toan cao cap chapter 1 ver1

Bi ging Ton Cao Cp PGS.TS L Anh V

Chng I Ma trn, nh thc & h phng trnh tuyn tnh Page 1

BI GING TON CAO CP (HIGHER MATHEMATICS)

PHN I: I S TUYN TNH V QUY HOCH TUYN TNH

(LINEAR ALGEBRAS AND LINEAR PROGRAMMING)

CHNG I. MA TRN, NH THC V H PHNG TRNH TUYN TNH

(MATRICES, DETERMINANTS AND SYSTEMS OF LINEAR EQUATIONS)

I.1. MA TRN (MATRICES)

Ni dung c bn

- Khi nim ma trn. Cc loi ma trn.

- Cc php ton i s trn ma trn.

- Ma trn bc thang dng v cc php bin i s cp dng.

- ng dng ma trn biu din cc d liu trong thc tin.

- Hng ca ma trn v cch tm hng ma trn.

Thut ng then cht (Vit Anh)

- Ma trn Matrix; - Ma trn vung Square Matrix;

- Ma trn n v Unit/Identity Matrix; - Ma trn khng Zero Matrix;

- Ma trn tam gic Triangular Matrix; - Ma trn cho Diagonal Matrix;

- Ma trn bc thang Echelon Matrix; - Bin i s cp Elementary Operations;

- Hng ca ma trn Rank of Matrix.

I.1.1. VI V D TRONG THC TIN 1. Bng cc ch tiu 2. Lu tr cc h phng trnh bc nht nhiu n I.1.2. KHI NIM V MA TRN V VI LOI MA TRN 1. Khi nim ma trn

Mt ma trn cp mn (matrix of size mn) (m, n t nhin dng) l mt bng gm m.n s aij c sp xp thnh m dng v n ct di dng

11 12 1

21 22 2

1 2

...

...

... ... ... ...

...

n

n

m m mn

a a a

a a aA

a a a

v c vit tt bi A = [aij]mn hay A = (aij)mn.

Phn t aij l phn t dng i v ct j ca ma trn A; i l ch s dng, j l ch s ct ca phn t aij . Ty vo cc phn t aij l s thc hay phc m ma trn A cng c gi l ma trn thc hay ma



trn phc. Trong sut gio trnh ny, ta ch yu ch xt ma trn thc nn ta s ch gi n gin l ma trn nu iu ny khng gy ra s hiu nhm no. Hai ma trn c xem l bng nhau nu chng cng cp v mi phn t tng ng u nh nhau.

Tc l ij m n

a = ij m nb aij = bij; i = 1, 2, , m; j = 1, 2, , n.

V d 1. 1 2 3

4 5 6A

l ma trn cp 23, y a13 = 3, a21 = 4, .

2. Vi loi ma trn a) Ma trn vung (square matrix): l ma trn c s dng m bng s ct n (m = n l s t

nhin dng), khi thay v ni ma trn cp nn ta ch ni l ma trn vung cp n.

V d 2. 1 3

5 7B

l ma trn vung cp hai.

Trong ma trn vung cp n, ngi ta gi cc phn t a11, a22, , ann l cc phn t thuc ng cho chnh ca ma trn.

b) Ma trn n v (identity matrix or unit matrix): l ma trn vung c tt c cc phn t thuc ng cho chnh u bng 1, cc phn t cn li u bng 0, k hiu l In hay ch n gin l I khi cp c ch r. Cng c khi k hiu ma trn n v l En hay E.

V d 3. 2 3

1 0 01 0

, 0 1 00 1

0 0 1

I I

l cc ma trn n v cp 2, cp 3.

c) Ma trn tam gic (triangular matrix): l ma trn vung c tt c cc phn t nm pha di, hoc pha trn ng cho chnh u bng 0.

V d 4. C =

1 2 3

0 4 5

0 0 6

,

D =

1 0 0 0

2 3 0 0

4 5 6 0

7 8 9 10

l cc ma trn tam gic.

d) Ma trn cho (Diagonal matrix)): l ma trn vung c tt c cc phn t nm ngoi ng cho chnh bng 0.

V d 5. E =

1 0 0

0 2 0

0 0 3

l ma trn cho.

e) Ma trn ct (column matrix or column): l ma trn ch c mt ct. f) Ma trn dng (row matrix or row): l ma trn ch c mt dng.

V d 6. F =

1

2

3

,

G = 1 2 3 4 ln lt l ma trn ct, ma trn dng.

g) Ma trn khng (zero matrix): l ma trn c tt c cc phn t u bng 0, k hiu l Omn hay ch n gin l O khi cp c ch r.

V d 7. O23 0 0 0

0 0 0

l ma trn khng cp 23.



Ch : tin, ta s dng cc k hiu Mat(m,n) v Mat(n) ch tp hp cc ma trn (thc) cp mn v ma trn vung cp n tng ng (m, n l cc s nguyn dng).

I.1.3. CC PHP TON TRN MA TRN 1. Php cng ma trn (matrix addition): Tng hai ma trn cng cp A = [aij]mn v B =

[bij]mn. l mt ma trn cng cp, k hiu A + B, c xc nh bi A + B:= [cij]mn vi cij = aij + bij; i = 1, 2, , m; j = 1, 2, , n.

V d 8. Cho A = 1 2 3

4 0 2

, B = 3 2 0

5 6 7

. Th th A + B = 2 0 3

1 6 9

.

Ch : Hai ma trn ch cng c vi nhau khi chng c cng cp.

2. Php nhn s vi ma trn (scalar multiplication): Cho s a v ma trn A = [aij]mn. Tch ca a vi ma trn A l mt ma trn cng cp, k hiu aA, c xc nh bi aA:= [bij]mn vi bij = a.aij; i = 1, 2, , m; j = 1, 2, , n.

V d 9. Cho ma trn a = 2, A = 1 2 3

4 0 2

. Th th 2A =2 4 6

8 0 4

.

3. Php nhn ma trn (matrix multiplication): Cho hai ma trn A = [aij]mn v B = [bjp]np.

Tch ca A vi B l ma trn, k hiu AB, c xc nh bi AB: = [cik]mp vi

1

k

ik ij jk

j

c a b

; i = 1, 2,

, m; k = 1, 2, , p.

V d 10. Cho hai ma trn A = 1 2 3

4 0 2

v B =

2 3

1 1

4 2

.

Th th AB = 11 12

21 22

c c

c c

l ma trn vung cp hai. Ta tnh cc phn t ca AB.

Ta c

11 12

21 22

1.2 ( 2).( 1) 3.4 16, 1.3 ( 2).1 3.2 7,

4.2 0.( 1) 2.4 16, 4.3 0.1 2.2 16

c c

c c

Vy AB = 16 7

16 16

.

Ch

- Hai ma trn ch nhn c vi nhau khi s ct ca ma trn u bng s dng ca ma trn th hai.

- Mun tm phn t dng i, ct j ca ma trn tch A.B, ta nhn cc phn t dng i ca ma trn A ln lt vi cc phn t ct j ca ma trn B ri cng cc tch li.

? Ti sao php cng hai ma trn v php nhn mt s vi mt ma trn nh ngha rt t nhin nhng php nhn hai ma trn li nh ngha kh phc tp nh trn?



4. Php chuyn v ma trn (transpose of a matrix) Cho ma trn A = [aij]mn. Ma trn thu c t A bng cch vit cc dng ca A ln lt thnh cc ct c gi l ma trn chuyn v ca A v k hiu l At. Khi At l ma trn cp nm.

V d 11. Cho ma trn 1 2 3

4 0 2A

. Th th

1 4

2 0

3 2

tA

.

Hin nhin ta c (At)t = A, tc l sau hai ln chuyn v ta li tr v ma trn ban u.

5. Ly tha mt ma trn vung (powers of a matrix) Khi A l mt ma trn vung, ta c thm php ton ly tha. C th, ly tha bc n (n nguyn

dng) ca A l ma trn tch ca n ma trn A, ngha l A

n: = A.A. A (n ln) . Tng t nh ly tha ca cc s thc, ta quy c A0 = I, trong A l ma trn vung cp bt

k v I l ma trn n v cng cp vi A.

V d 12. Cho ma trn 1 2

0 3A

. Khi

A0 =

1 0

0 1; A

2 =

1 8

0 9

; A3 =

1 26

0 27

; An =

1 3 1

0 3

n

n

; n l s t nhin.

? Hy kim chng cc kt qu nu trn.

Ch : Th t thc hin cc php ton trn ma trn tng t nh i vi cc s: nhn trc, cng sau. Php tr c xem l h qu ca php cng v php nhn vi mt s: A B: = A + ( 1)B.

