Baigiangds_DVV

7/13/2019 Baigiangds_DVV

1/79

i hc Quc gia TP.HCMTrng i hc Bch Khoa

B mn Ton ng dng

.

Bi Ging i S Tuyn Tnh

TS. ng Vn Vinh

E-mail: [email protected]

Website: www.tanbachkhoa.edu.vn/dangvanvinh

Ngy 14 thng 8 nm 2013


2/79

Mc tiu mn hc

Mn hc cung cp kin thc c bn ca i s tuyn tnh. Sinh vin cn nm vng kin thc nn tng vbit gii cc bi ton c bn: s phc, tnh nh thc, lm vic vi ma trn, gii h phng trnh tuyntnh, khng gian vc t, khng gian euclide, nh x tuyn tnh, tm tr ring - vc t ring, a dng tonphng v dng chnh tc.

Ti liu tham kho

1) Cng Khanh, Ng Thu Lng, Nguyn Minh Hng. i s tuyn tnh. NXB i hc quc gia.

2) Cng Khanh. i s tuyn tnh. NXB i hc quc gia.

3) Trn Lu Cng. i s tuyn tnh.NXB i hc quc gia.

Ghi ch:

Ti liu ny ch tm tc li bi ging ca Thy ng Vn Vinh. hiu bi tt, cc em cn i hc trn lpl thuyt v bi tp.Sinh vin to ti khong trn websitewww.tanbachkhoa.edu.vn/dangvanvinh, lm thm bi tp trc nghimtrn .V ni dung mi c son li nn khng th trnh sai st. Mi gp , sinh vin c th lin h trn dinn website hoc qua mail: [email protected].

1


3/79

Mc lc0.1 Dng i s ca s phc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.2 Dng lng gic ca s phc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1 Ma trn 111.1 Cc khi nim c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.2 Cc php bin i s cp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.3 Cc php ton ma trn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.4 Hng ca ma trn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.5 Ma trn nghch o . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2 nh thc 182.1 nh ngha nh thc v v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.2 Tnh cht nh thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.3 Tm ma trn nghch o bng phng php nh thc. . . . . . . . . . . . . . . . . . . . . . . 21

3 H phng trnh 233.1 H Cramer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.2 H thun nht . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4 Khng gian vc t 284.1 nh ngha v v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284.2 c lp tuyn tnh - ph thuc tuyn tnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294.3 Hng ca h vc t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314.4 C s v s chiu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.5 Ta vc t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364.6 Ma trn chuyn c s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.7 Khng gian con . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384.8 Tng giao hai khng gian con . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5 Khng gian Euclide 44

5.1 Tch v hng ca 2 vc t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445.2 B vung gc ca khng gian con. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475.3 Qu trnh Gram-Schmidt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 495.4 Hnh chiu vung gc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

6 nh x tuyn tnh 526.1 nh ngha v v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 526.2 Nhn v nh ca nh x tuyn tnh. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546.3 Ma trn ca nh x tuyn tnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

7 Tr ring - vc t ring 60

7.1 Tr ring - vc t ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 607.2 Cho ha ma trn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 637.3 Cho ha ma trn i xng thc bi ma trn trc giao. . . . . . . . . . . . . . . . . . . . . . 657.4 Tr ring - vc t ring ca nh x tuyn tnh . . . . . . . . . . . . . . . . . . . . . . . . . . . 677.5 Cho ha nh x tuyn tnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

2


4/79

HBK TPHCM

8 Dng ton phng 728.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 728.2 a dng ton phng v dng chnh tc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 738.3 Phn loi dng ton phng. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

T.S.ng Vn Vinh Trang 3


5/79

S phc

Ni dung

1) Dng i s ca s phc.

2) Dng lng gic s phc.3) Dng m s phc.

4) Nng s phc ln ly tha.

5) Khai cn s phc.6) nh l c bn i s.

0.1 Dng i s ca s phc

nh ngha 0.1 .

i) Si, c gi ln v o, l mt s sao cho i2 =1.

ii) Cho a, bl 2 s thc, il n v o. Khi z = a + bic gi l s phc.S thca:= Re(z)gi lphn thcca s phcz.S thcb:= I m(z)gi lphn oca s phcz.

iii) Tp tt c cc s phc dngz= 0 + ib,bR \ {0}gi ls thun o.

V d 0.1

i, 2i, 3il nhng s thun o.Tp hp s thc l tp hp con ca tp hp s phc, v:aR : a = a + 0.il mt s phc.nh ngha 0.2 2 s phc bng nhau khi v ch khi phn thc v phn o tng ng bng nhau

a1+ ib1= a2+ ib2

a1= b1,

a2= b2.

V d 0.2 cho z1 = 2 + 3i, z2= m + 3i. Tmmz1= z2.

z1= z2

2 =m,

3 = 3.

Php cng tr 2 s phc(a + bi) + (c + di) = (a + c) + (b + d)i

(a + bi) (c + di) = (a c) + (b d)i

V d 0.3 Tm phn thc v o caz= (3 + 5i) + (2 3i).

z= (3 + 5i) + (2 3i) = (3 + 2) + (5 3)i= 5 + 2i. =Re(z) = 5, Im(z) = 2.

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6/79

0.1. DNG I S CA S PHC HBK TPHCM

Php nhn 2 s phc(a + bi)(c + di) = (ac bd) + (ad + bc)i

V d 0.4 Tm dng i s caz = (2 + 5i)(3 + 2i).

z= (2 + 5i)(3 + 2i) = 2.3 + 2.2i + 5i.3 + 5i.2i= 6 + 4i + 15i + 10i2 = 6 + 10(

1) + 19i=

4 + 19i.

Ghi chKhi cng(tr) 2 s phc, ta cng(tr) phn thc v phn o tng ng.Khi nhn 2 s phc, ta thc hin ging nh nhn 2 biu thc i s vich i2 =1.

S phc lin hpS phc z= a bigi l lin hp ca s phc z = a + bi.

V d 0.5 Tm s phc lin hp caz = (2 + 3i)(4 2i).

Ta c z = (2 + 3i)(4 2i) = 2.4 2.2i + 3i.4 3i.2i= 8 4i + 12i + 6 = 14 + 8i= z= 14 8i.Tnh chtcho 2 s phc z , w

1) z+ zR.2) z.zR.3) z= zzR.4) z+ w= z + w.

5) z.w = z.w.

6) z= z.

7) zn =zn, nN.

Chia 2 s phcz1z2

=a1+ ib1a2+ ib2

= (a1+ ib1)(a2 ib2)(a2+ ib2)(a2 ib2) =

a1a2+ b1b2a22+ b

22

+ ib1a2 a2b1

a22+ b22

.

Ta nhn lin c t v mu cho lin hp mu.

V d 0.6 Thc hin php tonz= 3 + 2i5 i

Nhn c t v mu cho 5 + i, ta c

z=

(3 + 2i)(5 + i)

(5 i)(5 + i) =15 + 3i + 10i

2

25 + 1 =

13 + 13i

26 =

1

2+

1

2 i.

Ch : so snh vi s phcTrong trng s phc Ckhng c khi nim so snh. Biu thcz1< z2hay z1z2u khng c ngha trong trng s phc.



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0.2. DNG LNG GIC CA S PHC HBK TPHCM

0.2 Dng lng gic ca s phc

M uns phc z = a +bil mt s thc khng m c nhngha

mod(z) =|z|=

a2 + b2

Argumentca s phc z l gc v c k hiu l

arg(z) =

Gc c gii hn trong khong (0, 2)hoc (, ).

V d 0.7 Tm m un ca s phcz= 3 4i.

a= 3, b=4 = |z|= 32 + (4)2 = 5.Ch

Nu xem s phc z = a+bi l mt im (a, b)trong mtphng phc th

|z|=

a2 + b2 =

(a 0)2 + (b 0)2

lkhong ccht gc ta O(0, 0)n z .

Cho z = a + bi,w= c + dith

|z w|=|(a c) + (b d)i|=

(a c)2 + (b d)2

lkhong cch gia 2 im zv w.

V d 0.8

Tp hp cc s phc z tha|z (2 3i)|= 5l ng trn tm (2, 3)bn knh bng 5.

Cng thc tm argument

cos =a

r =

aa2 + b2

,

sin = b

r =

ba2 + b2

hoc tan = b

a.



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V d 0.9 Tm argument s phcz=

3 + i.

a=

3, b= 1. Ta tm gc tha

cos = a

r =

3

32

+ 12=

3

2 ,

cos = b

r =

1

32 + 12=

1

2.

= = 3

.

Dng lng gic s phc

z = a + bi=

a2 + b2

aa2 + b2

ba2 + b2

i

= z= r(cos + i sin ) gi l dng lng gic.

V d 0.10 Tm dng lng gic s phcz =1 + i3.

a=1, b= 3.M un:r=|z|= 1 + 3 = 2. Argumentcos =

a

r =1

2 ,

sin = b

r =

3

2

= = 23

.

Dng lng gic z = 2(cos2

3 + i sin

2

3 ).

S bng nhau ca 2 s phc dng lng gic

z1= z2

r1= r2,

1= 2+ k2.

Php nhn dng lng gic

z1z2= r1r2(cos(1+ 2) + i sin(1+ 2)).

M un nhn vi nhau, argument cng li.

V d 0.11 Tm dng lng gic s phcz = (1 + i)(1 i3).

z= (1 + i)(1 i3) = 2(cos4

+ i sin

4).2(cos

3

+ i sin

3 ) = 2

2(cos

12

+ i sin

12).

Php chia dng lng gic

z1z2

=r1(cos 1+ i sin 1)

r2(cos 2+ i sin 2) =

r1r2

(cos(1 2) + i sin(1 2)) , r2= 0.

M un chia cho nhau, argument tr ra.

V d 0.12 Tm dng lng gic s phcz = 2 i

12

3 + i .

z=2 i12

3 + i

= 4(cos

3 + i sin

3 )

2(cos 56

+ i sin 56

)= 2

cos(

3

56

) + i sin(

3 5

6 )

= 2

cos

76

+ i sin7

6

.

nh l Euler(1707-1783)

ei = cos + i sin .

Dng m ca s phc z = r.ei.



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V d 0.13 Tm dng m ca s phcz =3 + i.

Dng lng gic z = 2

cos5

6 + i sin

5

6

. Dng Mz = 2ei

56.

V d 0.14 Biu din s phc sau trn mt phng phcz = ea+3i, aR.

Ta c z = ea(cos 3 + i sin3).= 3khng i nn tp hp l na ng thng nm trong gc phn t th 2.

Php nng ly tha.z = a + bi, z2 = (a + bi)2 =a2 + (bi)2 + 2abi= (a2 b2) + 2abi,z3 = (a + bi)3 =a3 + 3a2bi + 3a(bi)2 + (bi)3 = (a3 3ab2) + ( 3a2b b3)izn =C0na

n + C1nan1bi + C2na

n2(bi)2 + + Cnn (bi)n :=A + Bi.

V d 0.15 Cho s phcz= 2 + i. Tnhz5.

z5

= (2 + i)5

=C052

5

+ C152

4

i + C252

3

i2

+ C352

2

i3

+ C452.i

4

+ C55 i

5

= 32 + 5.16.i + 10.8(1) + 10.4.(i) + 5.2.1 + i =38 + 41i.

Ly tha bc nca i.Ta phn tch n = 4p + r: rl phn d trong php chia n cho 4.

in =ir

V d 0.16 Tnhz = i2013.

Ta c 2013 = 503.4 + 1 =

z = i2013 =i1 =i.

V d 0.17 Cho s phcz= 1 + i. Tmz3 vz100.

a) z3 = (1 + i)3 = 1 + 3i + 3i2 + i3 = 1 + 3i 3 i=2 + 2i.b) Ta dng nh thc newton nh trn s rt di.

Cng thc De MoivreDng lng gicz = r(cos +i sin ) = zn =rn(cos n + i sin n)

Dng lng mz = rei = zn =rnein

M un mn ln, argument tngnln.

V d 0.18 S dng cng thc De Moivre, tnh

a) (1 + i)25. b) (1 + i3)200. c) (

3 i)17(

12 + 2i)20.

a) z= 1 + i=

2(cos

4+ i sin

4) =z25 = 225(cos25

4 + i sin

25

4) = 12

2(cos

4+ i sin

4).

b) Tng t.

c) Tng t.cn bc nca s phcCn bc n ca s phc z l s phc w tha wn =z, nN.



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Cng thc cn bc n.Cho dng lng gic z = r(cos + i sin ). Cng thc

n

z= n

r

cos

+ k2

n + i sin

+ k2

n

; k= 0, 1, . . . , (n 1)

Cn bc n ca z (z= 0)c ng n gi tr phn bit.V d 0.19 Tm cn bcnca cc s phc sau:

a) 3

8.

b) 4

3 + i.

c) 8

16i

1 + i.

d) 6

1 + i3 i .

e)

5 + 12i.

f)

1 + 2i.

Bi lm

a) 8 = 8(cos 0 + i sin 0) = 38 = 2

cos0 + k2

3 + i sin

0 + k2

3

; k= 0, 1, 2.

b) 4

3 + i= 4

2

cos

6+ i sin

6

=

2

cos

6

+ k2

4 + i sin

6

+ k2

4

; k= 0, 1, 2, 3.

c) Tng t

d) Tng t

e) Argument ca 5 + 12ikhng phi cung c bit. Ta s dng dng i s tnh

5 + 12inh sau

5 + 12i= a+bi5+12i= (a+bi)2 5+12i= a2b2+2abi

a2 b2 = 5,2ab= 12

a=3,b=2.

Vy:

5 + 12i=(3 + 2i)

nh l c bn i sMi a thc bc n c ngn nghim k c bi.

H qu:Cho P(z)l a thc h s thc.

p(a + bi) = 0 =

p(a

bi) = 0.

V d 0.20 Tm tt c cc nghim ca a thcP(z) =z4 4z3 + 14z2 36z+ 45, bit 1 nghim l2 + i.

Theo h qu: P(2 + i) = 0 =P(2 i) = 0.Do P(z)chia ht cho (z (2 + i))(z (2 i)) =z2 4z+ 5v c thng l z2 + 9.Ta vit P(z) = (z2 4z+ 5)(z2 + 9)c 4 nghim l 2 + i, 2 i, 3i, 3i.V d 0.21 Gii phng trnhz9 + i= 0.

z=

9

i= 9

cos

2 + i sin

2 = cos

2

+ k2

9 + i sin

2

+ k2

9 , k= 0, 1, 2, . . . , 8.



