N TP MN I S TUYN TNHBi ging in t

TS. L Xun iTrng i hc Bch Khoa TP HCM

Khoa Khoa hc ng dng, b mn Ton ng dngEmail: ytkadai@hcmut.edu.vn

TP. HCM 2013.TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 1 / 26

Cu 1.

Cho hai ma trn A =

2 2 12 5 32 3 5

vB =

3 1 21 2 42 6 3

. Tm ma trn X thaAX X = BT

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 2 / 26

AX X = BT (A I )X = BT X = (A I )1.BT

Vy X =

1 2 12 4 32 3 4

1 3 1 21 2 42 6 3

T = 20 9 106 2 55 4 2

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 3 / 26

Cu 2.Trong R4 cho khng gian conU =< (1, 1, 2, 2), (2,1, 1, 0) >, z = (1, 2, 3, 1).a) Tm m v = (1, 2,1,m) thuc U .b) Tm c s v s chiu U.c) Tm hnh chiu ca z xung U.

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 4 / 26

a) v U th , R :v = (1, 2,1,m) = (1, 1, 2, 2) + (2,1, 1, 0)

+ 2 = 1

= 22 + = 1

2 = m

H ny v nghim nn @m sao cho v U .

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 5 / 26

b) Tm c s v s chiu U. Vctx = (x1, x2, x3, x4) U nn x (1, 1, 2, 2) vx (2,1, 1, 0){

x1 + x2 + 2x3 + 2x4 = 0

2x1 x2 + x3 = 0C s ca U : e1 = (1,1, 1, 0) ve2 = (2,4, 0, 3). S chiu dim(U) = 2.

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 6 / 26

c) Tm hnh chiu ca z xung U.z = e1 + e2 + g , vi g (U).{< z , e1 >= < e1, e1 > + < e1, e2 >

< z , e2 >= < e1, e2 > + < e2, e2 >

{

3 + 6 = 0

6 + 29 = 7 =14

17, = 7

17Vy

hnh chiu ca z xung U lf =

14

17(1,1, 1, 0) 7

17(2,4, 0, 3) =

(0,14

17,14

17,21

17)

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 7 / 26

Cu 3.Trong R4 cho 2 khng gian con

U =< (1, 1,2, 1), (1, 2, 1, 0) >

V :

{x1 + 2x2 + 3x3 5x4 = 02x1 x2 + 2x3 + x4 = 0

a) Tm c s v s chiu ca U V .b) Tm c s v s chiu ca U + V

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 8 / 26

a) Tm c s v s chiu ca U V .x = (x1, x2, x3, x4) U V x U x V .x U (x1, x2, x3, x4) = (1, 1,2, 1) +(1, 2, 1, 0) = ( + , + 2,2 + , )x V

{ 8 + 8 = 02 + 2 = 0 = .

Vy x = (2, 3,1, 1). T suy ra (2, 3,1, 1)l tp sinh ca U V . Vct (2, 3,1, 1) c lptuyn tnh nn c s ca U V l (2, 3,1, 1).Dim(U V ) = 1.TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 9 / 26

Tm c s ca V(1 2 3 52 1 2 1

)(1 2 3 50 5 4 11

)C s ca V l (7,4, 5, 0) v (3, 11, 0, 5)U + V =< (1, 1,2, 1), (1, 2, 1, 0), (7,4, 5, 0), (3, 11, 0, 5) >

1 1 2 11 2 1 07 4 5 03 11 0 5

1 1 2 10 1 3 10 0 18 100 0 0 0

C s ca U + V l

(1, 1,2, 1), (1, 2, 1, 0), (7,4, 5, 0). Dim(U + V ) = 3.TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 10 / 26

Cu 4.Trong R2 : x = (x1, x2), y = (y1, y2). Xt tch vhng (x , y) = 2x1y1 + 2x1y2 + 2x2y1 + 3x2y2.Tnh khong cch gia 2 vct u, v viu = (2,1), v = (1, 3).

d(u, v) = ||uv || = < u v , u v > = 34.