CC TNH CHT Gi s cc php ton di y u thc hin c vi cc ma trn A, B, C v cc s a, b. Khi

ta c cc tnh cht sau y: A + B = B + A; A + O = O + A = A; A + ( A) = O; (A + B) + C = A + (B + C); (AB)C = A(BC); 1.A = A; I.A = A.I = A; (ab)A = a(bA);

(a + b)A = aA + bA; a(A + B) = aA + aB; (A + B)C = AC + BC; A(B + C) = AB + AC;

(A + B)t = A

t + B

t; (AB)

t = B

tA

t.

? Hy chng minh cc tnh cht nu trn.

I.1.4. MA TRN BC THANG DNG V CC PHP BIN I S CP DNG 1. Ma trn bc thang (dng) (echelon matrix): l ma trn tho mn ng thi hai iu kin sau y

- Dng c tt c cc phn t bng 0 (nu c) lun nm pha di dng c phn t khc 0 (nu c); - i vi hai dng bt k, nu tnh t tri qua phi, phn t khc 0 u tin (nu c) ca dng di

lun bn phi so vi phn t khc 0 u tin (nu c) ca dng trn.

V d 13. M =

1 2 3 4 5

0 6 7 8 9

0 0 10 11 12

0 0 0 0 0

; N =

1 0 0 0 0

0 0 2 3 0

0 0 0 0 4

l cc ma trn bc thang.

? Ma trn O (cp ty ), ma trn n v c phi l ma trn bc thang (dng) khng? Ti sao?

2. Cc php bin i s cp dng (BSC) trn cc ma trn (elementary row operations) l mt trong ba php bin i sau y trn mi ma trn



(E1): i ch hai dng cho nhau di dj.

(E2): Nhn mt dng vi mt s khc khng di a.di (a 0).

(E3): Thm (bt) vo mt dng mt bi ca dng khc di di + a.dj (a ty ). 3. Tnh cht quan trng: Mi ma trn khc khng, sau mt s hu hn cc php BSC, u

a c v mt ma trn bc thang m c gi l dng bc thang ca ma trn ban u. Ch : Dng bc thang ca mi ma trn khng duy nht v thng c nhiu cch BSC a mt

ma trn v dng bc thang. I.1.5. NG DNG MA TRN TRONG THC TIN (SV t tm hiu)

I.1.6. HNG MA TRN V CCH TM HNG 1. Mnh : i vi mi ma trn khc khng A, dng bc thang dng ca n d khng duy nht nhng s dng khc khng ca mi dng bc thang ca A lun bng nhau v ch ph thuc vo A ch khng ph thuc vo cch BSC thc hin trn cc dng ca A. 2. Hng ca ma trn (rank of a matrix): Cho ma trn A. Nu A = O th hng ca A bng s 0. Nu A khc O th hng ca A chnh l s dng khc khng ca mi dng bc thang ca A. Hng ca A thng c k hiu l rank(A) hay ch n gin l r(A). 3. Cch tm hng ca mt ma trn khc khng: Nh vy, i vi mi ma trn khc khng A, tm hng ca n trc ht ta BSC trn cc dng ca A a n v dng bc thang. Sau m s dng khc khng ca dng bc thang ta c hng ca A. Ch : Nu A l ma trn cp mn th r(A) l s t nhin khng vt qu s b trong hai s m, n. Tc l

0 r(A) min (m, n).

? Hy t tm hiu xem khi nim hng ma trn c vai tr g?

I.2. NH THC (DETERMINANTS)

Ni dung c bn

- Khi nim nh thc.

- Cc tnh cht ca nh thc.

- Phng php tnh nh thc.

Thut ng then cht

- nh thc cp n Determinant of order n;

- Ma trn kh nghch Invertible Matrix;

- Nghch o ca ma trn Inverse of a matrix. I.2.1. NHN LI NH THC CP 2, 3 1. nh thc cp 2

Cho A = 11 12

21 22

a a

a a l mt ma trn vung cp 2 bt k. nh thc (cp 2) ca A l mt s, k

hiu detA hay 11 12

21 22

a a

a a c xc nh bi detA =

11 12

21 22

a a

a a: = a11a22 a21a12.

Nhn xt: nh thc cp 2 c dng xc nh tch c hng ca hai vect, din tch hnh bnh hnh v din tch tam gic trong hnh hc.



+ + +

_ _ _

2. nh thc cp 3

Cho A =

11 12 13

21 22 23

31 32 33

a a a

a a a

a a a

l mt ma trn vung cp 3 bt k. nh thc (cp 3) ca A l mt

s, k hiu detA hay

11 12 13

21 22 23

31 32 33

a a a

a a a

a a a

c xc nh bi

detA =

11 12 13

21 22 23

31 32 33

a a a

a a a

a a a

: = 11 22 33 12 23 31 13 21 32 31 22 13 32 23 11 33 21 12a a a a a a a a a a a a a a a a a a .

nh nh ngha ny, ta dng cng thc Sarrus c minh ha bng s di y.

11 12 13 11 12

21 22 23 21 22

31 32 33 31 32

a a a a a

a a a a a

a a a a a

Nhn xt: nh thc cp 3 c dng xc nh tch hn tp ca ba vect, th tch hnh hp (xin) v th tch khi t din trong hnh hc.

I.2.2. NH THC CP N (DETERMINANT OF ORDER N) 1. Khi nim: Ta s nh ngha nh thc cp n tng qut bng quy np. a) nh thc (cp 1) ca ma trn A = [a11] vung cp 1, k hiu detA, chnh l s detA:= a11.

b) Gi s nh thc (cp n = k) ca mi ma trn vung cp n = k 1 c xc nh. Xt

ma trn vung cp n = k + 1 ty A = ij 1ka . nh thc (cp n = k + 1) ca A, k hiu detA,

l mt s c xc nh nh sau

detA =

11 12 1

21 22 2

1 2 ...

n

n

n n nn

a a a

a a a

a a a

: = 1 1 2 2 ...n n n n nn nna A a A a A ;

y, Anj l tch ca ( 1)n+j

vi nh thc cp k ca ma trn nhn c t A bng cch xa i dng n v ct j; j = 1, 2, , n.

Nh vy, theo nguyn l quy np, ta nh ngha c nh thc cp n ( 1) bt k.

2. V d V d 1

11 12

21 21 22 22 21 12 22 11 11 22 21 12

21 22

:a a

a A a A a a a a a a a aa a

(trng li nh ngha s cp!).



V d 2

11 12 13

21 22 23

31 32 33

a a a

a a a

a a a

: = 31 31 32 32 33 33a A a A a A

= 12 13 11 13 11 123 1 3 2 3 3

31 32 33

22 23 21 23 21 22

( 1) ( 1) ( 1)a a a a a a

a a aa a a a a a

= 11 22 33 12 23 31 13 21 32 31 22 13 32 23 11 33 21 12a a a a a a a a a a a a a a a a a a

(trng li nh ngha s cp!).

V d 3. Cho ma trn vung cp bn

2 0 3 1

1 2 0 3

2 1 2 1

0 3 1 2

A

.

Khi 41 42 43 44det 0 1. 23A A A A A . y

42 43 44

2 3 1 2 0 1 2 0 3

1 0 3 1; 1 2 3 15; 1 2 0 7.

2 2 1 2 1 1 2 1 2

A A A

Vy detA = 2.

I.2.3. CC TNH CHT CA NH THC 1. nh thc khng thay i qua php chuyn v: detA = det(At). 2. det(AB) = detA.detB vi mi cp ma trn A, B vung cng cp. 3. Nu c mt dng (hoc mt ct) khng th nh thc bng 0. 4. Nu c hai dng (hoc hai ct) ging nhau hay t l vi nhau th nh thc bng 0. 5. nh thc ca ma trn tam gic hay ma trn cho bng tch cc phn t thuc ng cho

chnh.

6. Nu i ch hai dng (hoc hai ct) bt k th nh thc i du. 7. Nu nhn mt dng (hoc mt ct) bt k vi mt s th nh thc cng c nhn vi s .

Ni cch khc, nhn t chung ca mt dng (hoc mt ct) c th em ra ngoi nh thc. 8. nh thc khng thay i khi thm hoc bt vo mt dng (hoc mt ct) mt bi ca mt

dng (hay ct) khc. 9. Cng thc Laplace khai trin nh thc theo mt dng hay ct bt k

11 12 1

21 22 2

1 2 ...

n

n

n nn

a a a

a a a

a a a

= 1 1 2 2 ...i i i i in ina A a A a A (Khai trin theo dng i)

= 1 1 2 2 ...j j j j nj nja A a A a A (Khai trin theo ct j)

y, Aij l tch ca ( 1)i+j

vi nh thc ca ma trn nhn c t A bng cch xa i dng i, ct j; Aij c gi l phn b i s ca phn t aij hay v tr (i, j); i, j = 1, 2, , n.



