1

I HC QUC GIA THNH PH H CH MINH

TRNG I HC BCH KHOA TP.H CH MINH

KHOA KHOA HC NG DNG

B MN TON NG DNG

--------*-------

BO CO BI TP LN

TI S: 4

GVHD: PHAN TRNG THC

Khoa: KHOA HC NG DNG

Lp : KU1301 KU1302

Nhm: 4

Nhm sinh vin thc hin:

H v tn MSSV H v tn MSSV

1. Nguyn Tr Dn K1300508 6. Nguyn ng Phc K1303037

2. Dng Bnh Nguyn Lm K1301997 7. inh Th M Linh K1302054

3. u Th Ngc Cnh K1300338 8. V Th M Ngn K1302504

4. Trn Th Hng Lin K1302045 9. Nguyn Hong Long K1302149

5. Nguyn Trung Hiu K1301186

Tp. HCM, thng 1 nm 2014

2

ti ny gm 2 phn:

1. Nhp h Vector E dng ma trn ct. Kim tra xem E c l c s hay khng? Nu c nhp ma trn ca nh x tuyn tnh f trong c s E v vector x. Tm f(x).

2. Nhp vo ma trn A. Kim tra xem A c vung v kh nghch hay khng? Nu c, hy tnh cc phn t b i s Aij, lp ma trn ph hp v suy ra ma trn nghch o. Khng ng

dung bt c lnh mc nh no tm ma trn nghch o.

M C L C Ph n 1: ...........................................................................................................3

I. ..................................................................................3 t .................................................................................3

1. c a m ............................................3 2. ........................................................3

ng thu ..........................................................................4 nh ..............................................................5 Ch y th .............................................................6

Ph n 2: ............................................................................................................8 I. ..............................................................................8 II. t...............................................................................8

1. Ma trn kh nghch .......................................................................... 8

2. Ma tr n ph h p ......................................................................9 III. ng thu ......................................................................9 IV. nh .......................................................................10

1. t t .........................................................10 2. Ch y th .......................................................11

u tham kh o ...........................................................................................12

3

PHN 1:

I.

Nhp h vc t E dng ma trn ct, Kim tra xem E c l c s hay khng? Nu c nhp ma trn

ca nh x tuyn tnh f trong c s E v vc t x. tm f(x).

II. t 1.

t K-kgv. T

kim tra, ta c nu:

B l tp sinh+ B LTT= B l c s ca E

Trong B l tp sinh ca E nu vi mi xE,

X= = . Ta cng ni E sinh bi

B v k hiu : E=Span(M)=

B LTT =

V d: B= vi i=(1,0,0),j=(0,1,0).k=(0,0 ,1) l mt c s ca khng gian

th x=.i + .j+ .k

xt =(0,0,0 .

Vy B l mt c s ca khng gian

: Cho A l K- KGV, dim(A) = n

tp c s vector ln hn n u PTTT => 1 tp LTT th s vector n

tp c s vector nh hn n u khng l tp sinh => 1 tp l tp sinh ca A th s vector

n

1 tp l tp c s ca A th s vector ca n phi bng n. Vy h vector E l c s ca A

th dng ma trn ct ca h vector E phi l ma trn vung

2.

Cho E K-kgv, f . Khi f hon ton oc xc nh bi cc vect f( vi

B={ } l mt c s ca E,

Nu f()= th ma trn

A= chnh l ma trn biu din nh x f trong c s B ca E

4

X= = Y= = , th ta c = hay = ,

Cc buc tm f(x):

Bc 1: kim tra xem h vect E c l c s khng

Buc 2: tm to vect x trong c s E l nghim ca h phng trnh x=

=(

Hay x = E * [x]E

Buc 3: tm f(x) : tm =A*[ ; ( ) *[ ( )]Ef x E f x

: cho nh x tuyn tnh f: bit ma trn ca nh x tuyn tnh f trong c s

E= l A= . Tm f (-1,5)

Ta xt ma trn E= suy ra E l c s, v E c lp tuyn tnh v E vung.

Ta c x=(-1,5)= =

T ta c =A = =

Vy f(-1,5)=-1(1,1)+6(-1.1)=(-7,5)

III. ng Thu

Bc 1: Nhp h vc t E di dng ma trn ct

(ma trn A = [ 1e 2e 3e ne ]

T )

Kim tra xem E c l c s hay khng bng 2 bc:

Bc 1.1: Kim tra A c l ma trn vung hay khng Nu A vung : tip tc bc 2 Nu A khng vung :kt lun E khng l c s v kt thc

Bc 1.2: Tnh hng ca A Nu r(A) = m xut ra mn hnh: h vector E l c s . Nu r(A) khc m th xut ra mn hnh : h vector E khng l c s , thot.

