Vdi AeMmxn> so hang ndm d dong thd i, cot thd j,

1 < i < m , 1 < j < n , cua A con dope ky hieu la [A]

\amldope goi la mot ma tran cap mxn, ky hieu A = (ajj)

A = | aH~| , trong do a

L 'JJmxn

thuf j cua ma tran A

Tap hop tat ca cac

mxn Vdi AeMmxn,

X -x -2x

267

= 4,

□

y +2y +3y +

i [a] 12

Giong nhu- cac khai niem khac trong toan hoc, ma tran co the bleu dien nhieu dbi tapng khac nhau trong tting bai toan tfng dung cu the Ve mat toan hoc, ta xet mot bleu dien quan trong ciia ma tran trong viec khao sat cac he phnong trinh tuyen tinh, mot he thong gom nhieu phirong trinh bac nhat theo nhieu an sb

Cu the, xet he phiicmg trinh

ky hieu A g M2x3,

[A]22 = 5 va [A]23

Chu y rang viec xb ly bang bang bieu gom nhieu dong, nhieu cot, la mot cong cu quen thuoc trong dd’i song Chang han, de ghi chu sb liipng ban mot mat hang trong mot ngay, ta dung mot sb Sb ItYcrng ban n mat hang trong mot ngay diipc bieu dien bang n sb ma ta con goi la mot vecto n - chieu, hay mot ma tran cap 1 x n Sb litqng ban n mat hang trong m ngay diipc bieu dien bang m vecto n - chieu, hay mot ma tran cap

m x n Trong xtf ly anh, mot bdc anh den trang co the bieu dien bbng mot ma tran cac bit 0, 1 Trong thong ke u’ng dung, khi khao sat mot bien phu thuoc theo k bien doc lap, ngiTbi ta thu thap n bo sb lieu, moi bo sb lieu gom k + 1 sb chi gia tri cua k bien doc lap va gia tri cua bien phu thuoc tiiong u’ng Mot

bo sb lieu nhiT vay tao thanh mot ma tran cap n x (k + 1),

111

Vai tro ky hieu ciia cac an x, y, z la khong cd y nghia quyet dinh Chang han, he phtrnng trinh nay cd the viet lai thanh

' 1-1-2

-123

z 1-1

1

1 6 hay (A|B)1

Vt du 2 Cho hai ma tran A,B e M2x3 ,

+ x3 + x3 + x3

1.2 Ma tran bang nhau.

Hai ma tran A va B dupe goi la bang nhau neu chung cd cung cap va cac sb hang tifcrng tfng cua chung bang nhau tifng ddi mot, nghia la [A] = [B]^ vdi moi i, j

r

i

Ngoai ra, ta cd the gom chung hai ma tran nay lai mot

tran cac he so md rong (hay ma tran bo

xi - -xi + -2X] +

vdi cac an la xT, x2 , x3

Ndi khac di, mot he phtfong trinh tuyen tinh duqc hoan toan xac dinh chi bang cac sb hang di kem theo cac an ma ta goi la cac he so va cac sb hang ve phai ma ta goi la cac he so

tg do Cu the, mot he phPOng trinh tuyen tinh gom m phi/ong trinh theo n an sb duttc xac dinh bang ma tran cap m x n cac

he so' va ma tran cap m x 1 cac he sb tii do Chang han, he phuPng trinh (1.1) hay (1.2) dupe xac dinh bdi

-123

ma tran, goi la

sung}.

-2 6 3

'1

0,2

1s

3 "

5-5?

la mot ma tran vuong cap 3

Cac so hang nam tren dilhng cheo chinh la : [A]n =1,

1.3 Cac ma tran dac biet.

i) Ma tran khong : la ma tran ma moi so hang cua nd deu la so 0 Ma tran khong cap m x n dupe ky hieu la 0mxn hay van tdt la 0

la ma tran khong cap 2x3

ii) Ma tran vuong : la ma tran co so dong va so cot bang nhau Ma tran vuong cap n x n dupe gpi tat la ma tran vuong cap n Tap hop tat ca cac ma tran vuong cap n dupe ky hieu la Mn Vdi ma tran vuong A e Mn, cac sb hang [A]n ,