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V d 0.22 Gii phng trnh

a) z5 + 1 i. b) z2 + z+ 1 = 0. c) z4 + z2 + 2 = 0. d) z2 + 2z+ 1 i= 0.

Bi lm

a) z= 51 + i= ...tng t nh trnb) =b2 4ac= 12 4.1.1 =3 = (i3)2 = =i3.

Nghimz1=b + 1

2a =

1 + i32

, z2=b + 2

2a =

1 i32

.

c) t w = t2

d) Lp v tnh

ri tnh nghim theo cng thc.

Bi tp

Cu 1) Rt gn biu thc

(a) (2 i)5 (b) (2 3i)5

i5(1 + i) (c)

(2 + 2i)9

(i

3 1)7 (d) (i

12 2)14

(1 i)19

Cu 2) Tnh

(a) 6

64

(b)

5 + 12i(c) 6

16i

(i 3)2

Cu 3) Gii phng trnh:

(a) z2 2z+ 5 = 0 (b) z2 + z+ 1 i= 0 (c) z4 + z2 + 4 28i= 0

Cu 4) Tnh 10

zbit (

3 + 2i)z+2 + 6i

1 + i = 3iz+ (3 + i)(2 i)

Cu 5) Gii phng trnh z4 4z3 + 17z2 16z+ 52 = 0bit phng trnh c mt nghim z1= 2 + 3iCu 6) a v dng lng gic

(a) z = sin + 2i sin2

2

(b) w= cos + i(1 + sin )



12/79

Chng 1Ma trn

Ni dung

nh ngha v v d.

Cc php bin i s cp. Cc php ton i vi ma trn.

Hng ca ma trn.

Ma trn nghch o.

1.1 Cc khi nim c bn

nh ngha 1.1 (Ma trn) .Ma trn cm nl mt bng s (thc hoc phc) hnh ch nht c mhng vnct.

A=

a11 . . . a1j . . . a1n. . . . . . . . . . . . . . .ai1 . . . aij . . . ain. . . . . . . . . . . . . . .

am1 . . . amj . . . amn

V d 1.1

A=

3 4 12 0 5

23

, B =

1 + i 23 i 4i

Al ma trn c 2 3c 2 hng v 3 ct. Cc phn t ca ma trn A:a11= 3, a12= 4, a13= 1, a21 = 2, a22 = 0, a32= 5.Bl ma trn c 2 2c cc phn t trong phc.

Ghi ch

Ma trnA c m nthng c k hiu bi A = (aij )mn. Tp tt c cc ma trn c mn trn trng s Kc k hiu Mmn(K).

Ma trn khng.

Ma trn khng c tt c cc phn t bng 0, k hiu l 0

023=

0 0 00 0 0

.

C v s ma trn0 ty theo c.

11


13/79

1.1. CC KHI NIM C BN HBK TPHCM

Phn t c sca mt hng l phn t khc 0u tinca hng k t bn tri sang.Hng ton s 0 th khng c phn t c s.

Ma trn bc thang

1. Hng ton s 0 (nu c) th nm di.2. Phn t c s hng di nm bn phi phn t c s hng trn.

V d 1.2

A=

2 1 0 10 0

1 0

0

-1 0 2

0 0 0 0

khng phi bc thang. B =

-2 1 0 10 0 0

2

0 0 0

-3

khng phi bc thang.

C=

2 1 0 0 20 0

3 2 0

0 0 0 0

-3

0 0 0 0 0

l ma trn bc thang. D=

1 2 0 10 0

-1 0

0 0 0

-4

l ma trn bc thang.

Ma trn chuyn vChuyn v caA = (aij )mnl ma trn AT = (aji )nmthuc tA bng cch chuyn hng thnh ct.

V d 1.3 A= 1 2 32 0 3 A

T =

1 22 0

3 3

Ma trn vungc s hng bng s ct.

Tp tt c cc ma trn vung trn trng s Kc k hiu l Mn[K].

ng cho chnhca ma trn vung A i qua cc phn t

a11, a22, . . . , ann

V d 1.4

Ma trn vung cp 4

1 2 3 42 1 2 00 2 -3 2

1 1 2 0

c cc phn t trn ng cho chnh l 1, 1, 3, 0.

Ma trn tam gic

i) Ma trn vung A = (aij )ngi ltam gic trnnu aij = 0, i > jCc phn t pha di ng cho chnh bng 0.

ii) Ma trn vung A = (aij )ngi ltam gic dinu aij = 0, i < jCc phn t pha trn ng cho chnh bng 0.



14/79

1.2. CC PHP BIN I S CP HBK TPHCM

Ma trn choc cc phn t nm ngoi ng cho chnh bng 0.Hay n va tam gic trn, va tam gic di.Ma trn vung, khng cng l ma trn cho.

Ma trn n vl ma trn cho vi cc phn t trn ng cho bng 1.

Ma trn i xngthaAT =A

Ma trn phn i xngthaAT =A

V d 1.5

Ma trn tam gic trn A =

1 2 30 2 0

0 0 2

. Ma trn tam gic di A =

1 0 03 0 0

3 2 2

.

Ma trn cho D = 1 0 00 0 0

0 0 3 .Ma trn n v cp 3 l I=

1 0 00 1 0

0 0 1.

Ma trn i xng A =

0 1 2

1 2 -32 -3 4

. Ma trn phn i xng A =

0 1 21 0 3

2 3 0

.

1.2 Cc php bin i s cp

Cc php bin i s cp theo hng

1) Nhn mt hng vi 1 s khc 0:hihi; = 0.2) Cng vo mt hng mt hng khc c nhn vi 1 s ty :

hihi+ hj , .3) i ch 2 hng: hihj .

Tng t ta c 3 php bin i theo ct.Cc php bin i s cp l cc php bin i c bn nht i vi ma trn.

nh lMi ma trn u c th a v dng bcbng cc php bin i s cp.

Khi dng php bin i s cp vi ma trn, ta thu c nhiu ma trn bc thang khc nhau.

V d 1.6 Dng php bin i s cp a ma trn sau v dng bc thangA=

1 1 1 2 12 3 1 4 53 2 3 7 4

1 1 2 3 1

A=

1 1 1 2 12 3 1 4 53 2 3 7 4

1 1 2 3 1

h2 h2 2h1h3 h2 3h1

h4h4+h1

1 1 1 2 10

1 1 0 3

0 1 0 1 10 2 1 1 2

h3h3+h2

h4h42h2

1 1 1 2 10

1 1 0 3

0 0

1 1 4

0 0 1 1 4

h4h4+h3

1 1 1 2 10

1 1 0 3

0 0

1 1 4

0 0 0 0 0

=r(A) = 3.



15/79

1.3. CC PHP TON MA TRN HBK TPHCM

1.3 Cc php ton ma trn

Hai ma trn bng nhaunu chng cng c v cc phnt tng ng bng nhau: aij =bij , i, j.

Cho 2 ma trn A, Bcng cv s.

Tng A + B: cng cc phn t tng ng.Nhn .A: nhn vo tt c cc phn t ca A.

V d 1.7 a)

1 2 12 1 0

+

3 2 11 0 3

=

4 0 03 1 3

.

b) 2.

1 2 12 1 0

=

2 4 24 2 0

.

c) 2.

1 2 12 1 0

3.

3 2 11 0 3

=

7 10 51 2 9

.

Tnh cht

i. A + B = B + A.

ii. (A + B) + C=A + (B + C).

iii. A + 0 =A.

iv. (A + B) =A + B.

v. (A) = ()A.

vi. ( + )A= A + A.

Php nhn hai ma trnCho A = (aij )mp, B = (bij )pm.Tch A.B = C= (cij )mn : cij =ai1b1j + ai2b2j + + aipbpj .

AB =

. . . . . . . . . . . . . . . . .ai1 ai2 . . . aip

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

.

. . . b1j . . .

. . . b2j . . .. . . . .. . . bpj . . .

=

. . . . .. . . cij . . .

. . . . .

iu kin php nhn AB :s ct caAbng s hng caB.cijl tch v hng hngi ca A v ct j ca B .

V d 1.8 Cho A=

2 1 44 1 0

; B =

1 2 23 0 1

2 4 3

. TnhAB.

c11=

2 1 413

2

= 2.1 + (1).3 + 4.2 = 7: tch v hng hng 1 ca A v ct 1 ca B .

Tng t, ta tnh c AB =

7 12 157 8 9

.

Tnh cht

i. A(BC) = (AB)C.

ii. A(B+ C) =AB + AC.iii. (B+ C)A= BA + CA.

iv. ImA= AIm = A.

v. (AB) = (A)B = A(B).

Ch :Nhn chungAB=BA; AB = AC =B = C, AB = 0 =A = 0 B = 0.



16/79

1.4. HNG CA MA TRN HBK TPHCM

Nng ly tha:Quy c:A0 =I An =A.A...A.A(nn ma trn A).

V d 1.9 Cho A=

2 13 4

vf(x) = 2x2 4x + 3.Tnhf(A).

Ta c f(A) = 2A2 4A + 3I.

f(A) = 2

2 13 4

2 4

2 13 4

+ 3

1 00 1

= 2

1 6

18 13

8 412 16

+

3 00 3

=

3 824 13

V d 1.10 TnhA200, vi

a) A=

1 30 1

. b) A=

2 30 2

. c) A=

1 11 1

.

Bi gii

a) A2 =

1 30 1

.

1 30 1

=

1 60 1

, A3 =

1 30 1

.

1 60 1

=

1 90 1

=A200 =

1 200.30 1

.

b) A= 2

1 32

0 1

=A200 = 2200

1 200.3

2

0 1

= 2200

0 3000 1

.

c) A2 =

1 11 1

.

1 11 1

=

2 22 2

= 2

1 11 1

= 2A=A200 = 2199.A=

2199 2199

2199 2199

.

Tm li

1 a0 1

n=1 na

0 1

,1 1

1 1n

= 2

n

11 1

1 1

.

1.4 Hng ca ma trn

Hng ma trnAl s hng khc0ca ma trn bc thangca A, k hiu l: r(A).

V d 1.11 Tm hng ca ma trnA=

1 2 1 12 4 2 23 6 3 4

.

A=

1 2 1 12 4 2 2

3 6 3 4

h22h1

h33h1

1 2 1 10 0 0 0

0 0 0 1

h2h3

1 2 1 10 0 0 1

0 0 0 0

=r(A) = 2.

Tnh cht

i) r(A) = 0 =A = 0.ii) A= (aij )mn =r(A)min{m, n}.

iii) NuA bin i s cpB =r(A) =r(B).



17/79

1.5. MA TRN NGHCH O HBK TPHCM

1.5 Ma trn nghch o

Ma trn nghch oMa trn vungAgi l kh nghch nu tn ti ma trn Bsao cho

AB = I=BA.

Khi ,B gi l nghch o ca A, k hiu l A1.

V d 1.12

a) Nghch o caA =

1 22 3

l3 2

2 1

. V

1 22 3

3 22 1

=

1 00 1

=

3 22 1

1 22 3

.

b) Cho A =

2 15 3

. Ta tm ma trn nghch o ca A c dng B =

a bc d

.

Ta c AB = I

2 15 3

a bc d

=

1 00 1

2a + c 2b + d5a + c 5b + d

=

1 00 1

2a + c= 1

2b + d= 0

5a + c= 0

5b + d= 1

a= 3

b=1c=5d= 2

=A1 =B =

3 15 2

.

c) Hy th tm ma trn nghch o caA =

1 22 4

.

Ch :Khng phi mt vung no cng c nghch o. C rt nhiu mt vung khng c nghch o.

S tn ti ma trn kh nghch

Cho ma trn vung A. Cc mnh sau tng ngi) Akh nghch (tn ti A1).

ii) r(A) =n:ma trn khng suy bin

iii) AX= 0X= 0.

iv) A Bsc theo hngI.

Ma trn s cp: Ma trn thu c tIbng ng 1 php

bin i s cp gi l ma trn s cp.V d 1.13 .

I=

1 0 00 1 0

0 0 1

h33h3E1=

1 0 00 1 0

0 0 3

, I=

1 0 00 1 0

0 0 1

h2h2+2h1E2 =

1 0 02 1 0

0 0 1

A=

1 2 34 5 6

7 8 9

h33h3

1 2 34 5 6

21 24 27

=

1 0 00 1 0

0 0 3

1 2 34 5 6

7 8 9

=E1.A.

A= 1 2 34 5 67 8 9

h2h2+2h1 1 2 36 9 127 8 9

= 1 0 02 1 00 0 1

1 2 34 5 67 8 9

=E2.A.Tng t:

I=

1 0 00 1 0

0 0 1

c1c3E3=

0 0 10 1 0

1 0 0

=A =

1 2 34 5 6

7 8 9

h3h1

3 2 16 5 4

9 8 7

=A.E3.



18/79

1.5. MA TRN NGHCH O HBK TPHCM

Mi php bin i s cp tng ng vi php nhn ma trn s cp tng ng.Bsc theohng =nhnbn tri. Bsc theoct =nhnbn phi.

Cch tm ma trn nghch o

[A|I] Bsc theo hng[I|A1]

V d 1.14 Tm ma trn nghch o A=

1 1 11 2 2

1 2 3

.

Bi gii

[A|I] =

1 1 1 1 0 01 2 2 0 1 01 2 3 0 0 1

h2h1

h3h1

1 1 1 1 0 00

1 1 1 1 0

0 1 2 1 0 1

h3h2

h1h2

1 0 0 2 1 00 1 1 1 1 0

0 0

1 0 1 1

h2

h3

1 0 0 2 1 00 1 0 1 2 10 0 1 0 1 1

=A

1

= 2 1 01 2 1

0 1 1

.

Tnh cht ma trn nghch oCho hai ma trn A, Bkh nghch. Ta c

i) (A1)1 =A ii) (AB)1 =B1A1 iii) (AT)1 = (A1)T.

Bi tp

Bi1. Cho A =

1 2 11 1 2

, B =1 20 2

1 1. Tnh3A 2BT

Bi2. Cho A =

1 2 11 1 2

, B =

1 20 2

1 1

,C=

2 1 01 1 1

0 2 1

. Tnh2AC (CB)T

Bi3. Cho A =

1 22 3

vf(x) =x2 4x 1. Tnh f(A)vA2013.

Bi4. Cho A = 2 13 1 vB =

23 . Tm ma trnXtha AX=B .

p s X=

1 15 12

.