Bc 2: - Nhp ma trn ca nh x tuyn tnh f trong c s E.

- Nhp vector x

- Ta ca vector x trong c s E c tnh bi: xE=E-1

*x .

- Ta ca nh x tuyn tnh f trong c s E c tnh bi: fE= E*xE.

- F(x) c tnh bi: f(x)=E*fE.

- Xut ra mn hnh : f(x)

5

nh :

E = input( 'nhap ho vector E duoi dang ma tran cot' );

[m, n] = size(E);

if m==n

if rank(E) == m

disp( 'ho vector E la co so' );

A = input( 'nhap ma tran anh xa tuyen tinh f trong co so E' );

x = input( 'nhap vector x' );

xE = inv(E)*x;

fE = A*xE;

f = E*fE;

disp( 'f(x)= ' );

disp(f)

else

disp( 'ho vector E ko la co so' )

end

else

disp( 'ho vector E ko la co so' )

end

6

Ch y th

1. Cho nh x tuyn tnh f : 2 2

, v h vector B = {(1,1),(-1,1)}. Kim tra xem B c l c

s hay khng ?

Nu B = {(1,1),(-1,1)} l c s, cho A = 1 1

0 2.l ma trn ca nh x tuyn tnh f

trong c s B. Tnh f(-1,5)

Chng trnh chy:

>> bai1

nhap ho vector E duoi dang ma tran cot[1 - 1;1 1]

ho vector E la co so

nhap ma tran anh xa tuyen tinh f trong co so E [1 - 1;0 2]

nhap vector x[ - 1;5]

f(x)=

- 7

5

2. Cho nh x tuyn tnh f : 3 3

, v h vector B = {(1,1,3),(-1,1,2),(3, 5, 8)}. Kim tra xem

B c l c s hay khng ?

Nu B = {(1,1,3),(-1,1,2),(3, 5, 8)} l c s, cho A =

3 4 2

1 6 7

9 2 1

.l ma trn ca nh x

tuyn tnh f trong c s B. Tnh f(-1,5,3)

y:

>> bai1

nhap ho vector E duoi dang ma tran cot[1 - 1 3;1 1 5;3 2 8]

ho vector E la co so

nhap ma tran anh xa tuyen tinh f trong co so E[3 4 2; 1 6 7; 9 2 1]

nhap vector x[ - 1; 5; 3]

f(x)=

- 113.0000

- 145.0000

- 230.000

7

3. Cho nh x tuyn tnh f : 3 3

, v h vector B = {(1,1,3),(-1,1,2),(3, 5, 8),(2 4 6)}.

Kim tra xem B c l c s hay khng ?

Nu B = {(1,1,3),(-1,1,2),(3, 5, 8), (2, 4 , 6)} l c s, cho A =

3 4 2

1 6 7

9 2 1

.l ma trn

ca nh x tuyn tnh f trong c s B. Tnh f(-1,5,3)

Chng trnh chy:

>> bai1

nhap ho vector E duoi dang ma tran cot[1 - 1 3;1 1 5;3 2 8;2 4 6]

ho vector E ko la co so

>>

4. Cho nh x tuyn tnh f : 3 3

, v h vector B = {(1,1,3),(-1,1,2),(2, 2, 6)}. Kim tra

xem B c l c s hay khng ?

Nu B = {(1,1,3),(-1,1,2),(2, 2, 6)} l c s, cho A =

3 4 2

1 6 7

9 2 1

.l ma trn ca nh x

tuyn tnh f trong c s B. Tnh f(-1,5,3)

Chng trnh chy:

>> bai1

nhap ho vector E duoi dang ma tran cot[1 - 1 2; 1 1 2;3 2 6]

ho vector E ko la co so

>>

H T PH N 1

-------****-------

8

PHN 2 I.

Nhp vo ma trn A. Kim tra xem A c vung v kh nghch hay khng? Nu c, hy tnh cc

phn t b i s Aij, lp ma trn ph hp v suy ra ma trn nghch o. Khng c dung bt c lnh

mc nh no tm ma trn nghch o.

II. t:

1. Ma tr n kh ngh ch

Cho ( )nA M K . Nu tn ti ma trn ( )nB M K sao cho AB BA I , trong I l ma trn

n v, th B gi l ma trn nghch o ca ma trn A v k hiu l B=A-1

. Trong trng hp ny ta

ni A l ma trn kh nghch.