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 11 / 26

Cu 4.Trong R2 : x = (x1, x2), y = (y1, y2). Xt tch vhng (x , y) = 2x1y1 + 2x1y2 + 2x2y1 + 3x2y2.Tnh khong cch gia 2 vct u, v viu = (2,1), v = (1, 3).d(u, v) = ||uv || = < u v , u v > = 34.

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 11 / 26

Cu 5.

Cho nh x f : R3 R3, bit ma trn ca ftrong c s B = {(1, 1, 0), (1, 0, 1), (1, 1, 1)} l

A =

1 2 13 2 01 3 4

. Tm f (4, 3, 6)

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 12 / 26

[f (4, 3, 6)]B = A[(4, 3, 6)]B = 1 2 13 2 01 3 4

. 21

5

= 14

25

.Vy f (4, 3, 6) =1(1, 1, 0) 4(1, 0, 1) + 25(1, 1, 1) = (22, 26, 21)

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 13 / 26

Cu 6.

Cho ma trn cp 3

A =

0 2 21 3 21 5 4

Tm mt ma trn B M3(R) sao cho B3 = A.

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 14 / 26

Xt

A() = |A I | = 2 21 3 21 5 4

= 0 ( + 1)( 2) = 0 1 = 1, 2 = 0, 3 = 2.

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 15 / 26

ng vi 1 = 1 ta xt hx1 + 2x2 + 2x3 = 0

x1 2x2 2x3 = 0x1 + 5x2 + 5x3 = 0

X1 = 01

1

, 6= 0.

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 16 / 26

ng vi 2 = 0 ta xt h0x1 + 2x2 + 2x3 = 0

x1 3x2 2x3 = 0x1 + 5x2 + 4x3 = 0

X2 = 11

1

, 6= 0.

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 17 / 26

ng vi 3 = 2 ta xt h2x1 + 2x2 + 2x3 = 0x1 5x2 2x3 = 0x1 + 5x2 + 2x3 = 0

X3 = 11

2

, 6= 0.

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 18 / 26

Vy ta c ma trn lm cho ha

S =

0 1 11 1 11 1 2

S1 =

1 1 01 1 10 1 1

D = 1 0 00 0 0

0 0 2

.Do A = SDS1 = B3.

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 19 / 26

Vy 1 ma trn B cn tm l 0 1 11 1 11 1 2

(1)1/3 0 00 01/3 00 0 21/3

1 1 01 1 10 1 1

= 0 21/3 21/31 21/3 1 21/31 24/3 + 1 24/3

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 20 / 26

Cu 7.

a dng ton phng sau v dng chnh tcbng php bin i trc giao, nu r php bin i

f (x1, x2, x3) = x212x222x234x1x2+4x1x3+8x2x3.

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 21 / 26

Ma trn ca dng ton phng

A =

1 2 22 2 42 4 2

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 22 / 26

A() = det(A I ) =1 2 22 2 42 4 2

= 0 1 = 7, 2 = 3 = 2.

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

P1 =

132323

.

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 23 / 26

A() = det(A I ) =1 2 22 2 42 4 2

= 0 1 = 7, 2 = 3 = 2.Xc nh ma trn trc giao. Vi 1 = 2, ta c

P1 =

132323

.TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 23 / 26

Vi 2 = 3 = 2, ta c P2 =

25

15

0

,P3 =

2

35

435

535

.

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 24 / 26

Do ma trn trc giao

P =

13 25 23523 15 43523 0

535

.Php bin i (x1, x2, x3)T = P(y1, y2, y3)T s adng ton phng f v dng chnh tcf = 7y 21 + 2y 22 + 2y 23 .

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 25 / 26

CHC CC EM T KT QU TT

TRONG K THI SP TI

TS. L Xun i (BK TPHCM) N TP MN I S TUYN TNH TP. HCM 2013. 26 / 26