Bi5. Tm hng ca ma trn

(a) A=

1 2 12 2 1

1 8 2

.

(b) A=

1 2 1 22 3 1 13 4 3 22 3 1 3

(c) A=

1 1 2 1 12 1 3 4 23 1 4 7 35 3 8 9 5

(d)

1 1 12 3 1

3 5 m

.

(e) A=

m 1 11 m 1

1 1 m

.

(f)

1 m 1 22 1 m 51 10 6 m

.

Bi6. Tm ma trn nghch o (nu c) caA =

1 1 12 3 1

3 4 1

, p n

1 5 41 4 3

1 1 1

.



19/79

Chng 2nh thc

Ni dung

nh ngha nh thc v v d.

Tnh cht nh thc. Dng nh thc tm ma trn nghch o.

2.1 nh ngha nh thc v v d

nh thc ma trn vung A = (aij )n lmt s, c khiu bi

det(A) =|aij |n =|A|.

B i sca phn taijl

Aij = (1)i+jnh thc thu c tAb i hng i, ct j

n1

nh ngha nh thcbng qui np.

i) k= 1 :A = [a11] = |

A

|= a11.

ii) k= 2 :A =

a11 a12a21 a22

= |A|= a11A11+ a12A12 = a11a22 a12a21.

...

iii) k= n : A =

a11 a12 . . . a1n. . . . . . . . .

n

= |A|= a11A11+ a12A12+ + a1nA1n.

V d 2.1 Tnh nh thc ca

1 2 32 3 03 2 4

.

Bi giidet(A) =a11A11+ a12A12+ a13A13= 1A11+ 2A12 3A13.

18


20/79

2.2. TNH CHT NH THC HBK TPHCM

A11= (1)1+13 02 4

= 12(tA, b hng 1 v ct 1).Tng t:det(A) = 1(1)1+1

3 03 4+ 2(1)1+2

2 03 4 3(1)1+3

2 33 2 = 12 16 + 15 = 11.

2.2 Tnh cht nh thcC th tnh nh thc bng cch khai trin theo mthng hoc 1 ct bt k

|A|=. . . . . . . . .ak1 ak2 . . . akn. . . . . . . . .

=ak1Ak1+ak2Ak2+ +aknAkn.

V d 2.2 Tnh nh thc

a)

1 2 12 1 30 0 3

. b)

2 3 3 23 0 1 4

2 0 3 24 0 1 5

a) Khai trin theo hng 3:

1 2 12 1 30 0 3

=3(1)3+31 22 1

=3(3) = 9.b) Khai trin theo ct 2

I=

2 3 3 23 0 1 42 0 3 24 0 1 5

=3(1)1+2

3 1 4

2 3 24 1 5

khai trin theo hng 1

= 3

3(1)1+1

3 21 5+ 1(1)1+2

2 24 5+ 4(1)1+3

2 34 1

= 3(51 + 18 40) = 87.

nh thc cama trn tam gicbng tch cc phn t nmtrn ng cho chnh.

V d 2.3 .

1 2 2 30 4 2 00 0 3 20 0 0 5

= 1.4.(3).5 =60.

Dng bin i s cp tnh nh thc

1. Nu A hihjB th|B|= |A|.

2. Nu A hi+hjB th|B|=|A|.

3. Nu A hihjB th|B|=|A|.



21/79

2.2. TNH CHT NH THC HBK TPHCM

Nguyn tc tnh nh thc s dng bin i s cp

1. Chn 1 hng (hoc 1 ct ty ).

2. Chn 1 phn t khc 0 ca hng (ct) . Dng bini s cp, kh tt c cc phn t khc.

3. Khai trin theo hng (hay ct) chn.

V d 2.4 .

a) I=

1 1 2 12 3 5 03 2 6 2

2 1 3 1

h2 2h1h3 3h1======

h4+2h1

1 1 2 10 1 1 20 1 0 10 3 7 1

khai trin======theo ct 1

1.(1)1+1

1 1 21 0 13 7 1

h3

3h1======

1 1 2

1 0 14 0 15

= 1.(1)1+2 1 14 15 =1(15 + 4) =19.

b)

3 2 1 12 3 2 0

3 1 4 24 1 3 1

h3+2h1======h4h1

3 2 1 12 3 2 03 5 2 01 1 4 0

khai trin

======theo ct 4

12 3 23 5 21 1 4

=2 3 25 8 05 5 0

= 25 85 5

=30.

Tnh cht nh thc: Cho AMn.

i) det(AT) = det(A).

ii)|A|= n|A|.

iii) det(AB) = det(A). det(B).

iv)|Am|=|A|m.

v) Ac 1 hng (hoc ct) bng 0 th|A|= 0.vi) Ac 2 hng (hoc ct) t l th|A|= 0.

Ch :nhn chung det(A + B)= det(A) + det(B).

V d 2.5 Cho A, BM3 tha|A|= 2, |B|= 3.

Ta c|2A3|= 23.|A|3 = 8.23 = 64. |3ABT|= 33|A||B|= 27.2.3 = 162.

iu kin kh nghchAkh nghch khi v ch khi|A| = 0.

V d 2.6 TmmA.B kh nghch. BitA=

1 2 10 1 2

0 1 3

, B =

2 1 30 1 1

m 2 1

.

Bi lmABkh nghch khi v ch khi det(AB)= 0

det(A). det(B)

= 0

1.(

4m

1)

= 0

m

=

1

4.



22/79

2.3. TM MA TRN NGHCH O BNG PHNG PHP NH THC. HBK TPHCM

2.3 Tm ma trn nghch o bng phng php nh thc.

nh ngha 2.1 (Ma trn ph hp) .Ma trn ph hpca ma trn vungAMn c nh ngha l

PA=

A11 A12 . . . A1n

A21 A22 . . . A2n. . . . . . . . . .An1 An2 . . . Ann

T

.

Cng thc tnh ma trn nghch o A1 = 1|A| .PA

V d 2.7 Tm ma trn nghch o A=

1 1 12 3 1

3 4 0

.

Bi lmdet(A) =2= 0 = A kh nghch.A11= (1)1+1

3 14 0

=4, A12= (1)1+2

2 13 0

= 3, A13= (1)1+3 =

2 33 4

=1.

Tng t:A21= 4, A22 =3, A23=1, A31 =2, A32= 1, A33= 1.

Ma trn nghch o A1 = 1

|A|PA= 1

2

4 4 23 3 1

1 1 1

(nh ly chuyn v).

Tnh cht

i)|A1

|= 1

|A|ii) PA=|A|n1.

iii) r(PA) =

n, nur(A) =n1, nur(A) =n 10, nur(A)< n 1

.

V d 2.8 Cho AM3 bit|A|=2. Tnhdet(2P2A).

Bi lmTa c:det(2P2A) = 23.|PA|2 = 8.(|A|31)2 = 8.(2)4 = 128.

V d 2.9 Cho A=

1 2 12 3 11 1 m

. Tmmr(PA) = 1.

Bi lm

A=

1 2 12 3 1

1 1 m

bdsc

1 2 10 1 3

0 0 m + 2

. r(PA) = 1r(A) = 3 1 = 2m =2

Bi tp

1. Tnh nh thc

(a)

2 1 1 33 2 1 24 1 0 1

3 3 2 2

. S: 59. (b)

4 1 1 03 2 4 1

2 1 3 15 1 2 3

. S: -161.



23/79

2.3. TM MA TRN NGHCH O BNG PHNG PHP NH THC. HBK TPHCM

(c)

1 0 1 + i0 1 i

1 i 2 + i 1

. S:2i.

(d)

1 2 2 2 22 1 2 2 2

2 2 1 2 22 2 2 1 22 2 2 2 1

. S: 9.

(e)

1 x x2 x3

1 a a2 a3

1 b b2 b3

1 c c2 c3

.

S: (c x)(b x)(a x)(c a)(c b)(b a)

(f)

1 1 1 . . . 1

1 1 x 1 . . . 11 1 2 x . . . 1. . . . . . . . . . .1 1 1 . . . n x

n+1

.

S:x(1 x)(2 x) . . . (n 1 x).

(g)

1 2 3 . . . n1 0 3 . . . n1 2 0 . . . n. . . . . . . . .

1

2

3 . . . 0

. S: n!.

(h)

3 2 2 . . . 22 3 2 . . . 22 2 3 . . . 2. . . . . .2 2 2 . . . 3

. S: 2n + 1.

(i)2 x 2 3x 2 3 40 0 7 60 0 5 3

. S:9(x2 + 4).

(j) Dn =

7 5 0 . . . 02 7 5 . . . 00 2 7 . . . 0. . . . . .0 0 0 . . . 7

.

HD: kt theo h1, suy ra Dn = 7Dn1 10Dn2.

(k) Dn =

4 4 0 . . . 0

1 4 4 . . . 00 1 4 . . . 0. . . . . .0 0 0 . . . 4

.

HD: kt theo h1, suy ra Dn = 4Dn1 4Dn2.

(l) Dn =

2 2 0 . . . 01 2 2 . . . 00 1 2 . . . 0. . . . . .0 0 0 . . . 2

.

HD: kt theo h1, suy ra Dn = 2Dn1

2Dn

2.

2. Tm ma trn nghch o

(a) A=

1 2 12 3 1

3 5 2

S:A1 =1

2

11 1 57 1 3

1 1 1

.

(b) A=

1 0 0 02 1 0 05 4 1 01 2 3 2

S: A1 =

1 0 0 02 1 0 0

13 4 1 017 5 3

212

.

3. Tm m ma trn kh nghch

(a) A=

1 1 2 12 1 5 35 0 7 m

1 2 3 3

. S: m= 9. (b) A=

1 2 12 3 m

3 2 1

1 1 12 3 2

5 7 5

. S: m.

4. Cho A =

1 1 12 3 1

3 3 5

. Tnh|A1|, |(5A)1|, |2PA|. S: 1

2,

1

250, 32.

5. Cho A, B

M3[R] :|

A|= 2,

|B

|=

3. Tnh

|(4AB)

1

|,|P

AB|. S:

1

384, 36.



24/79

Chng 3H phng trnh

Ni dung

H phng trnh tng qut.

H Cramer. H phng trnh tuyn tnh thun nht.

nh ngha 3.1 (h phng trnh tuyn tnh) H phng trnh tuyn tnh gmm phng trnh, nnc dng

a11x1 + a12x2 + . . . + a1nxn = b1a21x1 + a22x2 + . . . + a2nxn = b2. . . . . . . . . . . . . . . . .am1x1 + am2x2 + . . . + amnxn = bm

a11, a12, . . . , amn c gi lh sca h phng trnh.b1, b2, . . . , bm c gi lh s t doca h phng trnh.

Ta k hiu

A =

a11 a12 . . . a1na21 a22 . . . a2n. . . . . . . . . .am1 am2 . . . amn

, X =

x1x2. . .xn

, b =

b1b2

bm

, (A|b) =

a11 a12 . . . a1n b1a21 a22 . . . a2n b2. . . . . . . . . . . . . .am1 am2 . . . amn bm

mn

.

H phng trnh c vit li

A.X=bhoc vit gn (A|b).

Ch thch

Mt h phng trnh tuyn tnh c th:1)v nghim 2)c nghim duy nht 3) v s nghim.

Hai h phng trnh gi l tng ng nu chng cng tp nghim. gii h phng trnh, ta dng php bin i tng ng a v

h n gin.

23


25/79

HBK TPHCM

Php bin i tng ngMt php bin i c gi l tng ng nu n bin mt hphng trnh bt k thnh mt h phng trnh tng ng.Ta c 3 php bin i tng ng thng gp:

i) Nhn 2 v ca mt phng trnh vi 1 s khc 0.

ii) Cng vo mt phng trnh mt phng trnh khc cnhn vi mt s ty .

iii) i ch hai phng trnh.

Ch :

y l 3 php bin i quen thuc ph thng m chng ta bit. Nu ta k hiu h phng trnh dng ma trn m rng (A|b). Cc php bin i s cp i vi ma

trn tng ng vi cc php bin i tng ng i vi h phng trnh.

n c sca h phng trnh dng bc thang

n c s l n tng ng vi ct cha phn t c s.

n t do l n tng ng vi ct khng c phn t c s.

V d 3.1

1 1 1 2 12 2 3 5 63 3 4 1

1

bin i s cp

1 1 1 2 10 0 1 1 4

0 0 0 -6 8

x1, x3, x4 l phn t c s. x2 l phn t t do.

Cc bc gii h phng trnh

Bc 1: a ma trn A = [A|b] v dng bc thang bngbin i s cptheo hng.Kim tra h c nghim hay khng.

Bc 2: Gii h phng trnh t di ln.

V d 3.2 Gii h phng trnh

x1+ x2 x3+ 2x4= 12x1+ 3x2 3x3+ 3x4= 33x1+ 2x2 5x3+ 7x4= 5.

Bi lm

A=

1 1 1 2 12 3 3 3 3

3 2 5 7 5

h22h1

h33h1

1 1 1 2 10 1 1 1 1

0 1 2 1 2

h3+h2

1 1 1 2 10 1 1 1 10 0 -3 0 3

tx4= . pt (3):x3=

1. T pt (2):x2 = 1 + x3 + x4= . T pt(1):x1 = 1

x2 + x3

2x4=

3.

Vy nghim ca h l (x1, x2, x3, x4) = (3,, 1, ), R.



26/79

3.1. H CRAMER HBK TPHCM

nh l Kronecker CapelliNur(A|b)=r(A)th h AX=bv nghim.Nur(A|b) =r(A)th h AX=bc nghim.

i) Nur(A|b) =r(A) =s n th h AX=bc nghim duy nht.ii) Nur(A

|b) =r(A)


27/79

3.2. H THUN NHT HBK TPHCM

3.2 H thun nht

H thun nht

H AX=bgi l thun nht nu tt c cc h s t do

b1= b2= = bm = 0. H thun nht lun cnghim tm thng.

x1= x2 = = xn = 0

H thun nht c nghim duy nht khi v ch khi

r(A) =n =s n.

ChoAl ma trn vung. H thun nhtAX= 0cnghim

khng tm thng(nghim khc 0) khi v ch khi|A| = 0.

V d 3.6 Gii h phng trnh

x1+ x2 x3+ 2x4= 02x1+ 3x2 3x3+ 3x4= 03x1+ 5x2 5x3+ 4x4= 0.

