tutorial 2 mth 3201

29
Tutorial MTH 3201 Linear Algebras

Upload: drradz-maths

Post on 04-Jul-2015

437 views

Category:

Documents


15 download

TRANSCRIPT

Page 1: Tutorial 2 mth 3201

Tutorial MTH 3201Linear Algebras

Page 2: Tutorial 2 mth 3201

Tutorial 2

Page 3: Tutorial 2 mth 3201

( ) (2,2,2)a w

1 2

2 1 2 1 2

(b) (3,1,5)

(3,1,5) (0, 2, 2) (1,3, 1)

( , 2 3 ,2 )

w

k k

k k k k k

1.Given (0, 2,2), (1,3, 1)u v

1 2w k u k v(2,2,2)

1 2

1 2

Check!

2 3 2(2) 3(2) 2

2 2(2) 2 2

the vector is linear

combination with u and v

k k

k k

2

1 2

2

2 3 2

k

k k

1

1

1 2

2 3(2) 2

2( ) 4

2 2

k

k

k and k

1

1 2

2 5 3

4 3

k

k and k

2

1 2

1 2

3

2 3 1

2 5

k

k k

k k

1 2

1 2

Check!

2 3 2(4) 3(3) 1

2 2(4) 3 5

linear combination

k k

k k

Page 4: Tutorial 2 mth 3201

2 1 2 1 2

2

1 2

1

1 2

1 2

1 2

( ) (0,0,0)

(0,0,0) ( , 2 3 ,2 )

0

2 3 0

2 0 0

0 0

Check!

2 3 2(0) 3(0) 0

2 2(0) 0 0

linear combination

c w

k k k k k

k

k k

k

k and k

k k

k k

2 1 2 1 2

2

1 2

1 2

1 2

1 2

( ) (0,4,5)

(0,4,5) ( , 2 3 ,2 )

0

2 3 4

2 5

52 0

2

!

2( 2) 3(0) 4

2 2( 2) 0 4

(0,4, 4) (0,4,5)

Not a linear combination

d w

k k k k k

k

k k

k k

k or and k

Check

k k

w

Page 5: Tutorial 2 mth 3201
Page 6: Tutorial 2 mth 3201

1 2 3

1 2

1 2 3 1 2

1 2 1 2 1 2

( ) (3, 2, 1), (1, 2, 3)

( , , )

( , ) (3, 2, 1) (1, 2, 3)

(3 , 2 2 , 3 )

a u v ERO

Let b b b b

b k u k v

b b b k k

k k k k k k

1 2 1 2

3 1 3

2 32

3 31

2

2 2 3 23

3 1 3 1

3 3

83 32 2

4

3 1 3 2 1

3 1 1 3 1 3

22 2 2 2 0 4

1 3 3 1 0 8 3

1 3 1 32 2

0 1 0 12 4 2 4

0 8 0 03 2

R R R R

R R R

R RR

b bb

b b b b

b b b b

b b

b bb b

b b b b b

3 2 1

3

2 0, '

'

Not span

If b b b thelinear systemdoesn t have solution

thevectors don t span R

1 2 1

1 2 2

1 2 3

3

2 2

3

k k b

k k b

k k b

Page 7: Tutorial 2 mth 3201

1 2 3

1 2 3

1 2 3 1 2 3

1 2 3 2 3 1 3

1 1 2 3

2 2 3

3 1 3

( ) (1,0,3) , (3,1,0) , (1,2,3)

( , , )

( , , ) (1,0,3) (3,1,0) (1,2,3)

( 3 , 2 ,3 3 )

3

2

3 3

1 3 1

0 1 2 3(6

3 0 3

b u v w

Let b b b b

b k u k v k w

b b b k k k

k k k k k k k

b k k k

b k k

b k k

Let A A

1) 0 3(1 0)

3(5) 3

1830.A Thevectors spanR

Page 8: Tutorial 2 mth 3201

31 2

1 2 3 1 2 3

1 2 3 1 2 3 1 2 3

( ) (2,1, 2) , (6,3, 6) , ( 2, 1,2)

( , , ) (2,1, 2) (6,3, 6) ( 2, 1,2)

(2 6 2 , 3 , 2 6 2 )

2 6 2

Let A 1 3 1

2 6 2

A 1(12 12) 3(4 4) 1( 12 12)

0

c v v v

b b b k k k

k k k k k k k k k

3

A 0. Ais not invertible.