[Al , , [Al dupe goi la nbm tren ditdng cheo (chinh) cua A

Cac sb hang [A]n] , [A]n1 9 , ,[A]1n dupe gpi la nam tren

ditdng cheo phu cua A

A =

□

: la ma tran cheo cap n , ky

(5ij)i,j=ur,0 o • 1J

□

i0

0-7 0

khikhi

i = j i* j

010

' 10

Hi) Ma tran cheo cap

moi so hang khong n^m

Vi du 5 Ma tran

n : la ma tran vuong cap n ma

tren dncmg cheo chinh deu la so 0

va khi do, ma

la mot ma tran cheo cap 3

iv) Ma tran dctn vi cap n

hieu la In, ma moi so hang n&m tren diTbng cheo chinh deu bang 1 De bieu dien ma tran don vi, ngodi ta con dung ky hieu Kronecker :

I =

5ii =

v) Ma tran tam giac tren (diidi) : la ma tran vuong ma

cac phan td d phia dutfi (d phia tren) diidng cheo chinh deu bang 0

Vi du 6 Ma tran don vi cap 2 va cap 3 lan loot la

M O'

B =

c =

lcnl

1) la mot ma tran dong

ii) Ma tran B = la mot ma tran cot

Vi du 7 Ma tran

0

Vi du 8 i) Ma tran A = (5 3

z 1 '0

bnn )

la mot ma tran tarn giac dirdi

vi) Ma tran dong (cot) : Ma tran chi co mot dong diroc

gpi la mot ma tran dong, ma tran chi cd mot cot difpc goi la

mot ma tran cot.

Cac ma tran dong va ma tran cot con dapc xem nhuf la cac vecto va dirpc lan lilpt goi la cac vecto dong va vecto cot Khi do, mot ma tran cd the xem nhu- difpc tao bdi nhieu vecto dong hay tao bdi nhieu vecto cot Vdi ma tran A e Mmvn , dong thtf i cua A gom cac phan td [A]n , [A]i2, [A]in va dope ky hieu la [A],; cot thd j gdm cac phan td [A]1., [A]2j, , [A]mj,

ky hieu [A]j.

la mot ma tran tarn giac tren va ma tran

0 0 'c22 ••• 0

iii) Ma tran

C =

o 2);-1

001

h e R , ma.

m x n

f 1

2-1 ; [C]3 = 0 ; [C]

0

' 1

3 l-l

= (1 2

12

2 CAC PHEP TOAN TREN MA TRAN.

Tren tap hop cac ma tran, ngitdi ta xay ditng nhieu phep toan nhSm phuc vu cho nhieu muc dich khac nhau Cac phep toan nay bao gom :

+ Hai phep toan hai ngoi : Phep cong hai ma tran va nhan hai ma tran

+ Phep toan ngoai : Phep nhan ma tran vdi mot sb

+ Hai phep toan mot ngoi : Phep lay chuyen vi va cac phep bien dbi so cap

2.1 Phep cong hai ma tran va nhan mpt sb vbi mot ma tran.

2.1.1 Dinh nghia Vdi hai ma tran A, B e Minxn ,

tran tong ciia A va B , ky hieu A + B, la ma tran cap

xac dinh bdi [A + B] = [A] + [B] vdi moi i, j

Ma tran tick cua A vdi h&ng so h, ky hieu hA, la

tran cap m x n xac dinh bdi [hA] = h [A]., vdi moi i, j

Vi du 9 Vcri A = , B =

va, 2A =

1-1

-1

1

4-4

-4^

4 y

'-4, 4

3'6?

1

-1,

(i) A + B = B + A (tinh giao hodn).

(ii) (A + B) + C = A + (B + C) (tinh ket hop).

ma Iran khong cap m x n ).

