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

BQ GIAO DUG VA DAO TAO
TRUCJNG DAI HOC DAN LAP CtfU LONG
Nguydn Cao Dat - Nguyen Quoc Huy

Gido Trinh

DAI SO TUYEN TINH

TR$NGDAIHOCDA • i-AFCuULONG

THUVIEN
SodAogky:.....

Vinh Long - 2007

>1



Ldl DE TUA
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.


Hy vong rang, mot ngay khong xa TrUdng Dai hoc Dan
lap Chu Long se co day du mot Bo giao trinh hoan chinh dung
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
ThS. Nguyen Cao Dat


5

Chiiomg 1

MA TRAN - DINHTHLfC
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 m dong va n cot
all

a12

aln

a21

a22

a2n

\aml

am2

amn 7

A=

dope goi la mot ma tran cap mxn, ky hieu A = (ajj)
A = | aH~|

L 'JJmxn

mxn

hay

, trong do a ij chi so hang nam d dong thd i, cot

thuf j cua ma tran A .

Tap hop tat ca cac ma tran cap m x n dope ky hieu la
Mmx
„.. Vdi AeMmxn>
mxn
mxn, so hang ndm d dong thd i, cot thd j,
1 < i < m , 1 < j < n , cua A con dope ky hieu la [A]...


6

Vi du 1. Vol A =

1
4

2
5

ky hieu A g M2x3, ™ [Alll:

3
, ta co A la ma tran cap 2x3,
6

i. [a]12

2, [Ha

3. [A]21 = 4,

□

[A]22 = 5 va [A]23 = 6

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), ...

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
X

-x +
-2x +

y + z
2y + z
3y + z

trong do x, y, z la cac an so can tim.

2
6
7

(1.1)


7

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
xi

-

x2

+

x3

-xi

+

+

x3

-2X]

+

2x2
3x2

+

x3

2
6
7

(1.2)

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

-1
A = -1 2
^-2 3
z 1

r
i

va B = 6

II

Ngoai ra, ta cd the gom chung hai ma tran nay lai mot
ma tran, goi la ma tran cac he so md rong (hay ma tran bo
sung}.

-1
A = -i 2
^-2 3
' i

1 2^
1 6 hay (A|B)
1 7)

' 1 -1
-1 2
-2 3

1
1

1

2"
6
7/

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.

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


8
P Q
1 0

A=

4^
1
, B=
2?
s

3

4'

0

2,

Ta co A = B neu va chi neu p = 1, q = 3 va s = 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 .

Vi du 3. 02x3 -

'0

0

<° 0

0
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]In-1,2
n1 9 ,...,[A]
1n dupe gpi la nam tren
’ -■[A]ln

ditdng cheo phu cua A .
Vi du 4. Ma tran

'1
A= 0
,2

6

3"
5

3

-5?

-2

la mot ma tran vuong cap 3.

Cac so hang nam tren dilhng cheo chinh la : [A]n =1,
[^122

MM

w ’ lXiJ33 = -5.

Cac so hang n^m tren diTdng cheo phu la : [A],l31=2’

[A]22 = 6 ’ [A]13 = 3 ’

□


9
Hi) Ma tran cheo cap n : la ma tran vuong cap n ma
moi so hang khong n^m tren dncmg cheo chinh deu la so 0.
Vi du. 5. Ma tran

f5
A= 0

0
-7

<0

0

(p
0

□

la mot ma tran cheo cap 3.

iv) Ma tran dctn vi cap n : la ma tran cheo cap n , ky
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 :
5ii =

khi
khi

i
0

i=j

i* j

va khi do, ma tran don vi cap n difpc viet dudi dang

' 1
0

0

1

. (P

,0

o

• 1J

.

0

(5ij)i,j=ur

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

I2 =

M O'

fl

0

; i3 = 0

1
0

lo

0

□

p

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.