Thesystemincosistent.

The vectorsdoesn't span R

Page 9: Tutorial 2 mth 3201

1 2 3

1 2 3

1 2 3 1 2 3

1 2 3 1 2 3 1 3

( ) (3,2, 2) , (1,1,0) , ( 2,1,2)

( , , )

( , , ) (3,2, 2) (1,1,0) ( 2,1,2)

(3 2 , 2 , 2 2 )

3 1 2

A 2 1 1

0 2 2

A 2(3 4) 2(3 2)

2(7) 2(1)

16

A

d u v w

Let b b b b

b k u k v k w

b b b k k k

k k k k k k k k

Let

30. The vectorsspan R

Page 10: Tutorial 2 mth 3201

Question 2(e)

• Congratulation!!

• This is your first assignment

• Please submit in tutorial class next week

Page 11: Tutorial 2 mth 3201

1 2 3

4 22

1 2

22

3 4

1 2 3 4

1 2

1 2 3 4

3 4

1 2 3

0 0 1 1 1 1( ) , , ,

1 0 0 0 1 0

0 0;

0 1

0 0 1 1 1 1 0 0

1 0 0 0 1 0 0 1

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

a u u u

u v M

b bLet b M

b b

b k u k u k u k u

b bk k k k

b b

b k k

2 2 3

3 1 3

4 4

...............[1]

.............................[2]

.............................[3]

....................................[4]

b k k

b k k

b k

Page 12: Tutorial 2 mth 3201

1 2 2

1 22

1 22 3

1 23 2

2 1

3 1 2 1

1 3 2 1

4 4

1 2 3 4

[1] [2]: 2

2

From[2] :2

2

1 1

2 2

1 1From[3] : ( )

2 2

1 1

2 2

From[4] :

Thesystem hassolution.All vector b is the linear combination of

u ,u ,u and u .Henc

b b k

b bk

b bb k

b bk b

b b

b k b b

k b b b

k b

e the vector span the vector space v.

Page 13: Tutorial 2 mth 3201

1 2 3

4 22

1 2

1 2 3 4

3 4

1 1

2 2

1 0 0 2 0 0( ) , , ,

0 0 0 0 3 0

0 0;

0 4

1 0 0 2 0 0 0 0

0 0 0 0 3 0 0 4

................................[1]

2 ........................

b w w w

w v M

b bk k k k

b b

b k

b k

3 3

4 4

32 41 1 2 3 4

.....[2]

3 ..............................[3]

4 ..............................[4]

; ; ;2 3 4

vector span the vector space v.

b k

b k

bb bk b k k k

Page 14: Tutorial 2 mth 3201

2 2

1 2 3

2

0 1 2

1 1 2 2 3 3

2

0 1 2 1 1 2 2 3 3

2

1 2 3

2

1 2 2 3 3

2

1 2 3 2 3

0 1 2 3

1 2 3

2 3

1 2 3 0

1 1 2

( ) 1, 1, 1,

( )

( )

( )

(1) ( 1) ( 1)

( ) ( ) ( )

c p p x p x v p

b x b b x b x

b x k p k p k p

b b x b x k p k p k p

k k x k x

k k x k k x k

k k k k x k x

b k k k

b k k

b k

k k k b

k b b

0

1 0 1 2

2 1

3 2

0 1 2 1 2 2 3( ) ( )

The vectorsspan the vector space V

b

k b b b

k b

k b

b x b b b b p b p

Page 15: Tutorial 2 mth 3201

2

1 2 23

2

0 1 2

1 1 2 2 3 3

2

0 1 2 1 1 2 2 3 3

2

1 2 3

2

1 2 2 3 3 3

2

1 2 3 2 3 3

0 1 2 3

( ) 2, 1, 1;