2.1.2 Menh de Vdi moi ma trdn A,B,C g Mmxn

ta co

'2 1 4

,3 6 5

Chu y : Hai ma tran chi co the cong vdi nhau khi chung

cimg cap va ma tran tong co cap bang cap cua hai ma tran da cho Ma tran (-l).A, ky hieu -A, dupe goi la ma tran doi cua

ma tran A Tit do, ta dinh nghia dupe phep trif ma tran

A-B = A + (-B) = A + (-l).B

M 2

,4 5'2 4 6,8 10 12

[h(A + B)]irh[A + B] =h

= [hAjy + [hB]yHai ma tran h(A + B), hA + hB co cung cap va moi so hang tiiong ilng ciia chung bang nhau nen la cac ma tran b^ng nhau

Trong cbng thiic tinh so hang [AB]jk cua ma tran tich

AB, cac so hang [A]n , [A]j2 ,[A]in tao thanh dong thd i, [A], , cua ma tran A va cac sb hang [B]lk, [B]2k ,[B]nk tao thanh cot thd k , [B]k , ciia ma tran B Khi do, sb hang [AB]

cd the coi nhif la tich vd hirbng cua hai vecto [A], va [B]k

A + B, hA, hB, tran cap m x n Ngoai ra,

Do A,BcMmxn nen cac ma tran

hA + hB va h(A + B) deu la cac ma

vdi moi i = l,m , j = l,n , ta cd

Tap hop Mmxn cung vdi phep cdng hai ma tran va phep nhan ma tran vcri mot sb thba 8 tinh chat neu trong menh de 2.1.2 nen sau nay ta ndi rang nd cd cau true cua mot khdng gian vecto (xem chifong 3)

2.2 Phep nhan hai ma tran

2.2.1 Dinh nghia Cho hai ma tran A e Mmxn , B g Mnxp Ta

tran tich cua hai ma tran A,B la ma tran cap

m x p , ky hieu AB , xac dinh bbi

9 >

2^

1 , B =3,

Tcot k

cot k

[ABJn

[AB121 [ab]22 [Mi

Vi du 10 Cho

( 1

-1 2

n

i

b

xiGoi X = x2

dilpc viet lai thanh

r i

-i

1-2

X2 2x2 3x2

tran Ching han, trd lai vdi he phiTcmg trinh tuyen tinh

+ +

+

x3 x3 x3

vdri ma tran he so va ma tran cac he so tP do,

6

J;

mxn > Mnxp

mxn A,BGMIlxp, C(A + B) = CA + CB

=z DAmk [c]

k=l\ v j=l j

=[(AB)C1

va do do ta diroc A(BC) = (AB)C

(ii) Cac ma tran (A + B) C , AC + BC co cung cap m x p va

h(AB) = (hA)B = A(hB)

Chi'ing minh (i) Chu y rang vo'i A e Mmxn, B e Mnxp va CeMpxq thi ABGMmxp, BC g Mnxq nen ta co A(BC)GMmxq

va (AB)Cg Mmxq, nghia la cac ma tran A(BC) va (AB)C co cung cap m x q Hon ntfa, vdi i = 1, m , 1 = 1, q , ta co

( p >

tA (BC)],! = Z [ Ak [ BC 11 = Z [ H Z IB k Mki

\k = l >

P n ZZKM.JCL k=lj=l

j=l

= Zc«[A]ik j=l

= lCAlik '' L^-^Jik

va do do, ta ditoc C(A + B) = CA + CB

(iii) Ta co h(AB), (hA)B va A(hB) la cac ma tran cap

m x p va vdi moi i = 1, m , k = 1, p ,

i>E[AyB]jkj=l

Vdi CeMmxn, A,BeM

cac ma tran cung cap m

h[ABk [h(AB)] it

+ [Bk)

n

ZcdlAkj=l

+ Z cy [ B kj=l

[CA + CB]ik

Iran noi chung khong co tfnh giao

10

01

00

■'1

0

nA

[B] lk

n

j=l

Z[H[Bk = j=l

la

> In

Trong trifdng hop ca hai

va thoa dang thtfc AB = BA, ta noi hai ma tran A va B giao hoan vdi nhau Chang han, ma tran don vi I,

moi ma tran vudng A cap n va InA = AL

Tong quat, neu B

ImB = BIn = B , trong do Im

cap m va n

That vay, ta cd ImB , BI

i= l,m , k = l,n ,

ma tran tich AB va BA ton tai

Chu y i) De cd the nhan ma tran A vdi ma tran B, ta can dieu kien la sb cot cua ma tran A phai bdng sb dong cua

ma tran B va khi do :