I.2.4. CC PHNG PHP TNH NH THC 1. Dng cc php bin i s cp: tnh nh thc ca ma trn vung bt k, trc ht ta

BSC a ma trn v dng tam gic (trn), sau ly tch cc phn t thuc ng cho chnh (theo tnh cht 5). Tt nhin, trong qu trnh BSC, ta lun nh gi c s thay i gi tr ca nh thc (nh cc tnh cht 6, 7, 8).

V d 4.

1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4

2 3 4 1 0 1 2 7 0 1 2 7 0 1 2 7160

3 4 1 2 0 2 8 10 0 0 4 4 0 0 4 4

4 1 2 3 0 7 10 13 0 0 4 36 0 0 0 40

.

2. Dng cng thc Laplace: Nu pht hin thy nh thc c mt dng hay ct no cha nhiu s 0 th nn khai trin nh thc theo dng hay ct .

V d 5. Tnh D(x) =

2013

2013 2012

2012 2011

2011 2010

0 0 0

0 ( 1) 0

( 2) 1

( 3)

x

x x

x x x

x x x x

v tm n s thc x D(x) = 0.

Gii

D(x) =

2013

2013

2013 2012

2013 2012 2013 2013

2012 2011

2011

2011 2010

0 0 00 ( 1) 0

10 ( 1) 0( 2) 1 ( 1)

( 2) 1( 3)

( 3)

xx

xx xx x x x x

x xx x xx x x

x x x x

= 2013 2013 2 2014 2014( 1) ( 1)x x x x x x . D(x) = 0 x {0, 1}. 3. Phng php tng hp: Trong thc hnh, ta thng phi hp BSC vi khai trin. i khi cn phi bin i tinh t na.

V d 6.

0 1 2 3 0 1 2 31 2 3 1 2 3

0 3 7 8 0 3 7 8 1 15 3 7 8 5 0 1 1 5 25.

5 4 5 6 5 4 5 6 1 45 11 19 0 1 4

5 9 16 25 0 5 11 19

I.2.5. MA TRN KH NGHCH (INVERTIBLE MATRIX) 1. Khi nim: Ma trn vung A c gi l c nghch o hay kh nghch nu tm c mt ma trn B vung cng cp sao cho AB = BA = I (ma trn n v cng cp vi A, B). Lc B c gi l (ma trn) nghch o ca A (inverse of A) v k hiu l A1. Nh vy, nu A kh nghch th A A1= A1A = I 2. Nhn xt a) Ta ch xt n tnh kh nghch ca ma trn vung.

? Hy t l gii ti sao?

b) Ma trn vung khng O ng nhin khng kh nghch.



c) Khng phi ma trn khc khng no cng kh nghch.

d) C th chng minh c AB = I BA = I.

? Hy t chng minh khng nh ny.

V d 7. Ma trn 1 2

3 6 khng kh nghch.

? Hy t kim chng iu ny bng nh ngha.

3. Mnh (v iu kin kh nghch) i vi mi ma trn vung A, cc khng nh sau tng ng

(i) A kh nghch. (ii) detA 0. (iii) rank(A) ng bng cp ca A.

? Hy t chng minh mnh ny.

V d 8. Tm m ma trn A = 2

1 2 0

1 1

0 1

m

m

kh nghch.

Gii

detA = m2 + m + 2; detA = 0 m { 1, 2}. Vy A kh nghch khi v ch khi 1 m 2.

4. Thut ton tm ma trn nghch o Bi ton: Cho ma trn vung A. Tm nghch o ca A nu c.

a) Thut ton dng nh thc v phn b i s

Bc 1: Tnh D = detA. + Nu D = 0 th kt lun A khng kh nghch. Thut ton dng. + Nu D 0 th A kh nghch. Lm tip bc 2.

Bc 2: Tm ma trn ph hp PA ca A. Ma trn ph hp PA ca A l ma trn to thnh t cc phn b i s ca cc phn t ca A, tc l PA = [Aij]n, y Aij l phn b i s ca v tr (i, j); i, j = 1, 2, , n.

Bc 3: Xc nh ma trn nghch o A1 = 1 t

APD

, y tAP l chuyn v ca PA.

V d 9. Tm nghch o (nu c) ca ma trn

1 2 1

2 3 2

3 1 3

A

.

Gii + Ta c D = detA = 6 0. Do A kh nghch. + A11 = 11, A12 = 12, A13 = 7; A21 = 7, A22 = 6, A23 = 5. A31 = 1, A32 = 0, A33 = 1.

11 12 7 11 7 1

7 6 5 ; 12 6 0 .

1 0 1 7 5 1

t

A AP P



Vy 1

11 7 1

11 7 1 6 6 61 1

12 6 0 2 1 06

7 5 1 7 5 1

6 6 6

t

AA PD

.

b) Thut ton BSC Bi ton: Cho ma trn vung A. Tm nghch o ca A nu c.

Bc 1: Lp ma trn [A I] bng cch thm vo bn phi A ma trn n v cng cp.

Bc 2: BSC trn cc dng ca [A I] a n v dng [I B] (B l ma trn no ). + Nu khng th bin i c nh th, tc l trong qu trnh BSC, ma trn bn tri xut hin mt dng khng, th kt lun A khng kh nghch. + Nu bin i c nh th th kt lun A kh nghch vi A1 = B.

V d 10. Tm nghch o (nu c) ca ma trn A =

1 1 2

2 3 5

3 4 8

.

Gii + [A I] =

1 1 2 1 0 0

2 3 5 0 1 0

3 4 8 0 0 1

.

+ BSC (trn cc dng ca) ma trn ny ta c

1 1 2 1 0 0 1 1 2 1 0 0 1 1 2 1 0 0

2 3 5 0 1 0 0 1 1 2 1 0 0 1 1 2 1 0

3 4 8 0 0 1 0 1 2 3 0 1 0 0 1 1 1 1

1 1 0 3 2 2 1 0 0 4 0 1

0 1 0 1 2 1 0 1 0 1 2 1 .

0 0 1 1 1 1 0 0 1 1 1 1

Vy A kh nghch vi 14 0 1

1 2 1

1 1 1

A .

I.3. H PHNG TRNH TUYN TNH (SYSTEM OF LINEAR EQUATIONS)

Ni dung c bn

- Khi nim v h phng trnh tuyn tnh (PTTT).

- Dng ma trn ca h PTTT. iu kin c nghim.

- H Cramer v cng thc Cramer.

- H tng qut v phng php Gauss.

- H thun nht. iu kin c nghim khng tm thng.

- Lin h gia h tng qut v h thun nht.



Thut ng then cht

- H phng trnh tuyn tnh System of Linear Equations;

- H Cramer Cramer System;

- H phng trnh tuyn tnh thun nht Homogeneous System of Linear Equations.

I.3.1. KHI NIM 1. H phng trnh tuyn tnh tng qut m phng trnh, n n s: l h phng trnh c dng

(I)

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

... (1)

... (2)

..........

... (m)

n n

n n

m m mn n m

a x a x a x b

a x a x a x b

a x a x a x b

trong aij, bi l cc s cho trc m ln lt c gi l cc h s (ca n) v h s t do, xj l cc n s, i = 1, 2, , m; j = 1, 2, , n. Nghim ca h phng trnh tuyn tnh (I): l mt b gm n s c sp th t (a1, a2, , an) sao cho khi thay xj = aj (j = 1, 2, , n) vo tt c cc phng trnh trong h, ta c cc ng thc ng. H c th v nghim, c th c nghim duy nht hoc v s nghim. Gii mt h PTTT l vic i tm tp hp nghim ca h . 2. Dng ma trn ca h PTTT

Xt li h (I) nu trn. Ta s a vo mt s ma trn m cn cho vic gii h (I). A = [aij]mn l ma trn gm tt c cc h s ca n v c gi l ma trn h s.

B =

1

2

m

b

b

b

l ma trn gm cc h s t do v c gi l ct t do hay ct v phi.

X =

1

2

m

x

x

x

l ma trn gm cc n s v c gi l ct n (s).

Khi , h phng trnh (I) c vit dng ma trn: AX = B.

Ngoi ra, khi xt h (I), ma trn [A B] (m dng, n + 1 ct) nhn c bng cch ghp thm ct

t do B vo bn phi ma trn h s A s ng vai tr quan trng. Ma trn [A B] c gi l ma trn m rng hay ma trn b sung ca h (I).

V d 1. Xt h phng trnh tuyn tnh 1 2 3

1 2 3

2 3 1;

2 3 11.

x x x

x x x

y, ta c

+ Ma trn h s A = 1 2 3

2 3 1

; ma trn m rng [A B] = 1 2 3 1

2 3 1 11

;



+ Ct t do B = 1

11; ct n s X =

1

2

3

x

x

x

.

Thay x1 =3, x2 = 1, x3 = 2 vo hai phng trnh ca h ta c cc ng thc ng. Vy (3,1,2) l mt nghim ca h cho.