( )

( )

( )

(2) ( 1) ( 1)

2

(2 ) ( ) ( )

2 ..............

d q q x q x x v p

Let b x b b x b x

b x k q k q k q

b b x b x k q k q k q

k k x k x x

k k x k k x k x k

k k k k x k x k x

b k k k

1 2 3

2 3

1 2 2

2 1 2

.......................[1]

..............................................[2]

.....................................................[3]

[3] [2] :

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

b k k

b k

in b k b

k b b

0 1 1 2 2

0 1 1 2

0 1 21

0 11 2

0 12 1 1 2 2 2 3

.............[4]

[3],[4] [1] : 2 ( )

2 2

2

2

2

( ) ( ) ( )2

The vectorsspan the vector space V

in b k b b b

b k b b

b b bk

b bk b

b bb x b q b b q b q

Page 16: Tutorial 2 mth 3201

3

1 2 3

1 2 3

1 2 3

1 3

1 2

0 2 1

( ) 1 , 0 , 1 ;

3 3 0

Let

0 2 1 0

1 0 1 0

3 3 0 0

2 0

0

03 3

0 2 1

1 0 1

3 3

a u v w V R

k u k v k w o

k k k

k k k

k k

k k

Let A

0

0(0 3) 2(0 3) 1(3 0)

9

0,A has invertible.

The vector are linearly dependent.

A

A

-

Page 17: Tutorial 2 mth 3201

1 2 22

1 1 2 2

1 2

2

1 2 1

2

1 1

1 2

0 0 1 0( ) , ;

1 1 1 0

Let

0 0 1 0 0 0

1 1 1 0 0 0

0 0 0

0 0

0

0 0

0

b v v V M

k v k v o

k k

k

k k k

k

k k

k k

1 2The system has trivial solution, k 0

The vector arelinearlyindependent

k

Page 18: Tutorial 2 mth 3201

2

2

1 2

2 2

1 2

2 2

1 1 1 2 2

1 1

1 2 2

1 2

1 2

( ) 3 1, 1;

0

(3 1) ( 1) 0 0 0

3 0 0 0

3 0 0

0 0

0

The system has trivial solution, 0

The vectorsspan the vector space V

c p x x q x v p

k p k q

k x x k x x x

k x k x k k x k x x

k k

k k k

k k

k k1 2The system has trivial solution, k 0

The vector arelinearlyindependent

k

Page 19: Tutorial 2 mth 3201

431 2

31 1 2 2 3

1 2 3

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

2 3

1 2

1 2 3

1 2 3

( ) (0, 3,1,1) , (5,5,1,3) , ( 1,0,5,1);

0

(0, 3,1,1) (5,5,1,3) ( 1,0,5,1)

(5 , 3 5 , , 5 , 3 )

0 5

0 3 5

0

0 3

5 1 0 0

3 5 0 0

1 1

d x x x V R

k x k x k x

k k k

k k k k k k k k k k k k k

k k

k k

k k k

k k k

4 1

3 23

1 3 1 0

0 8 15 0

1 0 1 1 15 0

1 3 1 0 0 5 1 0

r r

r r

Page 20: Tutorial 2 mth 3201

3

1 3

3 4 2 3

4

2 3

2

5

11

8

1 3 1 0 1 3 1 0

0 8 15 0 0 8 15 0

0 2 4 0 0 1 2 0

0 5 1 0 0 5 1 0

1 3 1 0 1 3 1 0

0 8 15 0 0 1 2 0

0 1 2 0 0 8 15 0

0 0 11 0 0 0 1 0

1 3 1 0

0 1 2 0

0 0 31 0

0 0 1 0

r

r r

r r r r

r

r r3

4 2

3

3 2 1

4 2

22

3312

1 3 1 0

0 1 0 0

0 0 1 0

0 0 1 0

1 3 1 0 1 0 0 0

0 1 0 0 0 1 0 0

0 0 1 0 0 0 1 0

0 0 0 0 0 0 0 0

r

r r

r

r r r

r r

Page 21: Tutorial 2 mth 3201

211 2

2 1 2

1

1 1 1 2 2

1 2

1 2 1 2

1 2 1

1 2 2

1 1 2

1 1 2

(5) ( 1,3) , (2,1) are ( 1, 2)

and ( 3,5)element ( , )

( 1, 2) .