Sb dong ciia ma tran tich AB bang sb dong cua ma tran

A va sb cot cua ma tran tich AB bang sb cot cua ma tran B

Do do, vdi hai ma tran A, B cho trade, khong nhat thiet tich AB ton tai va khi tich AB ton tai, khong chac tich BA ton tai

ii) Tich cua hai ma

hoan, nghia la tong quat ta cd AB BA

00

ma tran cap m x n, ta cd lan loot la cac ma tran don vi

5

-1 2 3

6,

0 1 0

0 1 0

4 3 -1

0

1.

'-1

-11

Khi do, neu nhan ben trai hai ve cua ding thilc (1.4) cho

^0

-1

1222

4

3

-1

31

2 -1

2212

"-1-11

ii) Phep bien doi 2 : Nhan dong i vdi mot so

hieu A- (i):~g(i)

giai ditoc he phitong trinh tuyen tinh (1.3)

2.3 Cac phep bien doi sof cap tren dong.

2.3.1 Dinh nghia Xet ma tran A e Mmxn veri m vecto dong

[A]] , [A]2, , [A]m Cac phep bien doi tren dong nham muc

bien doi so cap tren dong nhit sau :

i) Phep bien doi 1 : Hoan vi hai dong i va j, ky hieu A -■—> A', nh&m doi cho hai dong i, j trong ma tran A,

Vi du 14.

^301

<5

1-1 0

210

1-10

1-1101-1^

0k0

0.0

4

1 ?

0 '1-2>

va [A'] =[A] + a

(3)p3)

Vi du 16.

'-10

k 0Chu y rang ma tran cuoi cung co cac phan th nam phia diTo'i difo'ng cheo chinh deu la so” 0 nen la mot ma tran tarn giac tren

Doi vdi ma tran tarn giac tren ma moi phan th nam tren diTdng cheo deu khac 0, thi cung bang cac phep bien doi so cap tren dong, ta cd the bien nd thanh ma tran don vi

Hi) Phcp bien doi 3 : Thay dong i bbi dong i cdng vdi

a lan dong j, ky hieu A —I1) t!1) + «(j)—> n}1^m ^ong thuf i cua A bhng dong do cdng vdi a nhan cho dong thd j cua

A, nghia la moi dong khac dong i cua A va A' bang nhau, dong thd i cua A' bang dong thu‘ i cua A cong vdi a lan dong thu' j ciia A,

nghia la moi dong khac dong i cua A va A' bang nhau, dong thu" i cua A' bang dong thu" i cua A nhan vdi a ,

[A'lk = [Ak khi k * i va [A'] = a ■ [A],

□

1-1

0

010

Vi du 17.

r-i

o

(T0

' 1

00

0 "

1

-L

"i0,0

(1) :=(l)+(3) (2) :=(2)+(3) = I3-

(2) :=-l.(2) (3) :=-l.(3)

0-1

1 )

I).0

Tong qudt : Ta co the dung cac phep bien doi so cap tren

dong de chuyen mot ma tran vuong ve mot ma tran tarn giac tren va khi cac phan td tren diTdng cheo chinh cua ma tran tarn giac nay khac khong, ta cd the tiep tuc bien doi ve ma tran don vi

2.3.2 Giai thuat bien ma tran vuong thanh ma tran tarn giac tren.

De chuyen mot ma tran vuong ve mot ma tran tarn giac tren, ta duyet cac cot, tif cot dau den cot cudi :

Tren moi cot, chon mot phan til ma ta goi do la phdn td

true xoay Sau do, diing cac phep bien doi so cap tren dong de

bien cac phan td nam phia diidi phan tuf true xoay ve sb 0 Doi vdi giai thuat chuyen ma tran vuong ve ma tran tarn giac tren, phan th true xoay tren titng cot du'pc chon nam tren duftng cheo Khi do*, ta cd cac kha nang sau :