Bi lm 1 1 1 2 02 3 3 3 03 5 5 4 0

1 1 1 2 00 1 1 1 00 2 2 2 0

1 1 1 2 00 1 1 1 00 0 0 0 0

t cc n t do lm tham sx3= , x4= .Pt(2):x2= x3+ x4= + . Pt(1):x1=x2+ x3 2x4=3.Vy nghim ca h l (x1, x2, x3, x4) = (3, + ,,).V d 3.7 Tmm h c nghim khng tm thng

mx1+ x2+ x3+ x4= 0

x1+ mx2+ x3+ x4= 0

x1+ x2+ mx3+ x4= 0x1+ x2+ x3+ mx4= 0.

Bi lmH c nghim khng tm thng khi v ch khi r(A)< n |A|= 0.

|A|=

m 1 1 11 m 1 11 1 m 11 1 1 m

= (m + 3)

1 1 1 11 m 1 11 1 m 11 1 1 m

= (m + 3)

1 1 1 10 m 1 0 00 0 m 1 00 0 0 m 1

= (m + 3)(m 1)3.

Vym =3 m= 1.



28/79

3.2. H THUN NHT HBK TPHCM

V d 3.8 Tmm h c v s nghim

x1+ x2+ 2x3 x4= 0x1+ 3x2+ mx3+ 2x4= 0

mx1 x2+ 3x3 2x4= 0

Bi lmVA l ma trn c 3 4nn r(A)3 < 4 =s n. Vy h lun c v s nghim.Ch :H thun nht c s phng trnh t hn s n th v s nghim.

Bi tp

Bi1) Gii h phng trnh

(a) x1+ 2x2+ x3+ 2x4= 0

2x1+ 4x2+ x3+ 3x4 = 03x1+ 6x2+ x3+ 4x4 = 0

.

S: (2 ,, , )

(b)

1 5 2 11 4 1 6

1 3 3 9

.

S: (18, 5, 4)

(c)

1 5 2 60 4 7 2

0 0 5 0

. S:

(

17

2 ,

1

2, 0)

(d)

1 1 1 00 1 2 5

.

S(5 , 5 + 2, ).

(e) 0 1 1 33 5 9 2

1 2 3 3

.S:(43, 11, 8)

(f)

0 3 6 6 4 53 7 8 5 8 9

3 9 12 9 6 15

S:(24 + 2 3, 7 + 2 2,,, 4)

(g)

1 1 1 1 22 1 3 0 13 4 2

2 5

2 3 1 1 3

. S:(

1

3 2,1

3 +

,, 43

).

Bi2) Tm tt c cc gi tr ca m h sau c nghim

(a)

1 1 1 12 3 1 4

3 4 m m + 1

. S m= 2. (b)

m 1 1 11 m 1 m

1 1 m m2

. Sm=2.

Bi3) Tm m h sau c nghim duy nht

(a)

2 3 1 4 01 1 0 m 2

2 m 1 4 m2

. S:m. (b)

1 1 1 1 12 1 3 1 23 4 2 0 6

2 1 0 m m 1

.mR.



29/79

Chng 4Khng gian vc t

Ni dung

nh ngha v v d

c lp tuyn tnh - ph thuc tuyn tnh Hng ca h vc t

C s v s chiu

Ta vc t

Khng gian con

Tng giao 2 khng gian con

4.1 nh ngha v v d

nh ngha 4.1 (Khng gian vc t) ChoV l tp hp khc rng v 2 php ton: cng 2 vc t v nhnvc t vi mt s tha mn 8 tin sau

i) x + y= y + x

ii) (x + y) + z= x + (y+ z)

iii)0V :x + 0 =xiv)

x

V,

(

x)

V :x + (

x) = 0

v) , K : ( + )x= x + xvi) K :(x + y) =x + y

vii) ()x= (x)

viii) 1.x= x

Khi , ta niV l mt khng gian vc t.

Ch :y l khi nim c m rng t khi nim vc t ph thng.Tp cc vc t trong mt phng (hoc khng gian) c gc O l mt khng gian vc t.

Tnh cht

i) Vc t khng l duy nht. ii) Vc t i (x)ca x l duy nht.

iii) 0.x= 0, xV iv) .0 = 0, K v)x=1.x, inV

V d 4.1

28


30/79

4.2. C LP TUYN TNH - PH THUC TUYN TNH HBK TPHCM

1. TpV1 ={(x1; x2; x3)|xiR; i= 1, 2, 3}vi php ton cng 2 vc t v nhn vc t vi s thc thngthng l mt khng gian vc t trn R. K hiu l R3.Tng t, ta c khng gian R2, R3, R4, . . . , Rn, . . .

2. TpV2={ax2 + bx + c|a,b,cR}vi php ton thng thng i vi a thc l mt khng gian vct. K hiu l P2[x].

Tng t, ta c khng gian P3[x], P4[x], . . . , P n[x], . . .

3. TpV3=

a bc d

|a,b,c,dR

vi php ton thng thng i vi ma trn l mt khng gian vc

t. K hiu l M2[R].Tng t, ta c cc khng gian Mmn[R], Mmn[R]cc ma trn c m ntrong thc v phc.

4. TpV4 ={(x1, x2, x3)|xiR x1+ 2x2 3x3 = 0}vi php ton i vi vc t thng thng l mtkhng gian vc t.Ch :C nhiu cch nh ngha php ton cho cc tp hp trn l mt khng gian vc t, minl tha 8 tin ca khng gian trn.

4.2 c lp tuyn tnh - ph thuc tuyn tnh

nh ngha 4.2 Trong khng gian vc tV, cho tp hp con gmmvc tM ={x1, x2, . . . , xm}

Vc txgi lt hp tuyn tnhcaM nu1, 2, . . . , mK tha

x= 1x1+ 2x2+ + mxm

1, 2, . . . , m khng ng thi bng 0tha

1x1+ 2x2+ + mxm = 0 =M ph thuc tuyn tnh.

Mgi lc lp tuyn tnhnu n khng PTTT. Tc l

1x1+ 2x2+ + mxm = 01= 2= = m = 0.

Ni cch khc:MPTTT nu c mt THTT khng tm thng bng khng.MLTT nu n ch c duy nht mt THTT bng khng l t hp tm thng (k = 0, k).

V d 4.2 TrongR3, cho h vc tM ={(1;1;1), (2;1;3), (1;2;0)}.a) Vc tx= (2; 1;3)c l t hp tuyn tnh caMhay khng?b) MLTT hay PTTT?

Bi lm

a) Xt x = (1; 1; 1) + (2; 1; 3) + (1;2;0)(2; 1; 3) = ( + 2+ ; + + 2; + 3)

+ 2+ = 2

+ + 2=

1

+ 3= 3

, (A

|b) =

1 2 1 21 1 2

1

1 3 0 3 =r(A) = 2< r(A|b) = 3.

H v nghim, tc l khng tn ti , . Vy x khng l THTT ca M.



31/79

4.2. C LP TUYN TNH - PH THUC TUYN TNH HBK TPHCM

b) Xt t hp bng 0(1; 1; 1) + (2; 1; 3) + (1;2;0) = 0( + 2+ ; + + 2; + 3) = 0

+ 2+ = 0

+ + 2= 0

+ 3= 0

, A=

1 2 11 1 2

1 3 0

= |A|= 0

H v s nghim nn tn ti nghim khng tm thng, do MPTTT.Cho tpM ={x1, x2, . . . , xm}v vc tx

1x1+ 2x2+ + mxm = 0AX= 0

H c nghim duy nht X= 0 = MLTT.

H c nghim khc khng = MPTTT.1x1+ 2x2+ + mxm = xAX=b

H c nghim = x l THTT ca M .

H v nghim = xkhng l THTT ca M.

V d 4.3 Trong khng gian vc tV, cho h M={x,y, 2x + 3y, z}.a) Vc t2x + 3y c l THTT cax, y, z hay khng?

b) MLTT hay PTTT?

Bi lm

a) Chn = 2, = 3, = 0 : 2x + 3y= 2.x + 3.y+ 0.z=2x + 3yl THTT ca x, y, z.b) Chn1= 2, 2= 3.3=1, 4 = 0 : 2.x + 3.y 1.(2x + 3y) + 0.z = 0 =MPTTT.V d 4.4 Trong khng gian vc tV, cho{x,y ,z}LTT.Hy chng t M={x + y+ 2z, 2x + 3y+ z, 3x + 4y+ z}LTT.Bi lmXt mt t hp bng khng ca M:

(x + y + 2z) + (2x + 3y + z) + (3x + 4y + z) = 0( + 2+ 3)x + ( + 3+ 4)y + (2 + + 1)z = 0.

Vx,y , zLTT nn

+ 2+ 3= 0

+ 3+ 4= 0

2 + + 1= 0

= 0

= 0

= 0

. Vy MLTT.

V d 4.5 Trong khng gianV, cho{x, y}LTT. Cc tp hp sau LTT hay PTTT?

a) M1={2x, 3y}. b) M2 ={x + y, 2x + 3y}. c) M3 ={x + y, x y, 2x + 3y}.

p n: a) LTT. b) LTT. c) PTTT.

V d 4.6 Trong khng gian V, cho{x, y} LTT v z khng l THTT ca{x, y}. Chng t{x,y ,z}LTT.Bi lmXtx + y + z = 0. Nu= 0th z =

x

y, mu thun vi gi thit, suy ra = 0.

Khi x + y = 0 x,y LTT = = 0. Vy{x,y ,z}LTT.



32/79

4.3. HNG CA H VC T HBK TPHCM

Du hiu LTT-PTTT

Nu h Mcha vc t khng th PTTT. Trong hM, c mt vc t l THTT ca cc vc t cn li th MPTTT.

Thm mt s vc t vo h PTTT, ta thu c 1 h PTTT.

Bt i mt s vc t ca h LTT, ta thu c 1 h LTT.

B c bnCho h vc t gm m vc t M={x1, x2, . . . , xm}.Cho h vc t gm n vc t N={y1, y2, . . . , yn}.Nu mi vc tykca Nl THTT caMvn > mthNPTTT.

V d 4.7 Trong khng gian vc tV, tp N ={2x + y, x + y, 3x 2y}LTT hay PTTT?

Cc vc t ca Nl THTT ca M={x, y}v s vc t ca Nln hn s vc t ca MnnNPTTT.V d 4.8 Trong KGVTV, cho M={x,y ,z}, N={x + y + z, 2x + 3y z, 3x + 4y + z}. Chng minh rnga) NuMLTT thN LTT.

b) NuNLTT thM LTT.

Bi lm

a) Xt t hp bng 0 ca N:(x + y + z) + (2x + 3y z) + (3x + 4y + z) = 0( + 2+ 3)x + ( + 3+ 4)y + ( + )z = 0

M LTT

+ 2+ 3= 0 + 3+ 4= 0

+ = 0

= 0= 0

= 0

.Vy NLTT.

b) Dng phn chng, gi sMPTTT. Khi c 1 vc t l THTT ca cc vc t cn li.Khng mt tnh tng qut, ta gi sz l THTT ca x, y.Ta c cc vc t ca Nl THTT ca Mv cng l THTT ca{x, y}.S vc t ca Nln hn s vc t ca{x, y}. Theo b c bn, NPTTT, mu thun vi gi thit.

4.3 Hng ca h vc t

nh ngha 4.3 Cho h vc tM ={x1, x2, . . . , xm, . . . } V.Ta ni hng caM l k0 nu tn ti k0 vc t LTT caMv mi tp con hnk0 vc t caMlun PTTT.

Hng ca h M l sti icc vct c lp tuyn tnh ca M.

V d 4.9 Trong KGVTV, cho M ={x, y}LTT. Tm hng ca cc h vc t sau:

a) M1={2x, 3y} b) M2 ={x,y, 2x + 3y} c) M3 ={x,y, 2x + 3y, 0}.

Bi lm

a) Kim tra{2x, 3y}LTT. Do r(M1) = 2.b) 2x + 3y= 2.x + 3.y=M2PTTT v{x, y}LTT =r(M2) = 2.



33/79

4.3. HNG CA H VC T HBK TPHCM

c) M3cha vc t 0 nn PTTT. D thy 4 h con gm 3 vc t ca M3u PTTT.C 1 h 2 vc t LTT l{x, y}. Vy r(M3) = 2.

Tnh cht hng ca h vc t

i) Hng ca h vct Mkhng i nu ta nhn mt vct

ca Mvi mt s khc khng.ii) Cng vo mt vct ca h M, mt vct khc c

nhn vi mt s th hng khng thay i.

iii) Thm vo hMvctxl t hp tuyn tnh ca Mthhng khng thay i.

iv) Bt i 1 vc t caMl THTT ca cc vc t khc thhng khng thay i.

V d 4.10 Cho h vc tM={(1;1;1;0), (1;2;1;1), (2;3;2;1), (1;3;1;2)}.Bi lmTa c(2; 3;2; 1) = (1;1; 1;0)+(1; 2;1; 1), (1; 3;1; 2) =(1;1;1;0)+2(1;2;1;1) =r(M) =r{(1;1;1;0), (1;2;1;1)}.Hn na, v{(1;1;1;0), (1;2;1;1)}LTT nn r(M) = 2.

nh l v hngCho A l ma trn c m ntrnK. r(A)bng vi hng ca h vc t hng. r(A)bng vi hng ca h vc t ct.

V d 4.11 Tm hng ca hai h vc t

a) M={(1;2;1), (2; 1;7), (1;3;0), (1;2;1)}vN={(1;2;1;1;), (2; 1;3;2), (1;7;01)}.b) P ={(1;1;1;0), (1; 1;1;1), (2;3;1;1), (3;4;0;2)}.

Bi lm

a) XtA =

1 2 1 12 1 3 21 7 0 1

c h vc t ct l Mv h vc t hng lN. Do r(M) =r(N) =r(A) = 2.

b) Hng ca Pbng hng ca ma trn

B =

1 1 1 01 1 1 12 3 1 13 4 0 2

. Vr(B) = 2nn r(P) = 2.

Tnh chtcho h vc t Mv vc t x

Hng Mbng s vc t th MLTT.

Hng Mb hn s vc t thMPTTT.

r(M, x) =r(M)th x l THTT ca M.