Nghim ny cn vit dng ct

3

1

2

. Ta c th th li bng cch xt tch cc ma trn tng ng:

AX =

31 2 3 1

. 12 3 1 11

2

= B.

I.3.2. IU KIN C NGHIM CA H PHNG TRNH 1. nh l Kronecker Capelli

( H phng trnh (I) c nghim ) ( rank(A) = rank([A B]) ).

V d 2. Xt h phng trnh V d 1 trn, ta c ma trn m rng:

1 2 3 1 1 2 3 1

2 3 1 11 0 1 7 13A B

Suy ra rank(A) = 2 = rank([A B]). Do h c nghim (ng nh ta thy v d 1).

V d 3. Xc nh gi tr ca tham s thc m h di y c nghim.

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

2 3 2 1;

2 5 2 3 2 8;

4 9 4 2 5 6;

5 11 7 4 6 .

x x x x x

x x x x x

x x x x x

x x x x x m

Gii Ma trn m rng ca h l [A B] =

1 2 3 2 1 1

2 5 2 3 2 8

4 9 4 2 5 6

5 11 7 4 6 m

.

y, ma trn bn tri l ma trn h s A, cn ct bn phi l ct t do B. Ta BSC nh sau:

1 2 3 2 1 1

0 1 8 7 4 6

0 1 8 6 1 2

0 1 8 6 1 5

1 2 3 2 1 1

0 1 8 7 4 6

0 0 0 1 5 4

0 0 0 1 5 11m

A B

m

1 2 3 2 1 1

0 1 8 7 4 6

0 0 0 1 5 4

0 0 0 0 0 7m

.



y l dng bc thang ca ma trn m rng vi dng bc thang ca ma trn h s A bn tri. R rng ta c

+ rank(A) = 3 khng ph thuc vo m.

+ rank([A B]) = 3 khi 7;

4 khi 7.

m

m

Do ( H cho c nghim ) ( rank(A) = 3 = rank([A B]) ) ( m = 7 ).

2. Nhn xt: i vi h (I), ta lun c

a) rank(A) rank([A B]) m (s phng trnh). Bi th khi bit rank(A) = m, ni ring m n

(s n), th chc chn c ng thc rank(A) = rank([A B]) v h c nghim.

b) Gi s rank(A) = rank([A B]) = r , 0 r min(m, n). + Nu r = n, ni ring n m, th h c nghim duy nht. + Nu r < n th h c v s nghim ph thuc n r tham s ty . Ta s thy r iu

ny trong cc v d v gii h PTTT.

I.3.3. H CRAMER V CNG THC CRAMER

1. H Cramer H PTTT n phng trnh, n n s vi ma trn h s kh nghch gi l h Cramer. 2. nh l Cramer Cho h Cramer n phng trnh , n n s vi dng ma trn AX = B. Khi h c nghim duy nht cho bi cng thc

X = A1

B =

1

21

n

D

D

D

D

(C)

y D = detA 0, Dj l nh thc nhn c t D khi thay ct j bi ct t do B, j = 1, 2, , n. Cng thc (C) c gi l cng thc Cramer.

? Hy lin h cng thc Cramer vi cng thc nghim ca h n phng trnh, n n s (n = 2, 3)

bit trong i s s cp.

V d 9. Gii h phng trnh

1 2 3

1 2 3

1 2 3

6

2 3 4 21

7 3 6

x x x

x x x

x x x

Ta c A =

1 1 1

2 3 4

7 1 3

; B =

6

21

6

; detA = 12;

1 2 3

6 1 1 1 6 1 1 1 6

21 3 4 0; 2 21 4 36; 2 3 21 36

6 1 3 7 6 3 7 1 6

.D D D

Vy nghim ca h phng trnh cho l x1 = 0, x2 = 3, x3 = 3.



3. Nhn xt: Tht ra cng thc Cramer ch c ngha l thuyt ch t ngha trong thc hnh khi n khng b (n 4).

I.3.4. GII H PHNG TRNH TUYN TNH BNG PHNG PHP GAUSS

tng c bn ca phng php Gauss l bin i tng ng kh dn n s cc phng trnh t trn xng di. Trong ngn ng ma trn, iu ny ng ngha vi vic BSC (trn cc dng) ca ma trn m rng a n v dng bc thang. Sau , gii h ngc t di ln trn bng cch th dn cc n t phi qua tri.

Bi ton: Gii h PTTT (tng qut) m phng trnh, n n s

(I)

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

...

...

... ... ...

...

n n

n n

m m mn n m

a x a x a x b

a x a x a x b

a x a x a x b

1. Thut ton gii bng phng php Gauss

Bc 1: Lp ma trn m rng [A B] ca h (A l ma trn h s, B l ct t do).

Bc 2: BSC (trn cc dng ca) ma trn m rng a n v dng bc thang. T

tnh c hng ca A v [A B].

+ Nu rank(A) < rank([A B]) th kt lun h v nghim. Thut ton dng.

+ Nu rank(A) = rank([A B]) = r th h c nghim. Lm tip bc 3.

Bc 3: T ma trn bc thang, vit li h mi tng ng vi h cho nhng n gin hn. Gi li v tri r n ng vi cc h s u tin khc khng trn mi dng khc khng ca ma trn bc thang v gi chng l cc n chnh (c ng r n chnh). Cc n cn li chuyn sang v phi lm n t do (c n r n t do). Sau xem cc n t do nh tham s v gn cho chng cc gi tr ty ri gii h ngc t phng trnh cui ln phng trnh u bng cch th dn dn cc n t phi sang tri, t di ln trn.

Bc 4: Tm tt kt qu v kt lun v nghim ca h.

2. Ch

+ Nu r = n (s phng trnh) th mi n u l n chnh (khng c n t do), h c nghim duy nht.

+ Nu r < n th h c v s nghim ph thuc n r tham s ty .

3. Cc v d minh ha V d 4. Gii v bin lun h phng trnh cho v d 3:

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

2 3 2 1;

2 5 2 3 2 8;

4 9 4 2 5 6;

5 11 7 4 6 .

x x x x x

x x x x x

x x x x x

x x x x x m

Gii Lp ma trn m rng ri BSC nh v d 3 ta c ma trn bc thang



1 2 3 2 1 1

0 1 8 7 4 6

0 0 0 1 5 4

0 0 0 0 0 7m

.

T rank(A) = 3 ( m), ( rank([A B]) = 3 ) ( m = 7 ). Suy ra h ch c nghim khi m = 7. Lc t ma trn bc thang ta vit c h mi tng ng vi h c nhng n gin hn nh sau:

1 2 3 4 5 1 2 4 3 5

2 3 4 5 2 4 3 5

4 5 4 5

2 3 2 1 2 2 1 3

8 7 4 6 7 6 8 4

5 4 4 5

x x x x x x x x x x

x x x x x x x x

x x x x

Xem x3, x5 l tham s v gn cho chng gi tr ty : x3 = a, x5 = b; a, b l hai s thc ty . Thay vo h v gii ngc t di ln trn bng cch th dn ta c:

1

21 2 4 3 5

32 4 3 5

44 5

5

53 19 71 ;

2 2 1 3 22 8 31 ;

7 6 8 4 ( , );

4 5 4 5 ;

.

x a b

xx x x x x a b

xx x x x a ba

xx x b

x b

.

Kt lun: Ta c tp nghim ca h cho l

1 2 3 4 5( (53 19 71 ; 22 8 31 ; ; 4 5 ; ) / ,; ; ; ; )x a b a b a b b a bx x x x .

Ta c th vit tp nghim dng ct

53 19 71

22 8 31

;

4 5

/ ,

a b

a b

a

b

b

a bX

.

Mi nghim X =

53 19 71

22 8 31

;

4 5

a b

a b

a

b

b

(hoc dng dng (53 19 71 ; 22 8 31 ; ; 4 5 ; )a b a b a b b ) c

gi l nghim tng qut ca h ang xt (ph thuc hai tham s a, b ty ). Khi ta gn cho a, b cp gi tr c th (nhng bt k) ta nhn c mt nghim ring ca h.

I.3.4. H PHNG TRNH TUYN TNH THUN NHT

(HOMOGENEOUS SYSTEM OF LINEAR EQUATIONS)

1. nh ngha: H phng trnh tuyn tnh thun nht l h phng trnh c tt c cc h s t do v phi bng 0:



(II)

11 1 12 2 1

21 1 22 2 2

1 1 2 2

... 0

... 0

..........

... 0

n n

n n

m m mn n

a x a x a x

a x a x a x

a x a x a x

y, ct t do B = O nn dng ma trn ca h l AX = O. Ta cng bo h (II) l h thun nht tng ng vi h PTTT tng qut (I) vi dng ma trn AX = B. Hai h ny c v tri ging ht nhau.