K/td ndng 1 Phan tu true xoay bang 0 va cac phan tuf o

phia dutfi phan tu- true xoay cung bhng 0 : Chuyen qua cot ke

Khd ndng 2 Phan tuf true xoay bang 0 va cd mot phan th

d phia diTdi nd khac 0 : Hoan vi hai dong thich hop de du'a phan tid khac 0 nay ve vi tri phan tir true xoay Chuyen qua kha nang 3

A =

ri

□5

0 -1

1

0

0

-1 0 3

0 -1 0 3 0 -1 1 0

0 1

0 0

2 1

<3

'1 0

2

1 2

2' 5 5 9,

'1 0

2^

i 22i;

2 A12

0

2

10

2'

133>

4 3 7

0

1°

12

0 0,0 0

dutfi phan th true xoay bang dong do cong vdi mot hhng sb thfch hop nhan vdi dong chda phan tuf true xoay de bien cac phan th phia dubi true xoay thanh 0 Chuyen qua cot ke

Vi du 18.

1

o E]

0 0 ,0 0

1 0

0 2-1

0 0 [i]

.0 0Ngoai ra, neu ma tran tarn giac tren nhan duPc co cac phan th tren dubfng cheo khac 0 thi cung vdi cac phan th true xoay nam tren difdng cheo, duyet tu1 cot dau tdi cot cudi va tren moi cot :

Nhan dong chufa phan th true xoay vdi mot hang so thich hop de bien phan th true xoay thanh 1,

Thay cac dong phia tren phan th true xoay bhng dong do cong vdi mot hdng sb thich hop nhan vdi dong chda phan td true xoay de bien cac phan th phia tren phan th true xoay thanh 0 Chdng han, vdi ma tran nhan difOc b vi du 18, ta bien dbi tiep tuc

12

b

0100

0

0

1 0

01

0

X2

2x23x2

va thiTc hien cac phep bien dbi so cap tren dong cho ma tran

sao cho ma tran cac he so A trb

Chu y rang, neu ma tran tarn giac tren cd mot phan td tren du’dng cheo chinh bdng 0 thi ta cd the dung cac phep bien doi so cap tren dong de chuyen ve mot ma tran cd mot dong gdm toan sb 0

Nhan xet rang khi ta thiTc hien cac phep bien doi tren dong cho ma tran cac he sb mb rong cua mot he phoong trinh tuyen tfnh, ta da thay dbi thd to cac phifong trinh trong he, nhan hai ve cua mot phitong trinh cho mot sb khac 0 hay thay mot phifong trinh bang phitong trinh do cong cho mot hang sb nhan cho mot phuPng trinh khac Do cac su1 thay dbi nho vay khdng lam thay dbi tap nghiem cua he phoong trinh tuyen tfnh

nen sau khi thitc hien cac phep bien doi so cap tren dong cho

ma tran cac he sb mb rong, ta nhan dope mot ma tran cac he

sb mb rong cua mot he phuPng trinh tuyen tinh mdi, tuPng duPng vbi he phuPng trinh tuyen tinh ban dau, nghia la tap nghiem cua chung bang nhau Chang han, trb lai vbi he phuPng trinh

-110

001

010

321

123

-111

123

-123

121

23,

'1

00

2"

8

3 /

icP8

'i 0 0

2^

6 7,

n

00

2^

811

r i

-i-2

X3 2x3 x3

x2

X2

(1) :=(l)-3(3) (2) :=(2)-2(3)

He nay de dang giai dittfc bang each giai tiTng phitong trinh th dutfi len tren, ta dutfc x3=3, x2 = 8 - 2x3 = 2 va X! = 2 + x2 - x3 = 1 Phifcfng phap giai he phitong trinh tuyen tinh nay diftfc goi la phtTc/ng phap Gauss

Hon niia, neu ta bien dbi tiep A' = (A'|B') de chuyen

ve ma tran dnn vi,

(2) :=(2H(i) (3) :=(3)+2(l)

ta nhan difpc he phifcmg trinh tifpng difcfng

283

'1 0

0 1

0 0

\

Vi du 19 Cac ma tran sau la

0 0 0

0 0

0

00

040

0 k0

6^

-13

7^

63

0 [8

0 0 ;

va ta cung lai nhan diioc nghiem ciia he phiiong trinh ban dau Phtfong phap giai he phifong trinh tuyen tinh nay dirge goi la phirong phap Gauss-Jordan He phtfong trinh tuyen tinh se dirge khao sat mot each co he thong trong chirong sau

