Dang Toan Phuong

DNG TON PHNG

TS. L Xun iTrng i hc Bch Khoa TP HCM

Khoa Khoa hc ng dng, b mn Ton ng dng

TP. HCM 2011.

TS. L Xun i (BK TPHCM) DNG TON PHNG TP. HCM 2011. 1 / 42

Nhng khi nim c bn nh ngha

nh nghaDng ton phng trong Rn l mt hm thcf : Rn R, x = (x1, x2, . . . , xn)T Rn :f (x) = xT .A.x , trong A l ma trn i xngthc v c gi l ma trn ca dng ton phng(trong c s chnh tc).

V df (x) = f (x1, x2) = 2x

21 + 3x

22 6x1x2 l dng ton

phng. Ma trn A c dng A =(

2 33 3

)



nh nghaDng ton phng trong Rn l mt hm thcf : Rn R, x = (x1, x2, . . . , xn)T Rn :f (x) = xT .A.x , trong A l ma trn i xngthc v c gi l ma trn ca dng ton phng(trong c s chnh tc).

V df (x) = f (x1, x2) = 2x

21 + 3x

22 6x1x2 l dng ton

phng. Ma trn A c dng A =(

2 33 3

)TS. L Xun i (BK TPHCM) DNG TON PHNG TP. HCM 2011. 2 / 42


Dng ton phng trong R3 thng c ghi dng f (x) = f (x1, x2, x3) =Ax21 + Bx

22 + Cx

23 + 2Dx1x2 + 2Ex1x3 + 2Fx2x3.

Ma trn ca dng ton phng lc ny l ma trni xng

M =

A D ED B FE F C

Khi f (x) = f (x1, x2, x3) = xT .M .x


Nhng khi nim c bn V d

V df (x) = f (x1, x2, x3) =

x21 2x1x2 + 4x1x3 + 2x2x3 x23 l 1 dng tonphng. Ma trn ca dng ton phng l

M =

1 1 21 0 12 1 1


Nhng khi nim c bn a dng ton phng v dng chnh tc bng bin i trc giao

Cho dng ton phng f (x) = xT .A.x , vix = (x1, x2, x3)

T . V A l ma trn i xng thcnn A cho ha c bi ma trn trc giao P vma trn cho D : D = PTAP A = PDPT . Khi f (x) = xT .P .D.PT .x = (PT .x)T .D.(PT .x).t y = PT .x = P1x x = Py . Ta c g(y) =

yTDy = (y1, y2, y3)

1 0 00 2 00 0 3

y1y2y3

. Vyf (x) = g(y) = 1y

21 + 2y

22 + 3y

23 .



nh nghaDng ton phng g(y) = yTDy c gi l dngchnh tc ca dng ton phng f (x) = xTAx .

nh lDng ton phng f (x) = xTAx lun lun c tha v dng chnh tc g(y) = yTDy bng cchcho ha trc giao ma trn A ca dng tonphng.



a dng ton phng v dng chnh tc bng php bin

i trc giao

Bc 1. Vit ma trn A ca dng ton phng(trong c s chnh tc)

Bc 2. Cho ha A bi ma trn trc giao P vma trn cho D.Bc 3. Kt lun: dng chnh tc cn tm lg(y) = yTDy . Php bin i cn tm x = Py .




i trc giao

Bc 1. Vit ma trn A ca dng ton phng(trong c s chnh tc)Bc 2. Cho ha A bi ma trn trc giao P vma trn cho D.

Bc 3. Kt lun: dng chnh tc cn tm lg(y) = yTDy . Php bin i cn tm x = Py .




i trc giao

Bc 1. Vit ma trn A ca dng ton phng(trong c s chnh tc)Bc 2. Cho ha A bi ma trn trc giao P vma trn cho D.Bc 3. Kt lun: dng chnh tc cn tm lg(y) = yTDy . Php bin i cn tm x = Py .



V da dng ton phng sau v dng chnh tcbng php bin i trc giaof (x1, x2, x3) = 4x1x2 4x1x3 + 3x22 2x2x3 + 3x23

Ma trn ca dng ton phng

A =

0 2 22 3 12 1 3



V da dng ton phng sau v dng chnh tcbng php bin i trc giaof (x1, x2, x3) = 4x1x2 4x1x3 + 3x22 2x2x3 + 3x23Ma trn ca dng ton phng

A =

0 2 22 3 12 1 3



A() = det(A I ) = 2 22 3 12 1 3

= 0 1 = 2, 2 = 3 = 4.

Xc nh ma trn trc giao. Vi 1 = 2, ta c

P1 =

261616

. Vi 2 = 3 = 4, ta cP2 =

15

25

0

, P3 =

230

130

530

.