2. Nhn xt: a) Khc vi h tng qut c th c nghim hoc v nghim, h thun nht lun c nghim t

nht mt nghim, l nghim X = O (ct khng). Ta gi nghim X = O l nghim tm thng. Nh vy, i vi h thun nht, vn ta quan tm khng phi l vic h c nghim hay khng m l h c nghim khc tm thng hay khng.

b) V h PTTT thun nht l mt h PTTT nn ng nhin cng gii c bng phng php Gauss. Tuy nhin v ct t do bng khng nn thay v BSC ma trn m rng, ta ch cn BSC ma trn h s.

3. iu kin c nghim khng tm thng ca h thun nht a) H thun nht AX = O c nghim khng tm thng khi v ch khi rank(A) nh hn s

n, hn na lc h c v s nghim khng tm thng. b) Tri li, nu rank(A) ng bng s n th h ch c nghim tm thng v ng nhin

l nghim duy nht ca h. 4. Tnh cht ca tp nghim ca h thun nht v h nghim c bn a) Tp nghim ca mi h thun nht c tnh cht rt p nh sau: + Tng (hiu) ca hai nghim li l mt nghim:

(X1, X2 l nghim) (X1X2 l nghim).

+ Bi ca mi nghim li l mt nghim: (a l s, X l nghim) (aX l nghim). + Gi s hng ca ma trn h s l r vi 0 < r < n ( s n). Khi nh ta bit, h c v s nghim ph thuc n r tham s (n t do). Hn na, ta lun tm c mt h n r nghim khng tm thng {X1, X2, , Xnr}sao cho tp

{X= a1X1 + a2X2 + + anr Xnr / l a1, a2, , anr cc s ty } chnh l tp nghim ca h thun nht ang xt. H {X1, X2, , Xnr} ni chung khng duy nht.

? Hy chng minh cc tnh cht ny!

b) H {X1, X2, , Xnr} nh trn gi l h nghim c bn ca h thun nht ang xt. Ni chung, mi h thun nht c v s h nghim c bn. Mi X = a1X1 + a2X2 + + anr Xnr gi l mt nghim tng qut ca h. Khi gn cho cc tham s a1, a2, , anr cc gi tr c th (nhng ty ) ta c nhng nghim ring ca h. n gin, chng ta s ch nu cch tm h nghim c bn trong v d.

V d 5. Tm nghim tng qut v h nghim c bn ca h phng trnh tuyn tnh thun nht

1 2 3 4

1 2 3 4

1 2 3 4

2 5 7 0

4 2 7 5 0

2 5 0

x x x x

x x x x

x x x x

Gii Lp ma trn h s A ri BSC ta c:



A =

2 1 5 7 2 1 5 7 2 1 5 7

4 2 7 5 0 0 3 9 0 0 1 3

2 1 1 5 0 0 4 12 0 0 0 0

Ta thy rank(A) = 2 < 4 (s n) nn h c v s nghim ph thuc 2 tham s. T ma trn bc thang ta vit h mi (tng ng vi h cho) v gii ta c:

1 2 3 4

3

1 3 2 4

4 3 4

1

2 5 7 0 2 5 2

3 0 = 3 3

4

8

7 2 ( , )

6

2

x x x x x x x

x x x x

a b

x aa b

b

b

x

x

x

x

.

Vy nghim tng qut ca h cho l (a + 8b, 2a, 6b, 2b) vi a, b l cp s thc bt k.

Cho a = 1, b = 0 ta c nghim ring dng ct X1 =

1

2

0

0

. Cho a = 0, b = 1 ta c X2 =

8

0

6

2

.

Ta c h nghim c bn ca h chnh l {X1, X2}.

? Hy kim chng iu ny!

5. Lin h gia nghim ca h tng qut v h thun nht tng ng

a) Xt h tng qut AX = B v h thun nht tng ng AX = O. Gi s Xr l mt nghim ring ca h tng qut. Xtq, Xtn ln lt l nghim tng qut ca h tng qut v h thun nht. Khi ta c: Xtq = Xr + Xtn. Ngha l: Nghim tng qut ca h tng qut bng tng ca mt nghim ring ca n vi nghim tng qut ca h thun nht tng ng.

b) Nhn xt: Nh tnh cht trn, nu bng cch no ta d c mt nghim ca h tng qut th ch cn gii h thun nht (m chc chn l n gin hn gii h tng qut), ta c th suy ra nghim ca h tng qut.

V d 6. Gii h tng qut di y bit (1, 1, 0, 1, 0) l mt nghim ring ca n.

2 3 2 4 3;1 2 3 4 5

3 5 3 2 1;1 2 3 4 5

2 3 4 5 4.1 2 3 4 5

x x x x x

x x x x x

x x x x x

Gii Trc ht ta gii h thun nht bng cch BSC ma trn h s.

A =

1 2 3 2 4 1 2 3 2 4 1 2 3 2 4

3 5 1 3 2 0 1 10 9 10 0 1 10 9 10

2 3 4 5 1 0 1 10 9 9 0 0 0 0 1

.

T ma trn bc thang ta vit h thun nht mi ri gii tip ta c:



1

21 2 3 4 5

32 3 4 5

45

5

17 16

2 3 2 4 0 10 9

10 9 10 0 ,

0

0

.

x a b

xx x x x x a b

xx x x x a ba

xx b

x

Vy nghim tng qut ca h cho l

1

2

3

4

5

1 17 16

1 10 9

,

1

0

.

x a b

x a b

x a ba

x b

x

I.4. MT S M HNH TUYN TNH TRONG KINH T

I.4.1. M HNH CN BNG TH TRNG (MARKET EQUILIBRIUM MODEL) Mt m hnh kinh t thng bao gm mt s i lng (ch tiu) trong kinh t v cc mi

quan h gia chng. Theo ngn ng ton hc, cc i lng kinh t l cc bin s, cn cc mi quan h gia cc i lng kinh t c biu din bi cc phng trnh. Mt m hnh tuyn tnh trong kinh t l m hnh kinh t m tp hp cc quan h c biu din bi mt h PTTT. 1. M hnh cn bng th trng (n gin) mt loi hng ha Khi phn tch mt th trng hng ha, cc nh kinh t hc lun s dng hm cung v hm cu

biu th s ph thuc ca lng cung v lng cu ca hng ha (c tnh trong mt n v thi gian no ) vo gi ca hng ha (trong gi thit cc yu t khc khng thay i). Trong m hnh ny, ta ch xt mt loi hng ha v ch quan tm n ba bin s di y:

Bin gi p (price): gi ca loi hng ha (tnh bng n v tin t).

Hm cung Qs (Quantity Supplied): lng hng ha m ngi bn bng lng bn.

Hm cu Qd (Quantity Demanded): lng hng ha m ngi mua bng lng mua. R rng Qs = Qs(p), Qd = Qd(p) l cc hm s ca bin gi p. Trong thc tin ta thy rng:

(i) Qs l hm tng theo gi p v khi p ln hn mt gi tr p0 > 0 no th Qs

mi dng. (ii) Qd

l hm gim theo gi p. (iii) Th trng trng thi cn bng khi Qs = Qd.

M hnh Qs(p) = Qd(p) c gi l m hnh cn bng th trng (n gin) mt loi hng ha. T thc tin v cng n gin, ta gi s Qs(p) v Qd(p) l cc hm bc nht, tc l c dng tuyn tnh Qs = a0 + a1p, Qd = b0 b1p, y a0, a1, b0, b1 l cc hng s dng. M hnh cn bng th trng lc ny c dng

0 1 0 1

0 1 0 1

0 1 0 1

s s

d d

s d

Q a a p Q a a p

Q b b p Q b b p

Q Q a a p b b p

Gii h phng trnh (vi n l p), ta tm c



+ Gi cn bng p = 0 0

1 1

a bp

a b; + Lng (cung v cu) cn bng 1 0 0 1

1 1

.s d

a b a bQ Q Q

a b

V d 1. Cho hm cung v hm cu theo gi ca mt loi hng ha l Qs = 5 + p, Qd = 55 3p.

a) Tm gi cn bng th trng. b) Tm lng (cung v cu) cn bng.

Gii Gi cn bng th trng l nghim ca phng trnh

a) Qs = Qd 5 + p = 55 3p p = 15. Vy gi cn bng l p = p = 15 (n v tin t).

b) Lng (cung v cu) cn bng l s dQ Q Q = 0 5 + 15 = 10 (n v loi hng ha).

2. M hnh cn bng th trng tng qut nhiu loi hng ha By gi ta xt th trng c n loi hng ha. Lc , gi ca hng ha ny c th nh hng n

lng cung v lng cu ca loi hng ha kia. Ta s dng cc k hiu bin s nh sau:

Bin gi pi: gi hng ha th i, i = 1, 2, , n.

Hm cung Qsi: lng cung hng ha th i, i = 1, 2, , n.