10

Vi du 7. Ma tran
^12

B=

0

t>22

0

0

-

bln

b2n

bnn )

la mot ma tran tarn giac tren va ma tran

c=

C11

0

...

0 '

C21

c22

•••

0

lcnl

cn2

cnnJ

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

1) la mot ma tran dong.

z 1 '
ii) Ma tran B = 0 la mot ma tran cot.


11
iii) Ma tran

1
2

2
-1
0

0
0
1

-b

tao bdi 3 vecto dong [C]1 = (1 2

0

i); [c]2 = (3

' 1
C= 3

l-l
[C]3=(-l 0 1

(q
[C]1

-1

o 2);

1), hay tao bcfi 4 vectcf cot

fOA
|2

3 ; [c]

l-ij

-1 ; [C]3 = 0 ; [C]|4

f 12

□

0

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 , h e R , ma.
tran tong ciia A va B , ky hieu A + B, la ma tran cap m x n
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 ma
tran cap m x n xac dinh bdi [hA].. = h [A]., vdi moi i, j.


12

Vi du 9. Vcri A =

A+ B

M 2 3'
,4 5

6?

, B=

'2 4 6
'2 1 4
, 2A =
,8 10 12
,3 6 5

-1

1

1
-1
va

4B =

1
thi
-1,
'-4
, 4

4
-4

-4^
.
4y

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.

2.1.2. Menh de. Vdi moi ma trdn A,B,C

g

Mmxn vd. h,ke R,

ta co

(i) A + B = B + A (tinh giao hodn).
(ii) (A + B) + C = A + (B + C) (tinh ket hop).
(Hi) A + 0 = A r 0 : ma Iran khong cap m x n ).

(iu) A + (-A) = O.

(v) h(kA) = (hk)A.

(vi) h(A + B) = hA + hB.
(uii) (h + k)A = hA + kA.
(uiii) l.A = A.
Cac tinh chat trong menh de tren ditoc kiem chilng mot
each de dang va dtroc coi nhif la bai tap. (5 day, ta chdng minh
tinh chat (vi) lam vi du :


13

Do

A,BcMmxn

nen cac ma tran

A + B,

hA,

hB,

hA + hB va h(A + B) deu la cac ma tran cap m x n . Ngoai ra,
vdi moi i = l,m , j = l,n , ta cd

[h(A + B)]irh[A + B]..=h ([AVMj^tAVh^
= [hAjy + [hB]y = [hA + hB]..
Hai 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.
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

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

m x p , ky hieu AB , xac dinh bbi

Da], Ph
[AL [B]lk + H2 [B]2k+... + [A]in [B]nk
vdi moi i - 1, m , k = 1, p .

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] ik

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


14
'ik

dong i

—*

0

’2k

dong i

cot. k

T
cot k

Vi du 10. Cho

( 1 2^
2 3^1
A = -1 1 , B =
<-2 1,
.2 3,

Ta co A g M 3x2, B g M2x2 va khi do, ta cd ma tran tich
AB g M3x2 vdi cac so hang lan liTOt la

[ABJn [A^ -[B]1

=1.2 + 2(-2) = -2,

[A^-fB]2 =1.3 + 2.1 =5,

[AB121

[A^-p]1 =-1.2 + l(-2) = -4,

[ab]22

[A]2.[B]2=-1.3+l.l = -2,

[Mi

[A]3 ■ [B]1 = 2.2 + 3(-2) =

[ABL

[A]3 ■ [B]2 = 2.3 + 3.1 = 9 ,

2,

Tom lai,
' -2 5
AB = -4 -2

1-2

9>

□


15
Chu y ring vdi phep nhan ma tran, ta co the bieu dien
mot he phirong trinh tuyen tinh bang mot phiTOng trinh ma
tran. Ching han, trd lai vdi he phiTcmg trinh tuyen tinh
xi

< -X1

-2x!