A() = det(A I ) = 2 22 3 12 1 3

= 0 1 = 2, 2 = 3 = 4.Xc nh ma trn trc giao. Vi 1 = 2, ta c

P1 =

261616

. Vi 2 = 3 = 4, ta cP2 =

15

25

0

, P3 =

230

130

530

.TS. L Xun i (BK TPHCM) DNG TON PHNG TP. HCM 2011. 9 / 42


Do ma trn trc giao

P =

26 1

5 2

3016

25 1

3016

0 530

.Php bin i (x1, x2, x3)T = P(y1, y2, y3)T s adng ton phng f v dng chnh tcf = 2y 21 + 4y 22 + 4y 23 .


Nhng khi nim c bn a dng ton phng v dng chnh tc bng bin i Lagrange

nh nghaPhp bin i x = Py c gi l php bin ikhng suy bin nu P l ma trn khng suy bin.

Ni dung ca phng php Lagrange l s dngcc php bin i khng suy bin a dng tonphng v dng chnh tc.



a dng ton phng v dng chnh tc bng bin iLagrange

Bc 1. Chn 1 tha s khc 0 ca h s ca x2k ,lp thnh 2 nhm: 1 nhm gm tt c cc h scha xk , nhm cn li khng cha xk .Bc 2. Trong nhm u tin: lp thnh tngbnh phng. Nh vy, ta s c 1 tng bnhphng v 1 dng ton phng khng cha xk .Bc 3. S dng bc 1, 2 cho dng tonphng khng cha xk .TS. L Xun i (BK TPHCM) DNG TON PHNG TP. HCM 2011. 12 / 42


Ch . Nu trong dng ton phng ban u ttc cc h s x2k u bng 0, th ta chn tha skhc 0 ca h s xixj . i bin k 6= i , j :xk = yk , xi = yi + yj , xj = yi yj .



V dDng phng php Lagrange a dng tonphng sau v dng chnh tcf (x1, x2, x3) = x

21 + 2x

22 7x23 4x1x2 + 8x1x3.

Ta cf (x1, x2, x3) = x

21 4x1(x2 2x3) + 2x22 7x23 =

[x21 4x1(x2 2x3) + 4(x2 2x3)2] + 2x22 7x23 4(x2 2x3)2 =(x1 2x2 + 4x3)2 2(x22 8x2x3) 23x23 =



V dDng phng php Lagrange a dng tonphng sau v dng chnh tcf (x1, x2, x3) = x

21 + 2x

22 7x23 4x1x2 + 8x1x3.

Ta cf (x1, x2, x3) = x

21 4x1(x2 2x3) + 2x22 7x23 =

[x21 4x1(x2 2x3) + 4(x2 2x3)2] + 2x22 7x23 4(x2 2x3)2 =(x1 2x2 + 4x3)2 2(x22 8x2x3) 23x23 =TS. L Xun i (BK TPHCM) DNG TON PHNG TP. HCM 2011. 14 / 42


= (x12x2 + 4x3)22(x22 8x2x3 + 16x23 ) + 9x23 == (x1 2x2 + 4x3)2 2(x2 4x3)2 + 9x23 .Vy dng php bin i y1 = x1 2x2 + 4x3;y2 = x2 4x3; y3 = x3 hay x1 = y1 + 2y2 + 4y3;x2 = y2 + 4y3; x3 = y3 ta a f v dng chnh tcf = y 21 2y 22 + 9y 23 .



Dng phng php Lagrange a dng tonphng sau v dng chnh tc f (x1, x2, x3) =x21 + 4x1x2 + 4x1x3 + 4x

22 + 16x2x3 + 4x

23 .

H s ca x21 khc 0 nn f c a v dngf = (x1 + 2x2 + 2x3)

2 + 8x2x3. Dng php bin iy1 = x1 + 2x2 + 2x3, y2 = x2, y3 = x3 hayx1 = y1 2y2 2y3, x2 = y2, x3 = y3 x1x2

x3

= 1 2 20 1 0

0 0 1

y1y2y3

ta a f v dng f = y 21 + 8y2y3



Dng phng php Lagrange a dng tonphng sau v dng chnh tc f (x1, x2, x3) =x21 + 4x1x2 + 4x1x3 + 4x

22 + 16x2x3 + 4x

23 .

H s ca x21 khc 0 nn f c a v dngf = (x1 + 2x2 + 2x3)

2 + 8x2x3. Dng php bin iy1 = x1 + 2x2 + 2x3, y2 = x2, y3 = x3 hayx1 = y1 2y2 2y3, x2 = y2, x3 = y3 x1x2

x3

= 1 2 20 1 0

0 0 1

y1y2y3

ta a f v dng f = y 21 + 8y2y3TS. L Xun i (BK TPHCM) DNG TON PHNG TP. HCM 2011. 16 / 42


i vi dng 8y2y3 v h s ca cc bnh phngu bng 0 nn ta ty1 = z1, y2 = z2 + z3, y3 = z2 z3 y1y2

y3

= 1 0 00 1 1

0 1 1

z1z2z3

ta a f v dng f = z21 + 8z22 8z23 .