Hm cu Qdi: lng cu i vi hng ha th i, i = 1, 2, , n. Trong m hnh ny, ta vn gi thit cc yu t khc khng thay i, cn cc hm cung v hm cu ph thuc tuyn tnh vo gi, tc l

Qsi = aio + ai1p1 + ai2p2 + + ainpn; i = 1, 2, , n. (1) Qdi = bio + bi1p1 + bi2p2 + + binpn; i = 1, 2, , n. (2)

By gi, m hnh cn bng th trng tng qut i vi n loi hng ha c biu din bi cc ng thc:

, 1,2,...,si diQ Q i n .

Thay vo ng thc trn cc biu din (1), (2) ca cc hm cung, cu; sau chuyn v v t

ik ik ikc a b , ta c h phng trnh tuyn tnh

11 1 12 2 1 10

21 1 22 2 2 20;

1 1 2 2 0

...

...

............

...

;

.

n n

n n

n n nn n n

c p c p c p c

c p c p c p c

c p c p c p c

Gii h phng trnh trn ta tm c gi cn bng ca tng loi hng ha, t tm c lng cung v cu cn bng ca n loi hng ha cho.

V d 2. Xt mt th trng gm ba loi hng ha. Hm cung, hm cu v gi ca chng tha mn cc iu kin sau

1 1 2 32 4sQ p p p ; 2 1 2 31 4sQ p p p ; 3 1 2 32 4sQ p p p ;

1 1 2 310 2dQ p p p ; 2 1 2 31 2dQ p p p ; 3 1 2 33 2 2dQ p p p .

a) Hy tm gi cn bng th trng ca tng loi hng ha. b) Xc nh lng cung v cu cn bng ca mi loi hng ha. Gii H phng trnh xc nh gi cn bng l

1 1 1 2 3 1 2 3

2 2 1 2 3 1 2 3

3 3 1 2 3 1 2 3

2 4 10 2

1 4 1 2

2 4 3 2 2

s d

s d

s d

Q Q p p p p p p

Q Q p p p p p p

Q Q p p p p p p



1 2 3 1

2 3 2

1 2 3 3

6 2 2 12 3;

6 2 2 1;

2 6 5 2.

p p p p

p p p

p p p p

Vy, gi cn bng mi loi l 1 2 13 , 1 , 2.p p p Ta cng gi b (3, 1, 2) l im cn

bng ca th trng. Suy ra, lng hng cn bng ca tng loi nh sau

1 1 2 2 3 37 , 4 , 4.s d s d s dQ Q Q Q Q Q

I.4.2. M HNH CN BNG KINH T V M

(MODEL OF MACROECONOMIC EQUILIBRIUM)

dng n gin, ta xt m hnh cn bng i vi nn kinh t ng, tc l nn kinh t khng c quan h kinh t i ngoi. Trong mi nn kinh t, ta lun xt cc i lng sau y:

Y (Income): tng thu nhp quc dn.

E (Expenditure): tng chi tiu ca nn kinh t.

C (Consumption): tng tiu dng ca dn c.

T (Tax): tng thu.

I (Investment): mc u t theo k hoch ca chnh ph cho nn kinh t.

G (Government): mc chi tiu ca chnh ph. Phng trnh cn bng trong nn kinh t ng l: Y = C + I + G.

Ta gi s u t theo k hoch ca chnh ph (t nht l trong mt khong thi gian khng qu ngn) l c nh I = I0. Hn na, chnh sch ti kha ca chnh ph cng c nh: G = G0. Cn tiu dng (ca dn chng) th ng nhin ph thuc vo thu nhp. Ta gi s hm tiu dng c dng bc nht: C = aY + b i vi bin thu nhp. y, 0 < b chnh l lng tiu dng ti thiu khi khng c thu nhp, cn 0 < a (< 1) biu th xu hng tiu dng cn bin, tc l lng gia tng ca tiu dng khi thu nhp tng thm 1 n v tin t (s hiu thm ngha ca a khi hc sang phn gii tch). Khi , m hnh cn bng kinh t v m dng n gin c quy v h phng trnh sau:

0 0Y C I G

C aY b

0 0Y C I G

aY C b

y Y, C l cc n cm tm, cn a, b, I0, G0 l cc s bit. Gii h ta xc nh c mc thu nhp cn bng v mc tiu dng cn bng ca nn kinh t (ng) v m:

0 0

1

b I GY Y

a; 0 0

( )

1

b a I GC C

a.

By gi, cho gn vi thc t hn, trc ht ta ch n thu thu nhp T. Lc thu nhp c tnh l thu nhp sau thu hay thu nhp kh dng (disposable income) Yd. V thu thng ph thuc vo thu nhp theo dng hm tuyn tnh T = d + tY, y 0 < d l thu ti thiu khi khng c thu nhp, 0 < t ( < 1) l t sut thu thu nhp hay thu cn bin (tc l s gia tng ca thu khi thu nhp tng ln 1 n v tin t). Ta c

Yd = Y T = Y d tY = d + (1 t)Y, C = aYd + b = ad + a(1 t)Y + b. M hnh cn bng nn kinh t v m gi y tr thnh h phng trnh tuyn tnh

( )

o o o oY C I G Y C I G

C a Y T b aY C aT b

T d tY tY T d



y Y, C, T l cc n s cn tm, cn a, b, d, t v oI , oG l cc s bit.

Gii h, ta tm c mc thu nhp quc dn, mc tiu dng v mc thu cn bng l

0 0

1 (1 )

b I GY Y

a t

ad; 0 0

(1 )( )

1 (1 )

b a t I G adC C

a t; 0 0

( ) (1 )

1 (1 )

t b I G a dT T

a t.

Nhn xt: Trong thc hnh, khi cho s liu c th ta c cc h PTTT n gin v gii d dng ch khng cn phi nh cc cng thc trn.

V d 3. Cho tng thu nhp quc dn Y, mc tiu dng C v mc thu T xc nh bi

15 0, 4( );

36 0,1 ;

;o o

Y C I G

C Y T

T Y

trong 500oI (triu USD) l mc u t c nh; 20oG (triu USD) l mc chi tiu c

nh. Hy xc nh mc thu nhp quc dn, mc tiu dng v mc thu cn bng. Gii Ta c

500 20 520

15 0, 4( ) 0,1 36

36 0,1 520 15 0, 4( 0,1 36)

Y C C Y

C Y T T Y

T Y Y Y Y

520 293, 4375

0,1 36 117, 34375

0, 64 520, 6 813, 4375

C Y C

T Y T

Y Y

.

Vy 813, 4375 ; 293, 4375 ; 117,34375.Y C T I.4.3. M HNH IS LM Trong kinh t v m, m hnh IS LM (Investment/Saving Liquidity preference/Money supply,

tm dch l u t/Tit kim Nhu cu thanh ton/Tin cung cp u i) do John Hicks (Anh) cng Alvin Hansen (Hoa k) a ra v pht trin. M hnh ny c dng phn tch trng thi cn bng ca nn kinh t trong c hai th trng: th trng hng ha v th trng ti chnh (tin t). mc trn, ta xt m hnh cn bng kinh t v m ca nn kinh t ng

0 0, ,

Y C I G

C aY b I I G G(0 < b, 0 < a < 1).

Vi s gp mt ca tin t, mt bin s c ngha quan trng cn c xem xt l li sut r (interest rate) v gi tr ca tin t thay i theo thi gian ty theo r. Khc vi m hnh cn bng kinh t v m ta gi thit tng u t khng i I = I0, xt nh hng qua li gia hai th trng hng ha v tin t, ta cn xem tng u t I thay i ph thuc vo li sut theo quy lut: li sut cng cao th u t cng gim. Ni cch khc, ta c hm u t I = b1 a1r (a1 > 0, b1 > 0). Lc ny phng trnh cn bng th trng hng ha l

1 1 0, ,

Y C I G

C aY b I b a r G G Y = aY + b + (b1 a1r) + G0

a1r = b + b1 + G0 (1 a)Y (IS)

Phng trnh biu th quan h gia li sut v thu nhp khi th trng hng ha cn bng (tng cung bng tng cu) nh trn c gi l phng trnh (IS). y, thu nhp Y cng tng th li sut r cng gim. Trn mt phng ta vi trc honh biu th thu nhp v trc tung l li sut th ng biu din I0 l ng thng dc i xung. ng thng gi l ng IS.



Trong th trng tin t, lng cu tin mt, k hiu L, ng bin vi tng thu nhp v nghch bin vi r. Gi s hm cu tin c dng tuyn tnh: L = b2Y a2r (a2 > 0, b2 > 0).

Gi lng cung tin mt l M0. iu kin cn bng trong th trng tin t l

M0 = L M0 = b2Y - a2r

a2r = b2Y M0 (LM)

Phng trnh trn biu th s cn bng ca th trng tin t v c gi l Phng trnh (LM). y, thu nhp Y cng tng th li sut r cng cng tng. Biu din trn mt phng ta vi trc honh biu th thu nhp v trc tung l li sut ta c ng LM, l ng thng dc i ln.