( 1, 2) ( 1,3) ( 2,1)

( 2 ,3 )

52 1

7

13 2

7

5 1

7 7

element is in span ( , )

v v R w

w span v v

w

Let w k v k v

k k

k k k k

k k k

k k k

w v v

w v v

Page 22: Tutorial 2 mth 3201

2 2 1 1 2 2

1 2

1 2 1 2

1 2 1

1 2 2

2 1 2

2 1 2

( 3,5) .

( 3,5) ( 1,3) ( 2,1)

( 2 ,3 )

132 3

7

43 5

7

13 4

7 7

element is in span ( , )

w Let w k v k v

k k

k k k k

k k k

k k k

w v v

w v v

Page 23: Tutorial 2 mth 3201

1 2

1 1 2 2

1

2

2

1

) , 123

0

10

120

3

1

Let A = ,12

3

Linearly dependent: A 0

10

4

1 1( )( ) 0

2 2

1

2

a x x

k x k x

k

k

1 2

1 1 2 2

1

2

2

2

2) ,

2

0

2 0

2 0

Let A = ,2

Linearly dependent: A 0

4 0

( 2)( 2) 0

2

b y y

k y k y

k

k

Page 24: Tutorial 2 mth 3201

1 2 31 2 3Assume = k x k x k x z

1

2

3

3

3

3

3

1 1 2 2

1 2

1 1 0

1 1 00

1 1 0

1 1 1

1 1 1

1 1 1 1

A ( 1) ( 1)

2 3

Linearly dependent A 0

2 3 0

3 2 0

2, 1

0

0, 0 linear

k

k Ak

k

k x k x

k k

1 2 3

ly dependent

0 linearly independent

A 0 linearly independent

k k k

Question 7

• Congratulation!!

• This is your second assignment

• Please submit in tutorial class next week

Page 25: Tutorial 2 mth 3201

second method

1 2 3

1 1 2 2 3 3

1

2

3

2

3

1 1

) , ,1 1

1 1

0

1 1 0

1 1 0

1 1 0

1 1

Let A = 1 1 ,

1 1

A 0

( 1) ( 1)( 1) ( 1)(1 )

a x x x

k x k x k x

k

k

k

3

3

( 1) ( 1 )

1 1 )

3 2

1,2

Page 26: Tutorial 2 mth 3201

1 2 3 1 2

1 2 3

3 1 2

3 1 1 2 2

1 2 3

1 2 3

(8) , is linearly independent span , ,

then, , , also linearly independent.

span ,

................................[1]

Assume that , , are linearly independent.

, ,

v v v v v

v v v

v v v

v v v

v v v

B B B IR

1 1 2 2 3 3

3

1 1 2 2

3 3 1 1 2 2

1 23 1 2

3 3

1 2 3

: , , 0

0,

0

contradiction with equation1

Hence v , v ,v is linearly independent

B v B v B v

B

B v B v

B v B v B v

B Bv v v

B B

Page 27: Tutorial 2 mth 3201

1 2 3 n 1 2 3 n

n

2 2 2

1 1 2 2 1 2 3

(9) Prove theorem 1.13 in chapter 1

Theorem 1.1.3 (Cauchy-Schwarz Inequality)

If u=(u ,u ,u ............u )and v=(v ,v ,v ...........v )

are vectors in R

then, u.v u v

......... ...n nu v u v u v u u u

1 12 2 2 2 22 2

1 2 3 3. ....nu v v v v

Page 28: Tutorial 2 mth 3201

Reference from Wikipedia(universal reference) ;p

Page 29: Tutorial 2 mth 3201

“Belajar dari kesalahan

membuatmu dewasa dan

belajar dari pengalaman

orang lain membuatmu

bijaksana…”

-dr Radz