Nh vy vi php bin i x1x2x3

= 1 2 20 1 0

0 0 1

y1y2y3

= 1 2 20 1 0

0 0 1

1 0 00 1 10 1 1

z1z2z3

= 1 4 00 1 1

0 1 1

z1z2z3

ta a f v dng chnhtc f = z21 + 8z22 8z23 .TS. L Xun i (BK TPHCM) DNG TON PHNG TP. HCM 2011. 18 / 42

Dng ton phng xc nh du nh ngha

nh nghaDng ton phng f (x) = xTAx c gi l

xc nh dng, nu x 6= 0 : f (x) > 0xc nh m, nu x 6= 0 : f (x) < 0na xc nh dng, nux : f (x) > 0,x0 6= 0 : f (x0) = 0.na xc nh m, nux : f (x) 6 0,x0 6= 0 : f (x0) = 0.khng xc nh du, nux1, x2 : f (x1) < 0, f (x2) > 0.


Dng ton phng xc nh du V d

V dKho st tnh cht xc nh ca dng ton phngf = x21 + 8x

22 + x

23 4x1x2 2x2x3

f c th a v dngf = (x1 2x2)2 + (x2 x3)2 + 3x23 . R rngf > 0, f = 0 khi v ch khi

x1 2x2 = 0x2 x3 = 0

x3 = 0

x1 = x2 = x3 = 0 nn dng

ton phng ny xc nh dng.



V dKho st tnh cht xc nh ca dng ton phngf = x21 + 8x

22 + x

23 4x1x2 2x2x3

f c th a v dngf = (x1 2x2)2 + (x2 x3)2 + 3x23 . R rngf > 0, f = 0 khi v ch khi

x1 2x2 = 0x2 x3 = 0

x3 = 0

x1 = x2 = x3 = 0 nn dng

ton phng ny xc nh dng.TS. L Xun i (BK TPHCM) DNG TON PHNG TP. HCM 2011. 20 / 42


Gi s dng ton phng a v dng chnh tcg(y) = 1y

21 + 2y

22 + . . . + ny

2n

Nu k > 0,k th DTP xc nh dngNu k < 0,k th DTP xc nh mNu k > 0,k,i = 0 th DTP na xc nhdngNu k 6 0,k,i = 0 th DTP na xc nhmNu i > 0, j < 0, i 6= j th DTP khng xcnh du


Dng ton phng xc nh du Lut qun tnh

Gi s dng ton phng a v dng chnh tcg(y) = 1y

21 + 2y

22 + . . . + ny

2n

nh nghaS cc h s dng c gi l ch s dng quntnh. S cc h s m c gi l ch s m quntnhC nhiu phng php khc nhau a dngton phng v dng chnh tc. c im chungca cc phng php ny l: s lng cc h sm v s lng cc h s dng l khng i.TS. L Xun i (BK TPHCM) DNG TON PHNG TP. HCM 2011. 22 / 42

Dng ton phng xc nh du Lut qun tnh

Lut qun tnh

nh lCh s dng qun tnh, ch s m qun tnh cadng ton phng l nhng i lng bt binkhng ph thuc vo cch a dng ton phngv dng chnh tc.


Dng ton phng xc nh du Tiu chun Sylvester

nh nghaCho ma trn A vung cp n. Tt c cc nh thccon to nn dc theo ng cho chnh c gil nh thc con chnh cp 1, 2, . . . , n. K hiu1,2, . . . ,n.

A =

a11 a12 a13 . . . a1na21 a22 a23 . . . a2na31 a32 a33 . . . a3n. . . . . . . . . . . . . . .

an1 an2 an3 . . . ann


Dng ton phng xc nh du Tiu chun Sylvester

Tiu chun Sylvester

nh lCho dng ton phng f (x) = xTAx

1 f (x) xc nh dng khi v ch khii > 0,i = 1, 2, . . . , n.

2 f (x) xc nh m khi v ch khi(1)ii > 0,i = 1, 2, . . . , n.



V dKho st tnh cht xc nh ca dng ton phngsauf (x1, x2, x3) = 5x

21 +x

22 +5x

23 +4x1x28x1x34x2x3

Ta c ma trn ca dng ton phng f l

A =

5 2 42 1 24 2 5

V 1 = 5 > 0,2 =

5 22 1 = 1 > 0,



3 =

5 2 42 1 24 2 5

= 1 > 0 nn theo tiuchun Sylvester dng ton phng cho xc nhdng.