X2

+

x3

2x2
3x2

+

x3

+

x3

2
6
7

(1.3)

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

ri
A = -i

1-2

-1
2
3

n
i

va B = 6 .

b

J;

xi

Goi X = x2

la ma tran cac an so. He phiTOng trinh (1.3)

dilpc viet lai thanh
AX = B

(1.4)

2.2.2. Menh de.
(i) Tinh ket hop : Vol moi ma trail A G M mxn >

Mnxp

va C g Mpxq,

A(BC) = (AB)C.

(ii) Tinh phdn bo : Vai moi ma tran A,BeMmxn va

Cg M
lvlnvp>
(A + B)C = AC + BC,

A,BGMIlxp,
va vol moi ma tran C g M mxn
C(A + B) = CA + CB.


16
(Hi) Vai moi ma Iran A e Mraxn -Be Mnxp vd h g R ,

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
n

>

( p

n

tA (BC)],! = Z [Ak [BC11 = Z [H \kZ= l IBk Mki >
j=l

j=l

P

n

= ZZ[H[
bLJc]m ZZKM.JCL
j=l k=l
k=lj=l
n

P

=z DAmk [c]

Ikl

k=l\vj=l

j

j=l

=[(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
vdi moi i = 1, m , k = 1, p , ta co

[(A+B)cL = Z!A-BMc]3k
j=l

j=l
n

n

j=l

+H[cM
n

= ZiAMck + Z[B]y[c]jk
j=i

M

HAC]ik+[BC]ik=[AC + BC]ik


17

va do do, (A + B) C = AC + BC.
Vdi CeMmxn, A,BeM nxp, ta co C(A + B), CA + CB la

cac ma tran cung cap m x p va vdi moi i = 1, m , k = 1, p ,
n

[c(A+B)k=Xc«[A+Bk = ZcdlAk + [Bk)
j=l

j=l

= Zc«[A]ik + Zcy[Bk
j=l

j=l

[CB]
= lCAlik ''+ L^-^Jik

[CA + CB]ik

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 ,

[h(AB)]it

h[ABk i>E[AyB]jk
j=l

n

Z(hH)[B]ljk=ZR.(h[B]jk)
j=l

j=l

Ma

Z(hlA]y)[B]Ijk ZMJB]., = [(hA)BL
j=l

j=l

va

ZH HUJ = Z[AMhBln< UA(hB)J
j=l

ik

j=1

nen ta suy ra h (AB) - (hA) B = A (hB).

TRtflKGDAiHOCDAM^

THU Vll
StfdAng.ky:....

•.

J

**•

•


18

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 Iran noi chung khong co tfnh giao
hoan, nghia la tong quat ta cd AB BA.
Vi du. 11. Vdi hai ma tran A =
■'1

AB =
0

°'l * BA =

°>

0 1
0 0
, ta co
, B=
0 0
1 0

0 0
0 1

□

Trong trifdng hop ca hai ma tran tich AB va BA ton tai
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,n giao hoan vdi
moi ma tran vudng A cap n va InA = ALn

A.

Tong quat, neu B la ma tran cap m x n, ta cd
ImB = BIn = B , trong do Im > In lan loot la cac ma tran don vi
cap m va n .
That vay, ta cd ImB , BI.n cd cung cap m x n va vdi moi
i = l,m , k = l,n ,
n

Z[H[
Bk = j=l

j=l
vi 5- = 1 khi i = j va Sjj = 0 khi i

[B]lk

j. Ttfcmg tif,


19
n

n

j=l

j=l

ML EPWiJk=En5

Jk -

[BJlk

vi 6jk = 1 khi j = k va 5jk = 0 khi j * k. Tom lai, ta difoc
= B = BIn •

Vi du 12. Cho

'1

2

3^

<4

5

6;

2

_ '1

2

5

’ ^4

5

A=
Ta co

i2a

=

.0

1
AI3 =
4

°Y
1
1 A4
2 3
5 6

6,

'i 0 (T
'1
0 1 0 =
<4
^0 0 1?

r

3"
6?
2
5

3'
6?

