tutorial 2 mth 3201
TRANSCRIPT
Tutorial MTH 3201Linear Algebras
Tutorial 2
( ) (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
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
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
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
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
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
Question 2(e)
• Congratulation!!
• This is your first assignment
• Please submit in tutorial class next week
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
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.
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
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
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
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
-
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
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
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
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
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
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
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
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
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
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
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
Reference from Wikipedia(universal reference) ;p
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