V dCho dng ton phng f (x1, x2, x3) =5x21 x22 mx23 4x1x2 + 2x1x3 + 2x2x3.Vi gi tr no ca m th dng ton phng f xcnh m

Ta c ma trn ca dng ton phng f l

A =

5 2 12 1 11 1 m



V (1)11 = (5) > 0,(1)22 = (1)2

5 22 1 = 1 > 0,

(1)33 = (1)35 2 12 1 11 1 m

= 2 + m. dng ton phng cho xc nh m thm 2 > 0 hay m > 2.


Kho st ng v mt bc hai nh ngha

nh nghang bc hai l ng c phng trnh dngax2 + 2bxy + cy 2 + dx + ey + f = 0,

a, b, c, d , e, f R.nh nghaMt bc hai l mt c phng trnh dngax2 + by 2 + cz2 + 2dxy + 2exz + 2fyz + gx +

hy + kz + m = 0, a, b, c, d , e, f , g , h, k,m R


Kho st ng v mt bc hai Cc ng v mt bc hai c bn

Ellipse x2

a2+

y 2

b2= 1

Hyperbol x2

a2 y

2

b2= 1

Parabol y 2 = 2px

Ellipsoid x2

a2+

y 2

b2+

z2

c2= 1

Hyperboloid 1 tng x2

a2+

y 2

b2 z

2

c2= 1

Hyperboloid 2 tng x2

a2+

y 2

b2 z

2

c2= 1


Kho st ng v mt bc hai Cc ng v mt bc hai c bn

Paraboloid Elliptic z = x2

a2+

y 2

b2

Paraboloid Hyperbolic z = x2

a2 y

2

b2

Mt nn 2 pha x2

a2+

y 2

b2=

z2

c2

Mt tr ellipse x2

a2+

y 2

b2= 1, z R

Mt tr parabol y 2 = 2px , z R


Kho st ng v mt bc hai Nhn dng ng v mt bc hai

Nhn dng ng v mt bc hai

Bc 1. a ng v mt bc hai v dng chnhtc bng php bin i trc giao (php quay)Bc 2. S dng php tnh tin a phngtrnh ca ng (mt) bc hai v ng (mt) bchai c bn.


Kho st ng v mt bc hai V d

V dNhn dng ng cong bc hai sau:3x2 + 2xy + 3y 2 + 8

2y 4 = 0.

Xt f = 3x2 + 2xy + 3y 2. Ma trn ca f l

A =

(3 1

1 3

). Phng trnh c trng ca A l

A() = det(A I ) = 3 11 3

= 01 = 2, 2 = 4.



Vi 1 = 2, ta c P1 =(

12

12

)

Vi 2 = 4, ta c P2 =(

1212

)Ma trn ca php bin i trc giao

P =

(12

12

12

12

)



Vi php bin i X = PY hayx =

12x +

12y

y = 12x +

12y

Vy thay vo phng

trnh ban u ta cx 2 + 2y 2 4x + 4y 2 = 0.S dng php tnh tin, ta vit phng trnh trndi dng (x 2)2 + 2(y + 1)2 = 8. t{

x = x 2y = y + 1

ta c x2

8+

y 2

4= 1. y l

ellipseTS. L Xun i (BK TPHCM) DNG TON PHNG TP. HCM 2011. 36 / 42


V dNhn dng mt bc hai sau:2x21 + 2x

22 + 3x

23 2x1x3 2x2x3 16 = 0.

Xt f = 2x21 + 2x22 + 3x23 2x1x3 2x2x3. Ma

trn ca f l A =

2 0 10 2 11 1 3

.



Phng trnh c trng ca A l

A() = |A I | =

2 0 10 2 11 1 3

= 0 1 = 1, 2 = 2, 3 = 4.



Vi 1 = 1, ta c P1 =

131313

Vi 2 = 2, ta c P2 =

12

12

0

Vi 3 = 4, ta c P3 =

1616

26



Ma trn ca php bin i trc giao

P =

13

12

16

13 1

216

13

0 26



Vi php bin i X = PY hayx1 =

13x 1 +

12x 2 +

16x 3

x2 =13x 1

12x 2 +

16x 3

x3 =13x 1 + 0.x

2 26x 3

Vy thay vo

phng trnh ban u ta cx 21 + 2x

22 + 4x

23 = 16 hay

x 2116

+x 228

+x 234

= 1.

y l ellipsoid



THANK YOU FOR ATTENTION


Nhng khi nim c bnnh nghaV da dng ton phng v dng chnh tc bng bin i trc giaoV da dng ton phng v dng chnh tc bng bin i LagrangeV d

Dng ton phng xc nh dunh nghaV dLut qun tnhTiu chun SylvesterV d

Kho st ng v mt bc hainh nghaCc ng v mt bc hai c bnNhn dng ng v mt bc haiV d

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