V d 4.12 Xt s LTT ca h vc t sau



34/79

4.4. C S V S CHIU HBK TPHCM

a) M={(1;1;1), (2;1;3), (1;2;0)}.b) N ={x2 + x + 1, 2x2 + 3x + 2, 2x + 1}.

c) P =

1 11 0

,

2 11 1

,

3 40 1

,

1 31 2

.

d) Q={(1;1;0), (1;2;1), (m;0;1)}.Bi lm

a) r(M) =r

1 1 12 1 3

1 2 0

= 2 =MPTTT (v hng b hn s vc t).

b) r(N) =r

1 1 12 3 2

0 2 1

= 3 =NLTT (v hng bng s vc t).

c) r(P) =r

1 1 1 02 1 1 13 4 0 11 3 1 2

= 4 =PLTT.

d) r(Q) =

1 1 01 2 1

m 0 1

1 1 10 1 1

0 m 1

1 1 10 1 1

0 0 m + 1

Num =1r(Q) = 2th Q PTTT.Num=1r(Q) = 3th Q LTT.

4.4 C s v s chiuTp sinhCho M={x1, x2, . . . , xm, . . . } V.Mgi l tp sinh ca Vnu mi vc t x ca V ul THTT ca M. Ta vit

V =< M >=< x1, x2, . . . , xm >

Ta cn ni Msinh ra V hayVc sinh bi M.

V d 4.13 Xt xem cc tp sau c l tp sinh trongR3 hay khng?a) M={(1;1;1), (1;2;1), (2;3;1)}.b) M={(1;1;1), (1;2;3), (3;2;1)}Bi lm

a)x= (x1; x2; x3)R3. Gi sx = (1; 1; 1) + (1; 2; 1) + (2;3;1)

+ + 2= x1

+ 2+ 3= x2

+ + = x3

, |A|=1 1 21 2 31 1 1

=1= 0.

H Cramer nn c nghimxR3. Do x l THTT ca M.VyMl tp sinh ca R3.



35/79


b)x= (x1; x2; x3)R3. Gi sx = (1; 1; 1) + (1; 2; 3) + (3;2;1)

+ + 3= x1

+ 2+ 2= x2

+ 3+ = x3

1 1 3 x11 2 2 x2

1 3 1 x3

1 1 3 x10 1 1 x2 x1

0 0 0 x3+ x1 2x1

Vix3+ x1 2x2= 0th h v nghim, ngha l tn ti x (v d nh(1; 0;0)) khng l THTT ca M.

VyMkhng l tp sinh ca R3

.

V d 4.14 Tp M={x2 + x + 1, 2x2 + 3x + 1, x2 + 2x}c l tp sinh caP2[x]hay khng?

Bi lmp(x) =ax2 + bx + cP2[x] :p(x) =(x2 + x + 1) + (2x2 + 3x + 1) + (x2 + 2x)

+ 2+ = a

+ 3+ 2= b

+ =c

1 2 1 a1 3 2 b

1 1 0 c

1 2 1 a0 1 1 b a

0 0 0 b + c 2a

.

Vib + c 2a= 0th h v nghim. Vy Mkhng l tp sinh ca P2[x].V d 4.15 Cho M=

{x,y ,z

}l tp sinh ca KGVTV. Tp no sau y l tp sinh caV?

a) M1={2x, x + y, z}. b) M2 ={x, x + y, x y}.

Bi lm

a) VMl tp sinh ca V nnvV :v = x + y + zv =

2 .2x + .(x + y) + .z . V v l THTT ca M1nn M1l tp sinh ca V.

b) Nu z l THTT ca x, y. Khi ta chng minh M2l tp sinh ca V. ??

Nuz khng l THTT ca x, y. Khi z ta chng minh c z khng l THTT ca M2. ??Do M2khng l THTT ca V.

C s v s chiu:Cho M={x1, x2, . . . , xm, . . . } V

M sinh ra V + M - LTT = M - l c s

C s c n vc t = S chiu caV l n:dim(V) =n

Vkhng c tp sinh hu hn th Vgi l KGVT v hn chiu.

V d 4.16 Cho M={x,y ,z}l c s caV. Xt xem tp no sau y l tp sinh, c s?a) M1={2x + y+ z, x + 2y+ z, x + y+ z}. b) M2 ={2x, 3y ,z ,x + y+ z}.

Bi lm

a) 1) Chng t M1l tp sinh ca V. 2)Chng t M1LTT. ??Suy ra M1l c s ca V.

b) Chng t M2l tp sinh ca V. ?? .D thy M2PTTT. Do M2l tp sinh nhng khng l c s ca V .

nh l c sCho Vl KGVT hu hn chiu Vc v s c s. S vc t trong mi c s u bng nhau.



36/79


C s chnh tc

i) dim(Rn) =nv c s chnh tc (c s n gin nht) l

E={(1; 0; . . . ; 0), (0;1; . . . ; 0), . . . , (0;0; . . . ; 1)}.

ii) dim(Pn[x]) =n + 1v c s chnh tc lE={xn, xn1, . . . , x , 1}.

iii) dim(Mn[R]) =n2 v c s chnh tc l

E=

1 0 . . . 00 0 . . . 0

. . . . . . .

n

,

0 1 . . . 00 0 . . . 0

. . . . . . .

n

, . . .

Tnh chtCho dim(V) =n

Mi tp con nhiu hn n vc t th PTTT. Mi tp con t hn n vc tkhngsinh ra V . Mi tp con LTT cng nvc t l c s. Mi tp sinh cng n vc t l c s. Mi tp c hng bng n l tp sinh.

V d 4.17 Kim tra tp sinh - c s trongR3.

a) M={(1;1;1), (2;3;1), (3;1;0)}. b) N ={(1;1;1), (2;0;1), (1;1;0), (1; 2;1)}.

Bi lm

a) Mc 3 vc t bng s chiu ca R3.

r(M) =r

1 1 12 3 1

3 1 0

= 3 =Ml c s ca R3.

b) Nc 4 vc t trong khng gian 3 chiu nn PTTT.

r(N) =r

1 1 12 0 11 1 01 2 1

= 3 =Nl tp sinh ca R3.

V d 4.18 Kim tra tp M={x2 + x + 1, 2x2 + x + 1, x2 + 2x + 2}c l c s caP2[x]?

Bi lmMc 3 vc t, bng s chiu ca P2[x].Ml c s khi v ch khi r(M) = 3.

r(M) =r1 1 1

2 1 11 2 2

= 2.=Mkhng l c s caP2[x].



37/79

4.5. TA VC T HBK TPHCM

4.5 Ta vc t

nh ngha 4.4 Cho E={e1, e2, . . . , en}l c ssp th tca K-KGVTV.B s(x1, x2, . . . , xn)gi l ta vc txtrong c sE. K hiu

[x]E=

x1x2. . .xn

x = x = x1e1+ x2e2+ + xnen.V d 4.19 Cho E={x2 + x + 1, x2 + 2x + 1, x2 + x + 2}l c s caP2[x].

a) Tmp(x)bit [p(x)]E=

32

5

. b) Cho q(x) =x2. Tm [q(x)]E.

Bi gii

a) [p(x)]E= 32

5

p(x) = 3(x2 + x + 1) 2(x2 + 2x + 1) + 5(x2 + x + 2)p(x) =5x + 2.

b) Gi s[q(x)]E=

q(x) =(x2+x+1)+(x2+2x+1)+(x2+x+2)x2 = (++)x2+(+2+)x+(++2)

+ + = 1

+ 2+ = 0

+ + 2= 0

= 3

=1=1

. Vy [q(x)]E=

311

.

Tnh cht ta

Cho El c s ca KGVT V :[x]E=

x1x2...

xn

[y]E=

y1y2...

yn

i) x= y

x1= y1

x2= y2

. . .

xn = yn

ii) [x + y]E=

x1+ y1x2+ y2

...xn+ yn

iii) [x]E=

x1x2

...xn

V d 4.20 Cho E={(1;1;1), (1;1;0), (1;0;1)}l c s caR3.

a) Tmx, bit [x]E=

12

1

. b) Cho x= (3; 1;2). Tm [x]E.

Bi lm

a) [x]E= 121x =1(1; 1; 1) + 2(1; 1; 0) + 1(1; 0; 1) = (2; 1; 0)

Ghi ch:

E.[x]E=

1 1 11 1 0

1 0 1

12

1

=

21

0

vit lix = (2; 1;0).



38/79

4.6. MA TRN CHUYN C S HBK TPHCM

b) Gi s[x]E=

x = (1; 1; 1) + (1; 1; 0) + (1;0;1)

+ + = 3

+ = 1

+ =2

[x]E=

452

Ghi ch:

E.[x]E=xT [x]E=E1xT =

45

2

.

Dng my tnh casio cho bi ton ta C s E={e1, e2, . . . , e3} MT ctE=

e1 e2 . . . e3

xT =E.[x]E [x]E=E1xT

ngha ca ta Cho E={e1, e2, . . . , en}l c s ca KGVT V .Mi vc t ca Vu biu din qua Edi dng ta .

Cc php ton ta ging nh cc php ton trong Rn.

=tt c cc khng gian n chiu u coi l Rn.

V d 4.21 Tm ta cap(x) = 3x2 + 4x 1trong c sE={x2 + x + 1, x + 1, 2x + 1}trongP2[x].

Bi lm

Lp ma trn E=1 0 01 1 2

1 1 1

.

Ta [p(x)]E=E1[p(x)] =

1 0 01 1 2

1 1 1

1

.

34

1

=

39

5

.

4.6 Ma trn chuyn c s

nh ngha 4.5 Cho 2 c s ca KGVTV: E={e1, e2, . . . , en}, E ={e1, e2, . . . , en}.

xV :x = x1e1+ x2e2+ + xnen = x1e1+ x2e2+ + xnen. (1)e1= a11e1+ a21e2+ + an1ene2= a12e1+ a22e2+ + an2en. . . . . . .en = a1ne1+ a2ne2+ + annenx= x1(a11e1+ a21e2+ + an1en) + x2(a12e1+ a22e2+ + an2en) + + xn(a12e1+ a22e2+ + an2en).

T (1), ta suy ra

x1x2. . .xn

=

a11 a12 . . . a1na21 a22 . . . a2n. . . . . . . . .an1 an2 . . . ann

x1x2. . .xn

.

Ma trnA=

a11 a12 . . . a1n

a21 a22 . . . a2n. . . . . . . . .an1 an2 . . . ann

gi l ma trn chuyn c s tEsangE.



39/79

4.7. KHNG GIAN CON HBK TPHCM

Ma trn chuyn c stEsangE

P =

[en]E [e

n]E . . . [e

n]E

|| || ||

=E1E(vit dng ct).

C tnh cht

[x]E=P.[x]E

.

Tnh cht

Ma trn chuyn c s Pkh nghch. Pchuyn c s tEsang Eth P1 l ma trn chuyn c

s tEsangE.

Pchuyn c s tEsang Ev Qchuyn c s tEsangEth P Ql ma trn chuyn c s tEsang E.

V d 4.22 TrongR3, cho 2 c sE={(1;1;1), (1;0;1), (1;1;0)}vE={(1;1;2), (1;2;1), (1;1;1)}.a) Tm ma trn chuyn c s tEsangE v ma trn chuyn c s tE sangE.

b) Cho x= (2; 1;3). Tm [x]E v [x]E.

Bi lm

a) E=

1 1 11 0 11 1 0

, E=

1 1 11 2 12 1 1

.

Ma trn chuyn c s tEsang E:P =E1E =1 1 11 0 1

1 1 0

1 1 11 2 12 1 1

= 2 2 10 1 01 0 0

Ma trn chuyn c s tEsang E:Q = E1E= P1 =

0 0 10 1 0

1 2 2

.

b) Ta c [x]E =E1xT =

1 1 11 2 1

2 1 1

21

3

=

11

2

.

[x]E=E1

xT

=1 1 1

1 0 11 1 02

13

= 2

11

.

Cch khc: [x]E=P[x]E =

2 2 10 1 0

1 0 0

11

2

=

21

1

.

4.7 Khng gian con

nh ngha 4.6 Trong KGVTV, nu tp conFvi cc php ton trong V lp thnh mt KGVT th taniF lkhng gian concaV.

nh l khng gian conTp con khc rng Fca KGVTVl mt khng gian con ca Vkhi vch khi hai iu kin sau tha

i)x, yF :x + yF. ii)xF, K :xF.



40/79


V d 4.23 Cho F ={(x1; x2; x3)R3|x1+ 2x2 x3= 0}.

a) Chng t Fl KG con caR3. b) Tm c s v s chiu ca F.

Bi lm

a) Sinh vin t kim tra 2 iu kin trong nh l.b)x= (x1; x2; x3)Fx1+ 2x2 x3= 0x3= x1+ 2x2.

x= (x1; x2; x3) =x = (x1; x2; x1+ 2x2) =x1(1; 0; 1) + x2(0;1;2).Suy ra E={(1;0;1), (0;1;2)}l tp sinh ca F.Kim traELTT. Vy El c s ca Fvdim(F) = 2.

V d 4.24 Cho F ={p(x)P2[x]|p(1) = 0 p(2) = 0}.


a) Sinh vin t kim tra 2 iu kin trong nh l.

b)p(x) =ax2 + bx + cFp(1) = 0 p(2) = 0

a + b + c= 0

4a + 2b + c= 0

a=

b=3c= 2

p(x) =x2 3x + 2= (x2 3x + 2).Suy ra E={x2 3x + 2}l tp sinh ca F.Hin nhin ELTT. Vy El c s ca Fvdim(F) = 1.

V d 4.25 Cho F = AM2[R]|A1 12 2 = 0.


Bi lm

a) Sinh vin t kim tra 2 iu kin trong nh l.

b)A=

a bc d

F

a bc d

1 12 2

= 0

a + 2b a 2bc + 2d c 2d

= 0

a + 2b= 0

c + 2d= 0

a=2bc=

2d

A=2b b

2d d

=b2 1

0 0

+ d

0 02 1

.

Suy ra E=2 1

0 0

,

0 02 1

l tp sinh ca F.

D thy ELTT. Vy El c s ca Fvdim(F) = 2.

nh lCho M={v1, v2, . . . , vp} VK hiuH :=Span(M) ={1v1+ 2v2+ + pvp|iR}.