M hnh IS-LM c biu th bi h hai phng trnh (IS) v (LM).

ISLM

1 1 0

2 2 0

(1 )a r b b G a Y

r b Y Ma

1 1 0

2 2 0

(1 )a r a Y b b G

a r b Y M (Y, r l hai n s)

Gii h trn, ta xc nh c mc thu nhp Y = Y v li sut r = r m bo cho s cn bng trong c hai th trng: hng ha v tin t. C th ta c

2 1 0 1 0

1 2 2

( )

(1 )

a b b G a MY

a b a a; 2 1 0 0

1 2 2

( ) (1 )

(1 )

b b b G a Mr

a b a a.

Nhn xt: Tt nhin, ta khng cn nh cc cng thc trn. Trong thc hnh, khi cc d liu c cho c th, vic gii h m hnh IS-LM ht sc n gin.

V d 4. Cho G0 = 250; M0 = 4500; I = 34 15r; C = 10 + 0,3Y; L = 22Y 200r. a) Lp phng trnh IS. b) Lp phng trnh LM. c) Tm mc thu nhp v li sut cn bng ca hai th trng hng ha v tin t.

Gii a) Ta c

Y= C + I + G0 Y = (10 +0,3Y) + (34 15r) + 250. Vy phng trnh IS l 15r = 294 0,7Y. b) Phng trnh LM c dng

L = M0 22Y 200r = 4500 200r = 22Y 4500. c) Mc thu nhp Y v li sut r cn bng l nghim ca h phng trnh

15 294 0, 7 15(0,11 22,5) 294 0, 7

200 22 4500 0,11 22,5

r Y Y Y

r Y r Y

2,35 631,5 268, 72

0,11 22,5 7, 06

Y Y

r Y r

.

Vy Y = 268,72; r = 7,06.

I.4.4. M HNH INPUT OUTPUT CA LEONTIEF (INPUT-OUTPUT MODEL) 1. M hnh Input Output Mc ny gii thiu m hnh Input-Output ca Leontief, cn gi l m hnh I/O hay m hnh

cn i lin ngnh. M hnh ny cp n vic xc nh tng cu i vi sn phm ca mi ngnh sn xut trong tng th nn kinh t a ngnh ca mt quc gia. Trong m hnh, khi nim ngnh kinh t c xem xt theo ngha thun ty sn xut. Hn na, m hnh c xt trong mt vi gi thit di y.

Mi ngnh kinh t ch sn xut mt loi hng ha.

Mi ngnh u s dng mt t l c nh ca cc sn phm ca ngnh khc lm u vo cho sn sut u ra ca mnh.



Khi u vo thay i k ln th u ra cng thay i k ln. Xt mt nn kinh t gm n ngnh kinh t (sn xut) gi quy c l ngnh 1, ngnh 2, , ngnh

n. tin cho vic tnh chi ph sn xut, ta s biu th lng cu ca tt c cc loi hng ha dng gi tr, tc l o chung tt c cc loi sn phm khc nhau vi n v khc nhau bng tin (vi n v tin t no ca quc gia hoc ngoi t mnh). Trc ht, ta a vo mt s khi nim v k hiu cn cho m hnh.

Cu trung gian xij: l gi tr hng ha ca ngnh i m ngnh j cn dng cho sn xut, cn gi l (lng) cu trung gian i vi sn phm ca ngnh i t ngnh j; i, j = 1, 2, , n.

Cu cui bi: l gi tr hng ha ca ngnh i cn cho lao ng, tiu dng, dch v v xut khu ca quc gia; i = 1, 2, , n.

Tng cu mi ngnh xi: l tng cu trung gian v cu cui ca ngnh i, i = 1, 2, , n. Hin nhin, ta c xi = xi1 + xi2 + + xin + bi; i = 1, 2, , n.

xi = 1 2

1 2

1 2

...i i in n in

x x xx x x b

x x x; i = 1, 2, , n. (1)

t aij := ij

j

x

x l t l (c nh khng i i vi mi i, j) ca cu trung gian i vi ngnh i t

ngnh j so vi tng cu ca ngnh j; i, j = 1, 2, , n. Hin nhin 0 aij 1, aij = 0 khi v ch khi hng ha ngnh i khng cn s dng cho sn xut ca ngnh j; i, j = 1, 2, , n. ngha ca cc h s aij nh sau: aij chnh l t phn chi ph m ngnh j phi tr cho ngnh i sn xut ra 1 n v gi tr hng ha ca ngnh j. lm r hn, ta gi s dng tin USD. Khi , tnh bnh qun trong 1 USD gi tr hng ha ca ngnh j c aij USD dng tr cho vic mua sn phm ca ngnh i. Chng hn, khi aij = 0,3 c ngha l tnh bnh qun sn xut ra 1 USD hng ha ca mnh, ngnh j cn phi mua (s dng) 0,3USD gi tr hng ha ngnh i. T cc h thc (1) ta c h PTTT

1 11 1 12 2 1 1

2 21 1 22 2 2 2

1 1 2 2

...

...

... ... ...

...

n n

n n

n n n nn n n

x a x a x a x b

x a x a x a x b

x a x a x a x b

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

(1 ) ...

(1 ) ...

... ... ...

... (1 )

n n

n n

n n nn n n

a x a x a x b

a x a x a x b

a x a x a x b

y, cc n s l x1, x2, , xn; cn cc h s aij, bi (i, j = 1, 2, , n) cho c nh i vi mt nn kinh t trong mt giai on nht nh. H ny gi l m hnh Input-Output hay m hnh cn i lin ngnh. Gii h ny ta s tm c tng cu x1, x2, , xn hay u ra ca mi ngnh trong nn kinh t. iu ny c ngha quan trng i vi vic lp k hoch sn xut, m bo cho nn kinh t vn hnh bnh thng, trnh tnh trng d tha mt hng ny hay thiu ht mt hng kia. Trong ngn ng ma trn, ta xt cc ma trn di y.

A:= [aij]n l ma trn gm cc h s t phn aij v c gi l ma trn (h s) k thut hay ma trn (h s chi ph) u vo ca nn kinh t.

B:= [bi]n1 l ma trn (ct) cu cui ca nn kinh t.

X:= [xi]n1 l ma trn (ct) tng cu (u ra) ca nn kinh t.

Lc ny, h trn c vit li dng ma trn nh sau: X = AX + B (I A)X = B. R rng nu I A kh nghch th li gii ca h l duy nht v cho bi X = (I A) 1.B. Cn nu

det(I A) = 0 th h c th v nghim, c th v s nghim.

2. Nhn xt



a) Trong nn kinh t hot ng bnh thng, ma trn h s u vo A = [aij]n cho ta nhng thng tin sau y:

+ Mi phn t aij dng i l t phn gi tr hng ha m ngnh i bn cho ngnh j lm hng ha trung gian sn xu. Chng hn aij = 0,2 tc l hng ha m ngnh i bn cho ngnh j lm hng ha trung gian chim 20% gi tr hng ha ca ngnh j (i, j = 1, 2, , n).

+ Tng cc phn t trn ct j chnh l t phn chi ph u vo m ngnh j phi tr cho vic mua hng ha trung gian tnh trn 1 n v gi tr hng ha ca mnh, do khng qu 1, tc l

1

n

ij

i

a

1; j = 1, 2, ..., n.

? Hy t l gii iu ny.

b) Hiu a0j: = 1 1

n

ij

i

a

1 chnh l h s t phn gia tng trong tng gi tr hng ha ca

ngnh j (cn gi l u vo c bit ca ngnh j), tc l bnh qun trong mi 1 USD gi tr

hng ha m ngnh j sn xut ra c aoj USD l gi tr tng thm, cn 1

n

ij

i

a

l tng chi ph u

vo c c 1 USD gi tr hng ha . Tnh trn ton b gi tr hng ha ca ngnh j, ta c t phn gi tr gia tng l 100aoj%, j = 1, 2, ..., n. V d 5. Cho ba ngnh kinh t vi ma trn h s u vo l

A =

0, 2 0,3 0, 2

0, 4 0,1 0, 2

0,1 0,3 0, 2

.

Bit nhu cu cui cng ca cc ngnh ln lt l 10, 5, 6. a) Gii thch ngha ca h s 0,3 dng 3, ct 2 ca ma trn u vo. b) Tm h s t phn gia tng a0j ca tng ngnh (j = 1, 2, 3). Gii thch ngha ca h s a01. c) Tm u ra cho mi ngnh. Gii a) tin ta gi s cc gi tr hng ha c quy v USD. Khi , h s a32 = 0,3 c ngha

sn xut ra 1USD gi tr hng ha ca ngnh 2 cn mua 0,3USD gi tr hng ha ca ngnh 3. b) Tng cc phn t trn mi ct ca ma trn A u nh hn 1. Ta c cc h s t phn gia tng

ca cc ngnh l a01 = 1 (a11 + a21 + a31) = 1 ( 0,2 + 0,4 + 0,1) = 0,3. a02 = 1 (a12 + a22 + a32) = 1 ( 0,3 + 0,1 + 0,3) = 0,3. a03 = 1 (a13 + a23 + a33) = 1 ( 0,2 + 0,2 + 0,2) = 0,4.