Tong quat, ta cd the dung cac phep bien ddi so cap tren dong de chuyen mot ma tran bat ky ve mot ma tran cd dang gan gid'ng nhir ma tran tarn giac tren ma ta goi la ma tran bac thang Trirdc het, ta can dinh nghia sau

2.3.3 Dinh nghia Ma tran bac thang theo dong la ma tran

ma ilng vdi hai dong bat ky, sb hang khac 0 dau tien cua dong dirdi ludn ludn nam ben phai sb hang khac 0 dau tien cua dong tren

Nhan xet : Trong ma

khdng (dong chiia toan sb hang 0), neu cd, phai nam diidi cac dong khac khdng (dong cd it nhat mot sb hang khac 0) Khi do,

cac sb hang’bang 0 dau tien tren moi dong tao thanh hinh bac

thang, moi bac thang chiTa it nhat mot cot

Chang han, vdi cac ma tran trong vi du 19, cac sb hang bbng 0 dau tien tren moi dong cd dang

ma tran bac thang theo

0

0

0 0

-2-131

000

42-6-2

210

5525

0 0

0 0

0

<0

4-10-10-50

5

1 -9 -10,

5 "l

-14-14-70,nam cf

do moi bac thang chda dung mot cot

Vdi mot ma tran A cap m x n bat ky, ta luon luon co the dung cac phep bien doi so cap de bien ma tran A thanh ma tran bac thang theo dong bang giai thuat sau :

2.3.4 Giai thuat chuyen ma tran bat ky ve ma tran bac thang theo dong.

De chuyen ma tran bat ky ve ma tran bac thang theo dong, ngudi ta thay doi each chon phan td true xoay trong giai thuat chuyen ma tran vuong ve ma tran tam giac tren Thay vi

vi tri phan td true xoay luon luon nam tren dudng cheo, ta chon

- Phan td true xoay cua cot 1 ndm b dong 1

- Neu sau khi bien doi xong mot cot ma phan td true xoay luc do khac 0 thi phan td true xoay cua cot ke nam d dong ke Ngiipc lai, neu phan td true xoay bang 0 (va moi phan td ndm dodi nd cung bang 0) thi phan td true xoay cua cot ke n& cung dong

Vz' du 20

31112

'1 0 0

^0

-2

5

0 0

0 >

Z[Al,[B]jk = j=l

ChiCng minh Ta lan lupt khao sat tiTng loai bien ddi so cap

+ A' : Khi do, vdi i = l,m,

2.3.5 Dinh ly Clio A € Mmxn vet Bg Mnxp Nou A lo mo iron nhqn duqc tit A quo cac phep bien ddi so cap tren dong

C thi ma iron A'B cung nhqn ditqc tit ma trqn AB qua cac phep bien ddi so cap tren dong C , nghia Id neu A—-— >A'

thi AB —>A'B

>A' : Khi do, vdi i = l,m,

[ABL •

va

n[AB}

Do do, AB

2.4 Ma tran chuyen vi

i = l,n , j = l,m

Chuyen vi cua A , ky hieu

[A] , vdi moi

j=i

j=i n

j=l

j=l n+ “Z[ A k[Bk

j=l+ “[ AB Lk

2.4.1 Dinh nghia Cho A e Mmxn.

At , la ma tran cap n x m xac dinh bdi KI,

Nhan xet Ma tran chuyen vi cua A co the nhan ditoc tut

A bang each bien dong cua A thanh cot cua AT (hay bien cot cua A thanh dong cua AT)

2.4.2 Tinh chat Vdi moi ma tran A g Mmxn,

Cluing minh Ta chi chtfng minh (iii) Cac phan con lai ditoc coi

nhif bai tap

4

J

3^

6 9?

det AineM2

l + n aln

3 DINH THLfC CUA MA TRAN VUONG

Xet ma tran vuong cap n

(-1) a

n

detA^^-1)

j=i / - \l + 2

+ (-!) a

g M3 th!