Hl mt KGVT c sinh bi S: H=< M >.

dim(H) =r(M). xHxl THTT ca Mr(S, x) =r(M)



41/79


V d 4.26 Tm c s v s chiu ca cc khng gian con sau

a) F =(1;1;1), (2;1;1), (3;1;1).b) F =x2 + x + 1, 2x2 + 3x 1, x2 + 2x 2.

c) F =1 1

2 1

,2 1

0 1

, 3 12 1 ,

1 0

2 0d) F =

(x1; x2; x3; x4)R3|x1+ x2+ x3= 0 x1 x2+ x4 = 0

Bi lm

a) A=

1 1 12 1 1

3 1 1

bdsc

1 1 10 1 1

0 0 0

=dim(F) =r(A) = 2v c s ca F l{(1;1;1), (0; 1; 1)}.

b) A=1 1 1

2 3 11 2 2

bdsc 1 1 1

0 1 30 0 0

=dim(F) =r(A) = 2v c s ca F l{x2 + x + 1, x 3}.

c) A=

1 1 2 12 1 0 13 1 2 11 0 2 0

bdsc

1 1 2 10 1 4 10 0 0 00 0 0 0

=dim(F) = 2v c s ca Fl

1 12 1

,

0 14 1

.

d) Gii h

x1+ x2+ x3= 0

x1 x2+ x4= 0

1 1 1 0 01 1 0 1 0

=

1 1 1 0 0

0 -2 1 1 0

tx3= 2, x4= 2, pt(2) :x2= 1

2

(

x3+ x4) =

+ , pt(1) :x1=

x2

x3=

xFx = ( ; + ; 2; 2) =(1; 1;2; 0) + (1;1;0;2)Suy ra E={(1; 1;2;0); (1;1;0;2)}l tp sinh ca F.D thy ELTT. Vy El c s ca Fvdim F = 2.

Tm c s v s chiu khng gian conTrong Rn, cho khng gian con F

TH1) Chotp sinh F =< v1, v2, . . . , vm >:

Lp ma trn hng A =

v1v2. . .

vm

bdscbc thang

dim(F) =r(A)v c s gm cc hng khc 0 ca ma trn bc thang.

TH2) Cho tp nghim cah thun nht AX= 0

Gii h: dim V + r(A) =n v c s c suy ra t nghim ca h.



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4.8. TNG GIAO HAI KHNG GIAN CON HBK TPHCM

V d 4.27 TrongR3, cho tp M={(1;1;1), (2 : 3 : 1), (1;0;2)}.a) x= (1; 2;3)thuc khng gian conspan(M)hay khng?b) Tmmx= (1; 0; m)span(M).

Bi lm

a) xthuc Kg con span(M)khi v ch khi x l THTT ca M. Ta lp ma trn ct

[M|x] = 1 2 1 11 3 0 2

1 1 2 3

bdsc

1 2 3 10 1 1 3

0 0 0 1

r(M) = 2< r(M|x) =x /span(M).

b) [M|x] = 1 2 1 11 3 0 0

1 1 2 m

bdsc

1 2 3 10 1 1 1

0 0 0 m 2

x

/

span(M)

r(M) =r(M

|x)

m = 2.

4.8 Tng giao hai khng gian con

nh ngha 4.7 (Tng giao 2 khng gian con) Cho hai khng gian conF vGca KGVTV.Giao 2 khng gian con

F G= xV|xF vxG .Tng 2 khng gian con

F+ G=

f+ g|fF vgG .Tnh cht

F GF, GF+ GV.

nh lF GvF+ Gl 2 khng gian con ca V v

dim(F G) + dim(F+ G) = dim F+ dim G.

nh ngha 4.8 (Tng trc tip) Khng gian conWgi l tng trc tip ca 2 khng gian conF vG,k hiuF G, nu

i) W =F+ G. ii) F G={0}.

nh lCho W =F G. Khi mi vc t xWc biu din duy nht di dng

x= f+ g|fF, gG.

Tch cht tng 2 khng gian conF =< f1, f2, . . . , f n > G=< g1, g2, . . . , gm >

=F+ G=< f1, f2, . . . , f n, g1, g2, . . . , gm >



43/79


V d 4.28 TrongR3, cho 2 khng gian conF ={(x1, x2, x3)R3|x1+ x2 2x3= 0}, G={(x1, x2, x3)R3|x1 x2+ x3= 0}.Tm c s v s chiu caF GvF+ G.

Bi lm

a) Tm c s v s chiu caF G.xF GxF xG

x1+ x2 2x3= 0x1 x2+ x3= 0

x1=

x2= 3

x3= 2

x = (, 3, 2) =(1;3;2).

Suy ra E={(1;3;2)}l tp sinh ca F G.Hin nhin ELTT do El c s ca F Gvdim(F G) = 1.

b) Tm tp sinh ca FvG.F =, G==F+G=

A=

1 1 02 0 11 1 0

1 0 1

bdsc 1 1 00 2 10 0 10 0 0

.=dim(F+ G) =r(A) = 3v c s E={(1;1;0), (0;2;1), (0;0; 1)}.Cch khc:ta cdim(FG)+dim(F+G) = dim F+dim G=dim(F+G) = dim F+dim Gdim(FG) = 2+21 = 3=F+ GR3, do c c s l{(1;0;0), (0;1;0), (0;0;1)}.

V d 4.29 TrongR3, cho 2 khng gian conF ={(x1, x2, x3)R3|x1+x2+x3= 0}, G=.Tm c s v s chiu caF GvF+ G.

Bi lmF+ Gtng t nh v d trn. Ta tm c s v s chiu ca F G.xF GxF xG.xGx = (1; 0; 1) + (2;3;1) = ( + 2; 3; + ).xFxtha iu kin ca F: + 2+ 3+ + = 0 =3.x= ( + 2; 3; + ) = (; 3; 2) =(1;3; 2).D dng suy ra E={(1;3; 2)}l c s ca F Gvdim(F G) = 1.

V d 4.30 TrongR3, cho 2 khng gian conF =< f1= (1; 0;1), f2= (1; 1;1)>, G=< g1= (1; 1;0), g2= (2; 1;1)>.

Tm c s v s chiu caF+ G vF G.Bi lmF+ Glm tng t v d trn. Ta tm c s v s chiu ca F G.xF Gkhi v ch khi x ng thi l THTT ca f1, f2vg1, g2:x= x1f1+ x2f2 = x3g1+ x4g2x1f1+ x2f2 x3g1 x4g2 = 0.

Vit li dng ma trn

1 1 1 2 00 1 1 1 0

1 1 0 1 0

bdsc

1 1 1 2 00 1 1 1 0

0 0 1 1 0

.

tx4 = =x3=x4 =.x= x3g1+ x4g2 =(1; 1; 0) + (2; 1;1) =(1;0;1).D dang suy ra c s ca F

Gl

{(1;0;1)

}vdim(F

G) = 1.



44/79


Bi tp

Cu1) Trong R4, cho U =(1, 2, 1, 1);(2, 1, 0, 2)vV =(1, 5, 3, 5);(3, 0, 1, m). Tmm U V.Cu2) Trong R4, cho Vl tp nghim ca h phng trnh

x1+ x2 x3= 02x1+ 2x2+ x3+ x4= 0

x1+ x2+ 2x3+ mx4= 0

(a) Tm m dim(V)ln nht.(b) Tm c s v s chiu caV vim cu a.

Cu3) Trong R4, cho U =(1, 2, 1, 0);(2, 1, 1, 1)V =(1, 1, 2, 1); (2, 0, 4, m)(a) Tm m dim(U V)ln nht.(b) Tm c s v s chiu caU+ V vU V.

Cu4) Trong R4, cho 2 khng gian di dng tp nghim ca h phng trnh

U :

1 1 2 0 01 1 1 2 0

, V :

1 2 2 2 01 0 1 m 0

(a) Tm m dim(U+ V)b nht.(b) Tm c s v s chiu caU+ V vU V.

Cu5) Trong R4, cho U={(x1, x2, x3, x4)R4 :x1+ x2 x3+ x4 = 0}VV =(1, 2, 1 2);(2, 1, 0, m)(a) Tm m dim(U V)ln nht.(b) Tm c s v s chiu caU+ V vU

V.



45/79

Chng 5Khng gian Euclide

Ni dung1) Tch v hng ca 2 vc t.

2) B vung gc ca khng gian con.

3) Qu trnh trc giao ha Gram-Schmidt.

4) Hnh chiu vung gc xung khng gian con.

5.1 Tch v hng ca 2 vc t

nh ngha 5.1 (Tch v hng) Tch v hng trong R-kgvtVl mt hm thc sao cho mi cp vctuvv thucV, tng ng vi mt s thc k hiu(u, v) tha 4 tin sau:

i) (u, vV) : (u, v) = (v, u).ii) (u,v,wV) : (u + v, w) = (u, w) + (v, w).

iii) (K, u, vV) : (u,v) =(u, v).iv) (uV) : (u, u)0; (u, u) = 0u = 0.

Khng gian hu hn hn chiu cng vi tnh v hng trn gi lkhng gian euclide.

Tch v hng chnh tctrnRnx= (x1; x2; . . . ; xn), y= (y1; y2; . . . ; yn)

(x, y) =x1y1+ x2y2+ + xnyn.

Tch v hng chnh tc tng t nh ph thng.

V d 5.1 TrongR2, cho php ton

x= (x1, x2), y= (y1, y2) : (x, y) =x1y1+ 2x1y2+ 2x2y1+ 10x2y2.

a) Chng t (x, y)l 1 tch v hng trnR2.

b) Tnh tch v hng ca 2 vc tu= (1; 2), v= (2;

1).

Bi lm

a) Sinh vin t kim tra 4 iu kin ca tch v hng.

b) (u, v) = 1.2 + 2.1.(1) + 2.2.2 + 10.2.(1) =12.

44


46/79

5.1. TCH V HNG CA 2 VC T HBK TPHCM

V d 5.2 TrongP2[x], cho tch v hng(p, q) =10

p(x).q(x)dx;p(x), q(x)P2[x].

a) Chng t (p, q) l 1 tch v hng trnP2[x].

b) Tnh tch v hng ca 2 vc tp(x) = 2x2 3x + 1, q(x) =x + 1.

Bi lma) Sinh vin t kim tra 4 iu kin ca tch v hng.

b) (p, q) =10

p(x)q(x)dx=10

(2x2 3x + 1)(x + 1)dx= 16

.

di vc t u c nh ngha bi

||u||=

(u, u)

Khong cch gia 2 vc t uvv c nh ngha bid(u, v) =||u v||

Gc gia 2 vc t uvv c xc nh bi

cos = (u, v)

||u||.||v||

Vc t c di bng 1 gi l vc t n v.

Chia 1 vc t khc 0 cho di ca n ta c vc t n v.

Qu trnh to ra vc t n v gi l chun ha.

Bt ng thc Cauchy-Schwatz

|(u, v)| ||u||.||v||,u, vV.

ng thc xy ra khi v ch khi u, vcng phng (hay PTTT).Bt ng thc tam gic

||u + v|| ||u|| + ||v||,u, vV.

ng thc xy ra khi v ch khi u, vcng hng.

V d 5.3 TrongR3 :x = (x1; x2; x3), y= (y1; y2; y3). cho tch v hng

(x, y) = 5x1y1+ 2x1y2+ 2x2y1+ 3x2y2+ x3y3

a) Chng t (x, y)l tch v hng.

b) Tm tch v hng ca 2 vc tu= (2; 1;0), v= (3; 2;4).c) Tm di vc tu= (3; 2;1).

d) Tm khong cch gia 2 vc tu= (1; 2;1)vv= (3;0;2).e) Tm gc gia 2 vc tu= (1;0;1) vv= (2; 1;0).

Bi lm



47/79

5.1. TCH V HNG CA 2 VC T HBK TPHCM


b) (u, v) = ((2; 1; 0), (3; 2; 4)) = 5.2.3 + 2.2.(2) + 2.1.3 + 3.1.(2) + 0.4 = 2.c)||u||= (u, u) = ((3;2;1), (3; 2; 1)) = 5.3.3 + 2.3.2 + 2.2.3 + 3.2.2 + 1.1 = 82d) d(u, v) =||u v||= (u v, u v) = ((2;2; 1), (2;2; 1))

=

5.(2).(2) + 2.(2).2 + 2(2).(2) + 3.2.2 + (1).(1) = 17.e) cos =

(u, v)

||u||.||v|| = 12

6,

31=

12168

= = arccos 12168

.

Nhn xt: di, khong cch, gc i vi tch v hng trn c gi tr khc vi tch v hng phthng.

V d 5.4 TrongP2[x], cho tch v hng(p, q) =11

p(x)q(x)dx; p(x), q(x)P2[x].

a) Chng t (p, q) l mt tch v hng trnP2[x].

b) Tnh tch v hng ca 2 vc tp(x) = 2x2 3x + 1, q(x) =x 3.c) Tm di ca vc tp(x) = 2x + 3.

d) Tnh khong cch gia 2 vc tp(x) =x2 + x + 2, q(x) =x2 2x + 3.e) Tnh gc gia 2 vc tp(x) =x2 + x, q(x) = 2x + 3.

Bi lm


b) (p, q) =1

1p(x)q(x)dx=

1

1(2x2 3x + 1)(x 3)dx=12.

c)||p||= (p, p) =

11

p(x).p(x)dx

11

(2x + 3)2dx=

62

3 .

d) d(p, q) =||p q||= (p q, p q) = (3x 1, 3x 1) =

11

(3x 1)2dx= 22.

e) cos = (p, q)

||p||.||q||

11

(x2 + x)(2x + 3)dx 1

1(x2 + x)2dx.

1

1(2x + 3)2dx

=5

310

124 .

Hai vc t vung gc

uv(u, v) = 0.

Vc t vung gc vi tp hp

uM(u, y) = 0, yM.

H trc giao:h vc t Mgi l trc giao nu

x, yM :xy.

H trc chun:h vc t Mgi l trc chun nu Mtrc giao v

xM :||x||= 1.



48/79

5.2. B VUNG GC CA KHNG GIAN CON HBK TPHCM

Mnh Cho khng gian conF =< f1, f2, . . . , f m>

xFxfk, k= 1, 2, . . . , m .

V d 5.5 TrongR3

vi tch v hng chnh tc, cho khng gian con

F ={(x1; x2; x3)R3 x1+ x2 x3= 02x1+ 3x2+ x3= 0

Tmmx= (2; 3; m)F.

Bi lmTp sinh ca F l{u= (4; 3;1)}.xFxu(x, u) = 02.4 + 3.(3) + m.1 = 0m = 1.

5.2 B vung gc ca khng gian connh ngha 5.2 Trong khng gian EuclideV, cho khng gian conF. Tp hp

F={xV|xF}

gi l b vung gc ca khng gian conF.

nh lCho Fl KG con ca KG Euclide V. Khi :

Fl khng gian con ca V vV =F F. dim F+ dim F= dim V.