2. Ma tr n ph h p

a) Ph i s

Cho A l ma trn vung cp n,nu ta b i dng i ct j ca ma trn th ta s nhn c ma trn

con cp n-1. Khi ij( 1) det

i j

ijA M gi l phn b i s ca phn t dng th I ct th j ca

ma trn A.

b) Ma tr n ph h p

Ma trn

11 1 1 11 1 1

1 1

1 1

A AT

j n i n

i ii inA j ii nj

n nj nn n in nn

A A A A

A A AP A A A

A A A A A A

c gi l ma trn phu

hp ca A

: cho ma trn

1 1 1

0 2 1

0 0 3

A

Khi

2

11

2 1( 1) 6

0 3A 312

0 1( 1) 0

0 3A 413

0 2( 1) 0

0 0A

3

21

1 1( 1) 3

0 3A 422

1 1( 1) 3

0 3A 523

1 1( 1) 0

0 0A

9

4

31

1 1( 1) 1

2 1A 532

1 1( 1) 1

0 1A 633

1 1( 1) 2

0 2A

Vy :

6 0 0 6 3 1

3 3 0 0 3 1

1 1 2 0 0 2

T

AP

III. ng thu

1. n ngh o :

Nu nh thc ca ma trn A kh nghch th ma trn nghch o A-1 c tnh bng cng thc :

1 1

det( )AA P

A

2. c gi i quy :

Bc 1 : Kim tra xem A c vung hay khng ?

Nu A vung, chuyn sang bc 2

Nu A khng vung, thng bo A khng vung v thot chng trnh.

Bc 2 : Tnh nh thc ca ma trn A

Nu det(A)=0 th A khng c ma trn nghch o A-1

.

Nu det(A) 0 th A c ma trn nghch o A-1

, chuyn sang bc 3

Bc 3: Tm ma trn ph hp ca A :

11 1 1 11 1 1

1 1

1 1

A AT

j n i n

i ii inA j ii nj

n nj nn n in nn

A A A A

A A AP A A A

A A A A A A

vi ij( 1) deti j

ijA M

Bc 4: tnh ma trn nghch o : 1 1 .

det( )AA P

A

10

: cho ma trn

1 2 4 1

1 1 2 1

2 3 1 4

2 1 2 1

A

Ta tnh c : ma trn ph hp

0 5 0 5

21 19 6 26

3 7 3 8

15 15 0 15

AP

Ma trn nghch o1

0 0.3333 0 0.3333

1.4000 1.2667 0.4000 1.7333

0.2000 0.4667 0.2000 0.5333

1.0000 1.0000 0 1.0000

A

IV. nh :

1.

A=input( 'nhap ma tran A' );

[m,n] = size (A);

if m==n;

disp( 'ma tran A vuong' );

d = det (A);

if d==0;

disp( 'ma tran A ko kha nghich' );

else

disp( 'ma tran A kha nghich' );

B=zeros(n);

[a , b] = size(B);

for a = 1:n

C = A;

for b = 1:m

C(a,:) = [];

C(:,b) = [];

B(a,b) = (( - 1)^(a+b))*det(C);

C = A;

end

end

B = B';

disp( 'ma tran phu hop cua A la' );

disp(B);

disp( 'ma tran nghich dao cua A la' )

C = 1/det(A) * B;

disp(C);

end

else

disp( 'ma tran A ko vuong' );

end

11

2. Ch y th

Tm ma trn nghch o (nu c) ca 3 ma trn sau:

1 2 3 6

1 4 5 3

2 1 5 7

1 4 1 2

A ;

1 2 5 7

3 1 3 4

4 2 1 4

2 4 10 14

B ;

1 2 3 1

3 2 1 4

4 2 1 2

5 2 1 5

6 3 1 7

C

a. Ma tr n A:

>> bai2

nhap ma tran A[ 1 2 3 6; 1 4 5 3;2 1 5 7;1 4 1 2]

ma tran A vuong

ma tran A kha nghich

ma tran phu hop cua A la

85 30 - 70 - 55

- 9 - 6 14 - 13

7 - 22 - 2 19

- 28 8 8 4

ma tran nghich dao cua A la

- 1.0625 - 0.3750 0.8750 0.6875

0.1125 0.0750 - 0.1750 0.1625

- 0.0875 0.2750 0.0250 - 0.2375

0.3500 - 0.1000 - 0.1000 - 0.0500

>>

b. Ma tr n B:

>> bai2

nhap ma tran A[1 2 5 7; 3 1 3 4; 4 2 1 4;2 4 10 14]

ma tran A vuong

ma tran A ko kha nghich

>>

c. Ma tr n C:

>> bai2

nhap ma tran A[1 2 3 1; 3 2 1 4; 4 2 1 2; 5 2 1 5; 6 3 1 7]

ma tran A ko vuong

>>

-------****-------

12

u tham kh o

1. n t i s tuy , TS. L Xun i, i hc Bch Khoa TpHCM, TpHCM,

2013

2. Tin H c ng d ng : ng d t, Nguyn Hoi Sn (Ch bin),

Nh xut bn i hc Quc gia TpHCM