H s a01 = 0,3 c ngha l t phn gi tr gia tng trong tng gi tr hng ha ca ngnh 1 l 30%.

c) Ta c I A =

1 0, 2 0, 3 0, 2

0, 4 1 0,1 0, 2

0,1 0, 3 1 0, 2

=

0,8 0, 3 0, 2

0, 4 0, 9 0, 2

0,1 0, 3 0,8

.

H Input Output y c dng ma trn l

(I A)X = B

1

2

3

10

5

6

0,8 0,3 0, 2

0, 4 0,9 0, 2

0,1 0,3 0,8

x

x

x

,

trong X l ma trn u ra, B l ma trn nhu cu cui cng. Tm ma trn nghch o ca ma trn I A, ta c



(I A) 1 =

0,66 0,3 0, 241

0,34 0,62 0, 240,384

0, 21 0, 27 0,6

.

Do

X =

1

2

3

x

x

x

= (I A) 1 B =

0,66 0,3 0, 241

0,34 0,62 0, 240,384

0, 21 0, 27 0,6

10

5

6

=

28,84

20, 68

18,36

.

Vy, u ra ca cc ngnh l 1x = 24,84; 2x = 20,68; 3x = 18,36.

BI TP CHNG I

I.1. Tnh

a)

3 4 2 2 4 7

5 5 0 7 4 3 8 6 ;

6 3 3 1 5 2

b)

2 1 7 3 0 1

4 4 3 3 3 5 1 4 .

1 0 4 1 4 3

t t

I.2. Thc hin php ton sau y i vi cc ma trn

a) 4 2 3 5 6 1

. 2 ;3 4 2 4 2 5

t

b)

4 2 2 54 3

3 5 6 1 3 . .2 6

1 7 4 0

I.3. Tnh

a)

33 2

;1 4

b)

22 1 2

3 1 0 ;

3 2 4

c*)

20131 3

;3 1

d*)

20133 8

2 3; e*)

1 3

2 2

3 1

2 2

n

; g*) 1 1

1 1

n

;

( )n .

I.4. Gii phng trnh ma trn

a) Tm ma trn X cp 23 sao cho 1 2 5 7 7

.3 5 13 18 17

X

b) Tm ma trn X cp 22 sao cho

1 1 -1 5 7 7.

2 3 4 13 18 17X

c) Tm cc s a, b, c, d sao cho

1 21 3 22

3 4 .4 5 64

5 6

a c

b d

I.5. BSC a cc ma trn v dng bc thang (dng) v tnh hng ca chng



a)

1 2 3 -1 -2 -3 1 1 2 -1 2 -3

2 5 8 1 -1 -5 5 5 13 10 18 -5 ; b)

5 11 18 10 9 6 3 3 7 5 9 4

8 18 29 10 6 -2

;

9 9 22 14 29 -4

c)

1 2 2 1 3 1 2 1 2 3 3 2 1 2

5 11 9 7 15 1 5 5 11 9 11 15 7 5; d) .

3 7 5 4 8 2 1 3 7 6 4 9 3 1

9 20 16 12 26 2 6 8 20 18 18 26 11 2

I.6. Tm hng (theo m nu c) ca cc ma trn di y. Vi m no th hng ln nht?

1 2 3 1 1 1 4 3 0 1

2 3 4 2 2 3 5 6 2 2) ; ) ; ) ;

0 1 3 5 1 4 9 3 0 2

3 2 4 8 2 8 15 5 0 7

m

m ma b c

m m m

m m

1 4 3 3 1 1 3 3

2 7 4 1 3 2 8 8) ; e) .

3 3 2 1 3 2 8 3

1 4 6 2 1 5

dm

m m

I.7. Tnh nh thc sau y

a)

1 2 3 2

2 1 0 3

2 3 4 1

4 3 0 4

; b)

1 3 4 2

3 0 2 0;

5 5 0 4

2 3 1 2

c)

1 2 3 4

2 3 0 1;

3 0 1 3

0 1 3 2

d)

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

.

e*)

2 2 1

2 2 1

1 1 1 ... 1 1

1 2 2 ... 2 2

... ... ...

1 ...

n n

n nn n n n

( 2 < n t nhin).

I.8. Tm ma trn nghch o ca cc ma trn sau y

1 2 0 1 3 2 1 1 1

) 2 3 1 ; ) 1 2 4 ; c) 1 0 3 .

0 4 3 3 1 3 2 1 2

a A b B C



I.9. Gii cc h phng trnh sau

a)

1 2 3 4

1 2 3 4

1 2 3 4

2 3 4 3

2 3 4 5 5

7 2

x x x x

x x x x

x x x x

b)

1 2 3 4

1 2 3 4

1 2 3 4

4 3 4 1

2 7 6 3 2

2 3 4 5 3

x x x x

x x x x

x x x x

c)

1 2 3

1 2 3

1 2 3

2 3 2 2

3 4 3 1

4 3 4 5

x x x

x x x

x x x

d)

1 2 3

1 2 3

1 2 3

2 3 2

2 3 4 1

3 2 7

x x x

x x x

x x x

I.10. Tm nghim tng qut v h nghim c bn ca h phng trnh tuyn tnh thun nht sau y

a)

1 2 3 4

1 2 3 4

1 2 3 4

2 4 3 0

2 3 2 4 0

2 4 3 0

x x x x

x x x x

x x x x

b)

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

2 2 0

2 2 3 3 5 0

3 2 3 0

x x x x x

x x x x x

x x x x x

I.11. Cho mt th trng gm hai loi hng ha. Hm cung, hm cu v gi ca chng tha mn cc iu kin sau

1 1 2 2

1 1 2 2 1 2

1 3 , 3 5

10 2 2 , 15 3 .

s s

d d

Q p Q p

Q p p Q p p

a) Hy tm im cn bng th trng. b) Xc nh lng cung v cu cn bng ca mi loi hng ha.

I.12. Cho mt th trng gm ba loi hng ha. Bit hm cung v hm cu l

1 1 2 3

2 1 2 3

3 1 2 3

1 1 2

2 1 2 3

3 2 3

15 8

10 12

6 10

20 4 3

40 2 6

30 2 6

s

s

s

d

d

d

Q p p p

Q p p p

Q p p p

Q p p

Q p p p

Q p p

a) Hy tm im cn bng th trng. b) Xc nh lng cung v cu cn bng ca mi loi hng ha.

I.13. Xt m hnh cn bng kinh t v m vi

50 0,6( )

12 0,3

o oY C I G

C Y T

T Y

,

800oI ; 55oG .

Hy xc nh mc thu nhp quc dn, mc tiu dng v mc thu cn bng. I.14. Xt m hnh IS-LM vi

0 075 ; 8160 ; 50 25

40 0,5 ; 28 400 .

G M I r

C Y L Y r

a) Xc nh cc phng trnh IS, LM.



b) Xc nh mc thu nhp v li sut cn bng.

1.15. Trong m hnh Input Output bit ma trn h s u vo ca ba ngnh l

0,2 0,2 0

0,3 0,1 0,3 .

0,1 0 0,2

A

v nhu cu cui cng ca cc ngnh tng ng l 40, 60 v 80. Hy xc nh u ra ca mi ngnh. I.16. Cho ba ngnh kinh t vi ma trn h s u vo l

0,4 0,2 0,2

0,2 0,3 0,4

0,3 0 0,1

A

a) Xc nh h s t phn gia tng ca mi ngnh. b) Xc nh u ra ca mi ngnh bit nhu cu cui cng ca cc ngnh tng ng l 40, 60, 80.

1.17. Gi s mt nn kinh t c ba ngnh: nng nghip, cng nghip v dch v. Bit rng sn xut mt n v u ra

- ngnh nng nghip cn s dng 10% gi tr ca ngnh, 30% gi tr ca cng nghip, 30% gi tr ca dch v;

- ngnh cng nghip cn s dng 20% gi tr ca ngnh, 60% gi tr ca nng nghip, 10% gi tr ca dch v;

- ngnh dch v cn 10% gi tr ca ngnh, 60% gi tr ca cng nghip, khng s dng gi tr ca nng nghip.

a) Lp ma trn h s u vo cho nn kinh t ny. b) Xc nh mc sn xut u ra ca mi ngnh tha mn nhu cu cui cng l 10, 8, 4.

toan cao cap chapter 1 ver1

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