V d 5.6 TrongR3, cho khng gian conF =< f1= (1; 1;1), f2= (2; 1;0), f3 = (1;0; 1)>.Tm c s v s chiu caF.

Bi lmx= (x1; x2; x3)FxFxfk, k= 1, 2, 3.

(x, f1) = 0

(x, f2) = 0

(x, f3) = 0

x1+ x2+ x3= 0

2x1+ x2+ 0x3= 0

x1+ 0x2 x3 = 0

1 1 1 02 1 0 0

1 0 1 0

(nhn xt cc hng ca ma trn)

Gii h suy ra x = (1; 2;1). C s ca Fl{(1; 2;1)}vdim F= 1.V d 5.7 TrongR3, cho khng gian con

F ={(x1; x2; x3)R3 x1+ x2+ x3= 02x1+ x2 x3= 0

Tm c s v s chiu caF.

Bi lmGii h suy ra tp sinh ca F:F =< u= (2;

1;3)>.

x= (x1; x2; x3)Fxu2x1 x2+ x3= 0.=F=.C s ca Fl{(1;2;0), (2;0; 1)}vdim F= 2.Ghi ch:



49/79

5.2. B VUNG GC CA KHNG GIAN CON HBK TPHCM

ChoF =< h1, h2, . . . , hm >.xFkhi v ch khi

(h1, x) = 0

(h2, x) = 0

. . .

(hm, x) = 0

h1h2. . .hm

.x= 0(ma trn hng ca F)

Do Fl tp nghim ca h phng trnh

h1 0h2 0. . .

hm 0

ChoF ={xRn|Ax= 0}.

Ax= 0 vit li theo vc t hng

h1h2

. . .hm

.x= 0

(h1, x) = 0

(h2, x) = 0

. . .

(hm, x) = 0

iu ny chng txF :xfk, k= 1, m.iu ny chng t Fc sinh bi cc vc t hng ca ma trn A.

F=< h1, h2, . . . , hm >, vihkl cc hng ca ma trn A

V d 5.8 Hy tm c s v s chiu caF trongR4, trong

a) F ={(x;x2; x3; x4)R4 x1+ x3+ x4= 02x1 x2+ 3x3+ x4= 0

b) F =

Bi lm

a) Ta c F=(ly t cc hng ca h phng trnh).Suy ra c s caFl{(1;0;1;1), (2; 1;3;1)}vdim F= 2.

b) VF =nn Fl tp nghim ca h phng trnh 1 1 2 1 02 1 1 0 0

Gii h suy ra tp nghim F=.Suy ra c s caFl{(1;1;1;0), (1;2;0;3)}vdim F= 2.

nh l

Mi tp trc giao, khng cha vc t khng th LTT.

ChoE=

{e1, e2, . . . , en

}l c s trc chun ca KG Euclide V.

xV lun c biu din duy nht dng

x= x1e1+ x2e2+ + xnen, vixk = (x, ek).



50/79

5.3. QU TRNH GRAM-SCHMIDT HBK TPHCM

V d 5.9 Trong khng gian EuclideV, cho c s trc chun

E=

1

6;1

6;2

6

,

1

2;

12

; 0

,

1

3;1

3;

13

Tm ta ca vc tx= (3; 2;1)trong c sE.Bi lmTa vit x = x1e1+ x2e2+ x3e3, trong x1 = (x, e1) = 3

6, x2= (x, e2) = 1

2, x3= (x, e3) = 6

3.

Vy ta ca x trong c s El[x]E=

36

12

63

5.3 Qu trnh Gram-Schmidt

nh l 5.3 (Gram-Schmidt) Cho E={e1, e2, . . . , em}l h LTT ca KGVTV. Khi tn ti mth trc giaoF ={f1, f2, . . . , f m}tha < e1, e2, . . . , em >=< f1, f2, . . . , f m> .

Thut ton Gram-Schmidt

f1= e1.

f2= e2 (e2, f1)(f1, f1)

f1.

f3= e3 (e3, f1)(f1, f1)

f1 (e3, f2)(f2, f2)

f2.

. . . . . . fk =ek (ek, f1)

(f1, f1)f1 (ek, f2)

(f2, f2)f2 (ek, fk1)

(fk1, fk1)fk1.

V d 5.10 Trc chun h vc tE={(1;0;1;1), (0;1;1;1), (1;1;1;1)}Bi lmChnf1= e1 = (1;0; 1;1).

f2 = e2 (e2, f1)(f1, f1)

f1= (0;1; 1;1) 23

(1; 0;1; 1) =

23

; 1;1

3;1

3

. Chnf2= (2;3;1;1).

f3 = e3

(e3, f1)

(f1, f1)f1

(e3, f2)

(f2, f2)f2= 2

5;2

5;1

5 ;1

5. Chnf3= (2; 2;1; 1).

H trc giao cn tm l F ={f1, f2, f3}.Chia mi vc t cho di ca n, ta c c s trc chun l

13

; 0; 1

3;

13

,

215

; 3

15;

115

; 1

15

2

10;

210

;1

10;1

10

.

V d 5.11 TrongR4 vi tch v hng chnh tc, cho khng gian con

F ={(x1; x2; x3; x4)R4 x1+ x2 x3+ x4= 02x1+ 3x2 x3+ 3x4= 0

Tm mt c s trc chun caF.

Bi lmGii h, tm mt c s ty ca F l{

(2;

1;1;0), (0;

1;0;1)}.

Dng Gram-Schmidt:f1= e1= (2; 1;1;0).f2 = e2 (e2, f1)

(f1, f1)f1= (0; 1;0;1) 1

6(2; 1;1; 0) = (1

3 ;5

6 ;1

6 ; 1). Chn f2= (2; 5;1; 6).

C s trc giao l F ={f1, f2}. C s trc chunl

26

;1

6;

16

; 0

,

2

66;

566

; 1

66;6

66;

.



51/79

5.4. HNH CHIU VUNG GC HBK TPHCM

5.4 Hnh chiu vung gc

nh ngha 5.4 (Hnh chiu vung gc) Trong KG EuclideV, cho khng gian conFv vc tv.Vc t c biu din duy nht di dng

v= f+ g; fF, gF.

Vc tfc gi lhnh chiu vung gccav xungf, k hiu: f= PrFv.Khong cchtv xungFc nh ngha ld(v, F) =||g||=||v PrFv .

V d 5.12 TrongR4 vi tch v hng chnh tc, cho khng gian con

F ={(x;x2; x3; x4)R4 x1+ x2 x3+ x4= 02x1+ x2 3x3+ 3x4= 0

v vc tx= (1;1; 0;1)

a) Tm hnh chiu caxxungF.

b) Tm khong cch txnF.

Bi lm

a) Chn 1 c s F =< f1= (2; 1;1;0), f2= (2;1;0;1)>.Vitx = f+ g= x1f1+ x2f2+ g, gF.Nhn ln lt f1, f2vo phng trnh theo ngha tch v hng, ta c

x1(f1, f1) + x2(f1, f2) = (x, f1)

x2(f2, f1) + x2(f2, f2) = (x, f2) 6x1 5x2 = 1

5x1+ 6x2=1

x1= 1

11

x2=1

11

PrFx = f=x1f1+ x2f2 = 1

11(2; 1;1; 0) +1

11f2= (2; 1; 0; 1) =

4

11;2

11;

1

11;1

11

b) d(x, F) =||g||=||x PrFx||=||(711

;13

11;1

11;

12

11)||= 3.

Bi tp

Cu1) Trong R2: x = (x1, x2), y= (y1, y2), cho tch v hng

(x, y) =x1y1 2x1y2 2x2y1+ 5x2y2.

(a) Tnh (x, y), ||x||, ||y||.(b) Tnh||x + y||, d(x, y).(c) Tnh (x, y).(d) Tm vc t u sao cho ux.

Cu2) Trong R4, cho KG conU =(1, 1, 0, 0);(2, 1, 1, 0), (2, 1, 0, 1)v vc t z = (7, 3, 0, 0).(a) Tm c s v s chiu caU.

(b) Tm P rU(z), P rU(z), d(z, U), d(z, U).(c) Tm mt c s trc chun caU.(d) Tm li P rU(z)theo c s trc chun.



52/79

5.4. HNH CHIU VUNG GC HBK TPHCM

Cu3) Trong R4, cho khng gian nghim ca h thun nht

U :

x1+ x2 x3+ x4 = 0,2x1 x2+ x3+ 2x4= 0.

V :

x1+ 2x2 x3 = 0,2x1+ 3x2 x4= 0.

(a) Tm c s v s chiu caW = (U

V).

(b) Tm c s trc chun EcaW.(c) Tm vc t e sao cho{E, e}l mt c s trc chun ca R4.



53/79

Chng 6

nh x tuyn tnh

Ni dung1) nh ngha v v d.

2) Nhn v nh ca nh x tuyn tnh.

3) Ma trn ca nh x tuyn tnh trong mt cp cs.

4) Ma trn chuyn c s v ng dng.

6.1 nh ngha v v d

nh ngha 6.1 (nh x) Cho 2 tp hp khc rngX, Y. nh xf tXnYl mt quy tc sao chomixthucX, tn ti duy nhty thucY. Ta vit

f : X Yx y= f(x).

nh xfgi ln nhnu: x1=x2=f(x1)=f(x2)nh xfgi lton nhnu:yY, xX :y = f(x)

nh xfgi lsong nhnu n nh v ton nh.

Hm s ph thng l v d v nh x.Cho nh x tc l ch ra quy lut, da vo vit nh ca mi phn t thuc X.C nhiu cch cho nh x: bng th, bng biu , bng biu thc i s, bng cch lit k,...

nh ngha 6.2 (nh x tuyn tnh) Cho V, Wl hai khng gian trn cng trng sK.nh xf :VW gi lnh x tuyn tnhnu tha:

i) f(v1+ v2) =f(v1) + f(v2), v1, v2V.ii) f(v) =f(v),

v

V,

K.

V d 6.1

a) f(x1; x2; x3) = (2x1+ x2 3x3; x1 4x2)l mt nh x tuyn tnh tR3 nR2. ?? .

52


54/79

6.1. NH NGHA V V D HBK TPHCM

b) Php quay trong khng gianOxyzquanh trc 0zmt gc 30o ngc chiu kim ng h nhn t hngdng ca trc 0z l mt nh x tuyn tnh tR3 nR3.

c) Tng t php i xng, php chiu,... qua cc ng thng v mt phng qua gc ta l nhng nhx tuyn tnh tR3 nR3.

ChoE={e1, e2, . . . , en}l tp sinh ca KGVT Vv axtt f :VW.Gi s ta bit f(e1), f(e2), . . . , f (en).xV :x = x1e1+ x2e2+ + xnen =f(x) =f(x1e1+ x2e2+ + xnen)f(x) =f(x1e1) + f(x2e2) + + f(xnen) =x1f(e1) + x2f(e2) + + xnf(en).nh x tuyn tnh c xc nh hon ton nu bit c nh ca mt tp sinh ca V.

V d 6.2 Cho nh x tuyn tnhf :R3 R2, bitf(1; 1; 0) = (2;1), f(1; 1; 1) = (1; 2), f(1;0;1) = (1;1)

a) Tmf(3;1;5). b) Tm f(x).

Bi lm

a) Vit (3; 1;5) =(1; 1; 0) + (1; 1; 1) + (1;0;1)

+ + = 3

+ = 1

+ = 5

=2= 3

= 2

f(x) =f((1; 1; 0) + (1; 1; 1) + (1; 0; 1)) =f(1; 1; 0) + f(1; 1; 1) + f(1;0;1).=f(3; 1;5) =2(2; 1) + 3(1; 2) + 2(1; 1) = (3; 10).

b) Lm tng t nh trn cho trng hp tng qutf(x1; x2; x3).Ta c th lm cch khc bng cch dng php bin i i s nh sau:f(0; 0;1) =f(1;1;1) f(1; 1; 0) = (1; 2) (2; 1) = (1;3).f(0; 1;0) =f(1;1;1) f(1; 0; 1) = (1; 2) (1; 1) = (2; 1).f(1; 0;0) =f(1;1;0) f(0; 1; 0) = (2; 1) (2; 1) = (0; 2)f(x1; x2; x3) =x1f(1; 0; 0) + x2f(0; 1; 0) + x3f(0; 0;1) =x1(0; 2) + x2(2; 1) + x3(1;3)f(x) = (2x2 x3; 2x1+ x2+ 3x3)

Ghi ch:Ta c th dng cc php bin i cho nh x tuyn tnh tm nh ca 3 vc t n v.Tuy nhin ta s gp kh khn tm ra php bin i trong trng hp tng qut.Ta c th vit nh x tuyn tnh di dng ma trn tm nh ca 3 vc t n v nh sau:

(theo hng) e1 f(e1)e2 f(e2)

e3 f(e3)

= 1 1 0 2 11 1 1 1 21 0 1 1 1

bsc 1 0 0 0 20 1 0 2 10 0 1 1 3

Kt hp vi ngha php nhn ma trn, ta c thut ton sau

Tm axtt cho nh ca c sCho c s E ={e1, e2, . . . , en}v nh x tuyn tnh thaf(ek) =fk

Theo hng E F bsc theo hng, tng ngnhn bn tri vi E1 I E1F

V d 6.3 Cho fl php i xng qua mt phng2x y+ 3z = 0 l nh x tuyn tnh trong khng gianOxyz. Hy tmf(x1; x2; x3).



55/79

6.2. NHN V NH CA NH X TUYN TNH HBK TPHCM

Bi lmfbin cp vc t ch phng thnh chnh n v vc t php tuyn thnh vc t ia1= (1; 2; 0) :f(1; 2;0) = (1; 2;0)a2= (0; 3; 1) :f(0; 3;1) = (0; 3;1)n= (2; 1;3) :f(2; 1; 3) = (2;1; 3).Vit dng ma trn

1 2 0 1 2 00 3 1 0 3 12 1 3 2 1 3

casio tnh E1F

1 0 0 37 27 670 1 0 2

767

37

0 0 1 67

37

27

f(x) =x1(37

; 27

;67

) + x2(27

; 67

; 37

) + x3(67

; 37

;27

) =1

7(3x1+ 2x2 6x3; 2x1+ 6x2+ 3x3; 6x1+ 3x2 2x3)

6.2 Nhn v nh ca nh x tuyn tnh

Cho nh x tuyn tnh f :VWNhnca fc nh ngha l

ker f={xV :f(x) = 0}

nhc nh ngha l

Imf={f(x)W|xV}

nh lCho nh x tuyn tnh f :VW

ker fl KG con ca V.