'-1

-1
Vi du 13. Cho A = -1 2 i va C = -1
1
1-2 3 i.

r 1

□
4

3
-1

-3"
-2
1 ?

Ta co

fl

0
AC = CA = 0 1 0 = I3.
0 1.

lo

va do do, hai ma tran A va C giao hoan vdi nhau. Thiic ra, ma
tran C nhif vay con diioc goi la ma tran nghich ddo cua A, ky

hieu A-1 (xem phan 4).
Khi do, neu nhan ben trai hai ve cua ding thilc (1.4) cho

C, ta dupe


20
C(AX) = CB.

Do C(AX) = (CA)X = IgX = X , d^ng there tren cho
"-1 4
X = CB = -1 3
1 -1

rr

-3" "2"

-2

6

2

i A7J

va do do, ta nhan difoc Xj = 1; x2 = 2; x3 = 3. Noi khac di, ta

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

dich thay doi cac dong cua ma tran A , bien no thanh ma tran
moi A'g Mmxn (A' cung cap vdi A). Ta dinh nghia 3 phep
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,
nghia la moi dong khac cac dong i, j cua A va A' bang nhau,
dong thtf i cua A' bang dong thdr j cua A va dong thu' j cua
A' b^ng dong thd i cua A ,

MM-

[A’]k = [A]k khi k*i,j, [A']. = [A], va [A']
Vi du 14.

^3
0
A=
1

2
1

<5

-1

3

1 S''
2 3
2 4
2

p
(1)~(3)

?

0

3
,5

3

1
2
-1

2 4^
2 3

1
2

5

□

0>

ii) Phep bien doi 2 : Nhan dong i vdi mot so a 0 , ky
hieu A- (i):~g(i) ^•A', nham nhan dong thu' i cua A vdi
a,


21

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
]k khi k * i va [A']. = a ■ [A],.

Vi d u 15.

"i

-s"

2

A= 0 1
^0 0

(3)p3)

4
5>

2

-> 0

1
k0 0

□

4

1 ?

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,

[A]k khi k

[A’]k

va [A']. =[A]. + a •[A]r

Vi du 16.

'-1
A=

1

1

0"

0

-1

1

4

0

-2?

(3)Hl)+(3)

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

,

0

.0

"-1
•» 0
k 0

1
-1
0

-1
1
0
1
-1^

0'
1

-2>

□

Chu 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.


22
Vi du 17.

r-i
o

1

-1
0

(l):=(l)+(2)

0"
1

-L

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

"i 0 -n
-> 0 1 -i

,0 0 i a

' 1 -i
0 i
0 o

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

0
-1
1 )

-> I)

.0

0 (T
1 0 = I30

□

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.


23
Kha ndng 3. Phan th true xoay khac 0 : Thay cac dong
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.

rs i
4

2

A=

-1
3 0
7 3

1
<3

'1

2'

0

5
5
9,

ri

1

0

2'

o Hl

1

3>

0

2

-1
0

.0

4

3

2^

1 0
0 2-1

0 0 [i] 2
.0 0

b

5

0

2 A

0
0 0

-1
1

1
2

,0 0

0

'1

1

3

1
2

□

Ngoai 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

1

0

2

1

0 [H

-1
1
0

1
2

o E]

^1
0

v0

0
0

0

0

,0

0

0
2

1
0

3A
2
1
2

'1

0

i
2

0

1

1
2
_1
2

2

0

0

0 2

0

0

2^

i;

1°

1,

24

'1

0 0

1

0

1

0

2
3
2

0

0

1

2

0

0

k°

'1 0 0 tn
0 1 0 o
0 0 1 0
0 0 1.

□

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
X1

X2

X3

= 2

“X1

2x2
3x2

X3

= 6

x3

= 7

-2x1

+

+

va thiTc hien cac phep bien dbi so cap tren dong cho ma tran
cac he so mo rong A = (A |B) sao cho ma tran cac he so A trb

thanh ma tran tarn giac tren,