3.1 Dinh nghia Cho A g

hay |A|, la sb thifc dirge dinh nghia bang quy nap theo n sau :

Vdi n = 1, nghia la A = (an), ta dat det A = an

, ta dinh nghia

'1 4

J

5 6A|

8 9.1

Chang han, khi n = 2 , nghia la A = , ta cd

a, a2

>i b2

■i.

b2 c2

bi q

b 3 c3

= al

a12 a22 >

+ a3 (b1c2 - b2c1)

- ajb3c2 - a2b1c3 - a3b2c1.

Trong thyc hanh ta cd the tinh dinh thde cap 3 bang each dung quy tac Sarrus nhif sau :

Viet theo thd ty hai cot 1 va 2 sau cot thd 3

Ba sb hang mang dau cong trong dinh thtfc la tich cac phan td nbm tren ba diTdng song song vdi diTdng cheo chinh

Ba so hang mang dau trh trong dinh thde la tich cac phan

td nam tren ba duftng song song vdi diidng cheo phu

di, det A chinh la tich cac sb hang tren duting cheo chinh trd

di tich cac sb hang tren diThng cheo phu

Vdi n = 3, ta cd edng thde tinh dinh thde cap 3 :

= aib2c3 + a2b3c1 +a3b1c2

I Khi do nxn

Cong thdc (3.1) cua dinh nghia 3.1 dupe goi la cong thtfc tinh det A bang each khai trien theo dong mot Thifc chat, dinh thde cua mot ma tran (vuong) khong dbi khi ta khai trien theo mot dong hay mot cot bat ky :

3.2 Dinh ly Clio ma tran A = (a: J

\ 1 / ;

(vdi 1 < i0, j0 < n )

Cong thifc (3.2) gpi la cong thifc khai trien theo dong i0

va edng thifc (3.3) la cong thifc khai trien theo cot j0

detA = ^(-1)lo + Jaloj detAioj

detA = ^M)^ aljo

i=l

3-2 0

-201

3 0

-21

-Tong quat hon, vdi ma tran A g Mn va vdi so nguyen k,

1 < k < n, ta chon trong A cac dong i1 < i2 < ••• < • Khi do, vdi moi bo k sb nguyen 1 < < 2 < < jk < n , ma tran vuong cap k nhan diipc tif A bang each giu' lai cac phan th nam tren cac dong , i2, , ik va tren cac cot j1, j2, , jk dupe ky hieu la A11,12 J2. .jk ma Vu6ng cap n -k nhan dupe

th A bbng each bo di cac dong i1} i2, , ik va cac cot jj , j2, , jk dupe ky hieu la Aipi2i jk • Chang han, vdi ma tran A trong vi du tren, ta co

'0 2 ,0 3

Khai trien theo dong 4, ta co

1 03

Khai trien theo dong 2 dinh thtfc d ve phai, ta dupe

1 -2

3 1

Ap dung dinh ly neu tren, ta co the tinh dupe dinh thde bang each khai trien theo mot dong hay mot cot bat ky Trong thuc te, ta lUa chon cac dong hay cot de khai trien sao cho sb cac phep tinh can thuc hien cang it cang tot ChAng han, bbng each khai trien theo dong hay cot chUa nhieu so' 0 nhat

0 3

0 2-2

3 2,0 3

ik ^n,

det A =

(2 + 4) + (l + 3)

+(2 + 4) +(2+3)

Vi du 26 Vcri ma tran trong vi du 25, ta khai trien theo

hai dong 2 va 4 (hai dong nhieu so 0 nhat),

3.3 Dinh ly Laplace V ol A e Mn, vd 1 < i1

ta co

i) Vdi ba ma tran A,B,C e Mn sao c/io

n.J = 2:(-ir(a[A]io])|Aioj j=l

dong thil i0 Dong thit i0 cua C bdng tong cdc dong thil i0 cua

A vd B tkl | C | = | A| + |B|.