Imfl KG con ca W.

dim(ker f) + dim(Imf) = dim V

V d 6.4 Cho axttf :R3 R2 thaf(x1; x2; x3) = (x1 x2; x1+ 2x2 x3).Tm c s v s chiu caImf vker f.

Bi lm

a) x

Kerf =

f(x) = 0

(x1

x2; x1+ 2x2

x3) = 0

x1 x2= 0x1+ 2x2 x3= 0

x1= x2

x3= 3x2.

=x = (x2; x2; 3x2) =x2(1; 1;3) =ker f= .C s ca ker fl{(1;1;3)}vdim(ker f) = 1.

b) Imfgm tt c cc f(x):f(x1; x2; x3) = (x1x2; x1+2x2x3) =x1(1; 1)+x2(1; 2)+x3(0; 1) =I mf=.C s ca I mf l{(1; 1), (0;3)}vdim(Imf) = 2.Cch khcdim(Imf) = dim(R3) dim(ker f) = 3 1 = 2 =I mfR2.C s ca I mf l{(1; 0), (0;1)}.

nh lCho axtt f :V

W

nh ca tp sinh l tp sinh ca nh:

V =< e1, e2, . . . , en >=I mf=< f(e1), f(e2), . . . , f (en)>



56/79

6.3. MA TRN CA NH X TUYN TNH HBK TPHCM

V d 6.5 Cho axttf :R3 R3 bit nh ca mt tp sinhf(1; 1;1) = (1; 2;1), f(1; 1; 2) = (2; 1;1), f(1;2;1) = (5;4; 1).Tm c s v s chiu caImf vker f

a) Theo nh l: I mf=.C s ca I mf l

{(1;2;1), (0;1;1)

}vdim(Imf) = 2.

b) E={(1;1;1), (1;1;2), (1;2;1)}l tp sinh ca R3.Vitx = x1e1+ x2e2+ x3e3=f(x) =x1f(e1) + x2f(e2) + x3f(e3)f(x) =x1(1; 2; 1) + x2(2; 1;1) + x3(5; 4;1) = (x1+ 2x2+ 5x3; 2x1+ x2+ 4x3; x1 x2 x3)

xker ff(x) = 0

x1+ 2x2+ 5x3 = 0

2x1+ x2+ 4x3 = 0

x1 x2 x3= 0

x1=x2=2x3=

.

=x =(1;1;) 2(1; 1; 2) + (1; 2;1) =(2;1;4).Vy c s ca ker f l{(2;1;4)}vdim(ker f) = 1.

6.3 Ma trn ca nh x tuyn tnhMa trn ca nh x tuyn tnhcho axtt f :VW.E={e1, e2, . . . , en}l c s ca V. F ={f1, f2, . . . , f m}l c s ca W.Ma trn

AE,F =

[f(e1)]F [f(e2)]F . . . [f(en)]F

|| || ||

mn

gi l ma trn ca nh x tuyn tnh ftrong cp c s E , F.

Ch : [f(ei)]F =F1f(ei). Do

AE,F =

F1f(e1) F1f(e2) . . . F 1f(en)| | |

mn=F1f(E).

V d 6.6 Cho axttf :R3 R2 bitf(x1; x2; x3) = (x1+ 2x2 3x3; 2x1+ x3).Tm ma trn caftrong cp c sE={(1;1;1), (1;0;1), (1;1;0)}, F ={(1;3), (2;5)}.Bi lmf(1; 1; 1) = (0; 3) =[f(1;1;1)]F =

6

3

.

f(1; 0;1) =(

2;3)=

[f(1;0;1)]F =

16

9.

f(1; 1; 0) = (3; 2) =[f(1;1;0)]F =11

7

.

Ma trn cn tm l AE,F =

6 16 113 9 7

.

Cch khc

AE,F =F1f(E) =

1 23 5

10 2 33 3 2

dng casio======

6 16 113 9 7

.



57/79


nh l

i) Cho axttf :VW. Khi tn ti duy nht ma trn AE,Fcm nsao cho

[f(x)]F =AE,F[x]E,

viE , Fl 2 c s ca V vWtng ng.

ii) 2. Cho ma trn A = (aij )mntrn trng s K. Khi tnti duy nht mt nh x tuyn tnh f :Kn Km tha

[f(x)]F =AE,F[x]E.

Ch :

Mi mt nh x tuyn tnh t khgian hu hn chiu vo KG hu hn chiu tng ng duy nht mtma trn v ngc li.

Ta coi nh x tuyn tnh l ma trn. Thng thng khng phn bit hai khi nim ny.

V d 6.7 Cho axttf :R3 R2 bit ma trn caftrong cp c sE={(1;1;1), (1;0;1), (1;1;0)}, F ={(1;1), (2; 1)}lAE,F =

2 1 30 3 4

.

a) Tmf(3;1;5). b) Tm f(x).

Bi lm

a) [(3; 1; 5)]E=E131

5

=

1 1 11 0 1

1 1 0

31

5

=

32

2

.

Dng cng thc [f(x)]F =AE,F[x]E

[f(3;1;5)]F =

2 1 30 3 4

322

= 142

=f(3; 1; 5) = 14(1; 1) 2(2; 1) = (10; 12).

b) [(x1; x2; x3)]E=E1x1x2

x3

= 1 1 11 0 11 1 0

x1x2x3

= x1+ x2+ x3x1 x2x1 x3

.Dng cng thc [f(x)]F =AE,F[x]E

[f(x1; x2; x3)]F =

2 1 30 3 4

x1+ x2+ x3x1 x2x1 x3

= 4x1+ x2+ 5x3

7x1 3x2 4x3

=f(x1; x2; x3) = (4x1 + x2 + 5x3)(1;1) (7x1 3x2 4x3)(2; 1) = (10x1 5x2 3x3; 3x1 2x2 + x3).

Ma trn trong 1 c scho axtt f :V

V.E={e1, e2, . . . , en}l c s ca V .Ma trn nh x ca ftrong cp c s E , El

AE=E1f(E).



58/79


V d 6.8 Cho axttf :R3 R3 c ma trn trong c sE={(1;1;1), (1;0;1), (1;1;0)}l

AE=

1 1 12 3 3

1 2 4

.

a) Tmf(2;3;

1).

b) Tm c s v s chiu caker f.

c) Tm c s v s chiu caImf.

Bi lm

a) Tng t v d trn: f(2;3; 1) = (12; 6; 2).

b) Gi sx = x1e1+ x2e2+ x3e3[x]E=

x1x2x3

.

xker ff(x) = 0[f(x)]F =AE[x]E= 01 1 12 3 3

1 2 4

x1x2x3

= 0 x1= 6x2=5x3=

.

x= 6(1;1;1) 5(1; 0; 1) + (1; 1;0) =(2;7;1)=ker f=. C s ca ker fl{(2;7;1)}vdim(ker f) = 1.

c) [f(1;1;1)]E=

1 1 12 3 3

1 2 4

10

0

=

11

1

=f(1; 1; 1) = (1; 1; 1) + (1; 0; 1) (1; 1;0) = (4; 2;3).

Tng t[f(1;0;1)]E=

13

2

=f(1; 1; 1) = (1; 1; 1) + 3(1; 0; 1) + 2(1; 1; 0) = (6; 3; 4).

[f(1;1;0)]E=

13

4

=f(1; 1;1) =(1; 1; 1) + 3(1; 0; 1) + 4(1; 1; 0) = (6; 3; 2).

Imf =< f(1;1;1), f(1;0;1), f(1;1;0)>. C s ca I mfl{(4;2;3), (0;0;1)}.Tnh Cht 6.3 (Mi lin h gia 2 ma trn trong c s khc nhau)

Cho axttf :VW.Cho 2 c s ca V l:E={e1, e2, . . . , en}, E ={e1, e2, . . . , en}.Cho 2 c s ca W l:F ={F1, f2, . . . , f n}, F ={f1, f2, . . . , f n}.

Pl ma trn chuyn c s tEvoE: [x]E=P[x]E

Ql ma trn chuyn c s tFvo F:[y]F =Q[y]FTa c

[f(x)]F =AE,F[x]EQ[f(x)]F =AE,FP[x]E =[f(x)]F =Q1AE,FP.[x]EKhi ,Q1AE,FPl ma trn ca ftrong cp c s E, F.Ta tm tc bng s sau

E A F

P QE

Q1AP F

Trong trng hp c bit: V

W, E

F, E

F, ta c kt qu tng t

E A E

P PE P

1AP E



59/79


V d 6.9 Cho axttf :R3 R3 cho bif(x1; x2; x3) = (x1+ 2x2 3x3; 2x1+ x2+ x3; 3x1 x2+ 2x3).Tm ma trn caftrong c sE={(1;1;1), (1;1;0), (1;0;1)}.Bi lm

Ma trn caftrong c s chnh tc E0l A = E10 f(E0) =f(E0) =

1 2 32 1 13

1 2

Ma trn chuyn c s tE0sang El P =E10 .E= E=

1 1 11 1 0

0 1 1

.

S ctc A ctcP P

E P1AP E

Ma trn cn tm P1AP =E1AE=

1 1 11 1 00 1 1

1

1 2 32 1 13

1 2

1 1 11 1 00 1 1

=

1 4 72 8 100

4

5

.


A=

1 0 12 1 4

1 1 3

.

Tm ma trn caftrong c sE={(1;2;3), (2;3;5), (5;8;4)}.Bi lmS

E A E

P PE P

1AP EMa trn chuyn c s tEsang El P =E1E.Ma trn caftrong c s El

P1AP =E1EAE1E=1

9

59 40 22153 37 206

5 4 23

.


A=

1 0 12 1 41 1 3

.Tm ma trn caftrong c s chnh tcF ={(1;0;0), (0;1;0), (0;0;1)}. T suy raf(x).Bi lmS

ctc A ctcP P

E P1AP E

Ma trn chuyn c s tEsang F l P =E1

.F =E1

.Ma trn caftrong c s chnh tc l

B = P1AP =EAE1 =

1 1 12 1 1

1 2 1

1 0 12 1 4

1 1 3

1 1 12 1 1

1 2 1

1

dng casio======

18 4 620 4 7

27 6 9



60/79


=f(x) = (18x1 4x2 6x3; 20x1 4x2 7x3; 27x1 6x2 9x3).??

nh ngha 6.4 (Hai ma trn ng dng) .Hai ma trn vungA, B gi l ng dng nu tn ti ma trn kh nghchP tha

P1AP =B.

Mnh cho axtt f :VV.Al ma trn ca ftrong c s E.Bl ma trn ca ftrong c s F.Khi A vB ng dng.



61/79

Chng 7Tr ring - vc t ring

Ni dung1) Tr ring - vc t ring ma trn

2) Cho ha ma trn

3) Cho ha ma trn i xng thc

4) Tr ring - vc t ring nh x tuyntnh

5) Cho ha nh x tuyn tnh

7.1 Tr ring - vc t ring

Tr ring - vc t ringca ma trn vung A.Sgi ltr ringca ma trnAnu tn ti vc txkhc khngtha

Ax= x

Khi ,x gi lvc t ringng vi tr ring ca ma trn A.

x= 0l VTR ca A nu Ax cng phng vi x.

V d 7.1 Cho A=

1 65 2

vu=

65

, v=

32

.

Ta c:Au =

1 65 2

65

=

2420

=4

65

=4u: u l VTR ng vi TR =4.

Av=

1 65 2

32

=

911

:khng cng phng vi x, d x khng l VTR ca A.

V d 7.2 Cho A=

3 46 5

, 1=1, 2 = 3. S no l TR caA?

Bi lm

60


62/79

7.1. TR RING - VC T RING HBK TPHCM

a) Xt h phng trnh Ax = 1x

3 46 5

x1x2

=1

x1x2

3x1+ 4x2=x16x1+ 5x2=x2

4x1+ 4x2= 0

6x1+ 6x2= 0

x1=

x2=.

Nh vy x =

, = 0l cc VTR ng vi TR =1ca A.

b) Xt h phng trnh Ax = 2x

3 46 5

x1x2

= 3

x1x2

3x1+ 4x2= 3x1

6x1+ 5x2= 3x2

4x2 = 0

6x1+ 2x2= 0

x1= 0

x2= 0.

V h c nghim duy nht x = 0nn 2= 3khng phi TR ca ma trn A.

nh ngha 7.1 (Cc khi nim c bn)

Gi s0l Tr ca ma trn vung A x0= 0 :Ax0= x0Ax0 0x= 0(A I)x= 0.V h thun nht c nghim khc khng nn

det(A

I) = 0 : gi l phng trnh c trngca A.

a thc PA() = det(A I)gi la thc c trngca A.Tm TR-VTRca ma trn vung

Bc1) Lp phng trnh c trng det(A I) = 0.Bc2) Gii phng trnh c trng tm tr ring.

Bc3) Vi mi TR i, gii h (A iI)x= 0:Tm VTR ng vi TR i.

nh ngha 7.2 .

i) Bi i sca tr ringi l bi nghim cai trong phng trnh c trng.

ii) Khng gian con ringca tr ringi l khng gian nghim ca h(A i)x= 0, k hiu lEi.iii) Bi hnh hccai l s chiu caEi: BH H= dim(Ei).

nh l 7.3 Cho Al ma trn vung.

i) C s ca cc KG con ring lp thnh mt h c lp tuyn tuyn tnh.

ii) 1BHHBScho tt c cc tr ringi.

Chng minh: Theo di bi ging trn lp.

V d 7.3 Cho A=

1 42 3

. Tm tt c cc TR, c s v chiu ca KG con ring tng ng.

Bi lmPhng trnh c trng det(A I) = 0

1 42 3 = 0(1 )(3 ) 2.4 = 0

2

4

5 = 0

1=1nghim n: BS=12= 5nghim n: BS=1

.

1=1, gii h (A 1I)x= 0

1 (1) 4 02 3 (1) 0

2 4 02 4 0



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7.1. TR RING - VC T RING HBK TPHCM

x =2

=

21

. C s ca E1l

21

vBHH= dim(E1) = 1.

2= 5, gii h (A 2I)x= 0

2 4 02 4 0

x =

=

11

.

C s ca E5l 11vB HH= dim(E5) = 1.

V d 7.4 Cho A=

3 1 12 4 2

1 1 3

. Tm tt c cc TR, c s v c

Baigiangds_DVV

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