[BL =“ [ A J 1 „i ™ [Bly =[A V vi;£i 0

n|A|, vdi moi

HoJ L Jioj L

| A | Do do, neu ma trail B nhan dilac til A bdng each nhan mot dong cua A cho hdng so a, cdc dong hhac gid nguyen, thi | B | = cc | A | Dae diet,

AeMn

Chilng minh i) Khai trien dinh thu’e cua C theo dong i0 vdi

nhan xet r^ng Ci()j = Aioj = Bioj va [C].= [A]ioj + [B].,., ta cd

[A]io j I Aio j

la ma tran

cota

va menh de 3.4 dilpc chtfng minh

TiT menh de 3.4, ta dupe sir lien he gida dinh thhc va cac phep bien ddi so cap tren dong sau :

3.5 Dinh ly Vdi cac ma Iran vuong A, B cap n, ta co

i) Neu A —LilizSii*—> B thi | B |

ii) Ma tran cd 2 dong ti le thi co dinh thitc bdng 0.

>B thi | B |

(i):=(i) + a(i’)

Cong thufe | aA | = a” | A | dupe chdng minh bang quy nap

theo n Hien nhien cong thde nay dung khi n = 1 Gia su

| ccA | = an 11 A | dung vdi moi ma tran vuong cap n - 1, vdi

n > 1 Xet ma tran A vuong cap n va khai trien dinh thde theo dong 1, ta co

= an

kAi=£(-ir

j=l n

A =

vi) Dinh thitc cila ma tran tam giac bhng tick cdc so hang ndm tren dilang cheo chinh.

vii) | AB | = | A | x | B |

Chilng minh Ket qua i) chinh la phan ii) cua menh de 3.4 De

chdng minh ii), ta dung quy nap tren cap n cua ma tran A Khi

De chting minh iii), xet cac ma tran :

A' la ma tran nhan dutfc tb A bang each gid nguyen moi hang, trit hang thu" i dupe cong them cho hang thuf i' cua nd.B' la ma tran nhan dupe tu1 B b^ng each giu’ nguyen moi hang, trif hang thu" i dupe cong them cho hang thuf i' cua nd.Khi do, do dieu i), menh de 3.4 va dieu i) neu tren, ta cd

|A| = |A'| va |B| = | B'[

’oi I •

a

^ctathi | A | = a x ab - b x aa = 0 va do do ii) dung vdi n = 2

Gia sd ii) dung vdi n > 2 Xet ma tran vuong A cap n + 1

cd hai dong ty le nhau Goi i0 la dong khac hai dong ty le vifa neu va dung cong thufe khai trien dinh thde theo dong i0,

n + l

I ai ^ h ) 10 ^

j=i

v) Cung dung quy nap tren n Hien nhien v) dung khi

n = 2 Gia sd v) dung vdi n > 2 Xet ma tran vuong A cap

n + 1 Dung cong thtfc khai trien dinh thtfc theo dong 1 ddi vdi

A va theo cot 1 vdi AT, ta dupe

la cac ma tran vuong cap n

Chu y rang hai ma tran A' va B' co moi dong bdng nhau trif dong thu1 i', trong do tong cac dong thu' i' cua A' va B' chinh la dong thu i cua A' (va cung la cua B') Do do, vdi ma tran C nhan dupe bang each giu' nguyen moi dong khac dong i' cua A' (va cung la cua B') va dong i' cua C bang tong cac dong i' cua A' va B', ta co C la ma tran co hai dong bang nhau nen |C| = 0 Mat khac, do dieu i), menh de 3.4, ta cd

| C | = | A' | + | B' | va do do

iv) Xet ma tran C nhan dupe th A bang each giu" nguyen moi dong khac dong i Rieng dong i diipc thay bang a lan dong i' cua A Ma tran nay cd hai dong ty le nen do i), | C | = 0 Mat khac, ma tran B chinh la ma tran nhan dupe bang each gid nguyen moi dong khac dong i cua A (va cung la cua C) va dong thdf i nhan dupe bang each lay tong hai dong thu1 i cua A va 0

Do do |B| = |A| + |C| = |A|

Aj1 | do gia thuyet