The document provides examples and explanations of linear algebra concepts including: 1. Checking if vectors are linear combinations of other vectors by setting up systems of equations and solving for coefficients. 2. Determining if vectors span a vector space by setting up a matrix with the vectors as columns and checking if it is invertible. 3. Examples involve vectors in R2 and R3, linear systems, and matrix operations to check linear dependence and independence.

1. Tutorial MTH 3201
Linear Algebras

2. Tutorial 2

3.  
1.Given u (0, 2, 2), v (1,3, 1)

(a) w (2, 2, 2) 
(b) w (3,1,5)

w k1u k2v (3,1,5) k1 (0, 2, 2) k2 (1,3, 1)
(2, 2, 2)
(k2 , 2k1 3k2 , 2k1 k2 )
k2 3
k2 2 2k1 3k2 1
2k1 3k2 2 2k1 k2 5
2k1 3(2) 2 2k1 5 3
2(k1 ) 4
k1 4 and k2 3
k1 2 and k2 2
Check! Check!
2k1 3k 2 2(2) 3(2) 2 2k1 3k2 2(4) 3(3) 1
2k1 k2 2(2) 2 2
2k1 k2 2(4) 3 5
the vector is linear
linear combination
 
combination with u and v

4. 
(c) w (0, 0, 0) 
(d ) w (0, 4,5)
(0, 0, 0) (k2 , 2k1 3k2 , 2k1 k2 ) (0, 4,5) (k2 , 2k1 3k2 , 2k1 k2 )
k2 0 k2 0
2k1 3k2 0 2k1 3k2 4
2k1 0 0 2k1 k2 5
5
k1 0 and k2 0 k1 2 or and k2 0
2
Check!
Check !
2k1 3k2 2(0) 3(0) 0 2( 2) 3(0) 4
2k1 k2 2(0) 0 0 2k1 k2 2( 2) 0 4
linear combination 
(0, 4, 4) w (0, 4,5)
Not a linear combination

6. 
(a) u 
(3, 2, 1), v (1, 2, 3) ERO

Let b (b1 , b2 , b3 ) 3k1 k2 b1

b ku k v 2k1 2k2 b2
1 2
(b1 , b2b3 ) k1 (3, 2, 1) k2 (1, 2, 3) k1 3k2 b3
(3k1 k2 , 2k1 2k 2 , k1 3k 2 )
3 1 b1 1 3 b3 1 3 b3
R1 R2 2 R1 R2
2 2 b2 R3
2 2 b2 3 R1 R3
0 4 2b3 b2
1 3 b3 3 1 b1 0 8 3b3 b1
b3 b3
1 3 1 3
R2 2b3 b2 8 R2 R3 2b3 b2
4
0 1 0 1
2 4 2 4
0 8 0 0
3b3 b1 b3 2b2 b1
Not span
If b3 2b2 b1 0, the linear system doesn ' t have solution
the vectors don ' t span R 3

7. 
(b) u  
(1, 0, 3) , v (3,1, 0) , w (1, 2, 3)

Let b (b , b , b )
1 2 3

b 
k1u  
k 2 v k3 w
(b1 , b2 , b3 ) k1 (1, 0, 3) k2 (3,1, 0) k3 (1, 2, 3)
(k1 3k2 k3 , k 2 2k3 , 3k1 3k3 )
b1 k1 3k2 k3
b2 k2 2 k3
b3 3k1 3k3
1 3 1
Let A 0 1 2 A 3(6 1) 0 3(1 0)
3 0 3
3(5) 3
3
A 0.Thevectors span R 18

8.  
(c) v1 (2,1, 2), v2 
(6,3, 6), v3 ( 2, 1, 2)
(b1 , b2 , b3 ) k1 (2,1, 2) k2 (6,3, 6) k3 ( 2, 1, 2)
(2k1 6k2 2k3 , k1 3k2 k3 , 2k1 6k2 2k3 )
2 6 2
Let A 1 3 1
2 6 2
A 1(12 12) 3(4 4) 1( 12 12)
0
A 0. A is not invertible.
The system incosistent.
The vectors doesn't span R 3

9. 
(d ) u  
(3, 2, 2) , v (1,1, 0) , w ( 2,1, 2)

Let b (b1 , b2 , b3 )
 
b ku k v k w  
1 2 3
(b1 , b2 , b3 ) k1 (3, 2, 2) k2 (1,1, 0) k3 ( 2,1, 2)
(3k1 k2 2k3 , 2k1 k2 k3 , 2k1 2k3 )
3 1 2
Let A 2 1 1
0 2 2
A 2(3 4) 2 (3 2)
2(7) 2(1)
16
A 0. The vectors span R 3

10. Question 2(e)
• Congratulation!!
• This is your first assignment
• Please submit in tutorial class next week

11.  0 0  1 1  1 1
( a ) u1 , u2 , u3 ,
1 0 0 0 1 0
 0 0
u4 ; v M 22
0 1
 b1 b2
Let b M 22
b3 b4

b 
k1u 
k 2u 
k3u 
k 4u
b1 b2 0 0 1 1 1 1 0 0
k1 k2 k3 k4
b3 b4 1 0 0 0 1 0 0 1
b1 k2 k3 .............................[1]
b2 k2 k3 .............................[2]
b3 k1 k3 .............................[3]
b4 k4 ....................................[4]

12. [1] [2]: b1 b2 2k 2
b1 b2
k2
2
b1 b2
From[2] : b2 k3
2
b1 b2
k3 b2
2
1 1
b2 b1
2 2
1 1
From[3] : b3 k1 ( b2 b1 )
2 2
1 1
k1 b3 b2 b1
2 2
From[4] : k4 b4

The system has solution.All vector b is the linear combination of
   
u1 ,u 2 ,u 3 and u 4 .Hence the vector span the vector space v.

13. 1 0 0 2 0 0
(b) w1 , w2 , w3 ,
0 0 0 0 3 0
0 0
w4 ; v M 22
0 4
b1 b2 1 0 0 2 0 0 0 0
k1 k2 k3 k4
b3 b4 0 0 0 0 3 0 0 4
b1 k1................................[1]
b2 2k 2 .............................[2]
b3 3k3 ..............................[3]
b4 4k 4 ..............................[4]
b2 b3 b4
k1 b1 ; k 2 ; k3 ; k4
2 3 4
vector span the vector space v.

14. (c ) 
p1 1,  2
p x 1,  3
p x2 1, v p2

b ( x ) b0 b1 x b2 x 2
b ( x) k 
 p k 
1 p 1 k 
p 2 2 3 3
(b0 b1 x b2 x 2 ) k1 
p1 k2  2
p k3  3
p
k1 (1) k2 ( x 1) k3 ( x 2 1)
k1 k2 x k2 k3 x 2 k3
( k1 k2 k3 ) (k2 x) ( k3 x 2 )
b0 k1 k2 k3
b1 k2 k3
b2 k3
k1 k2 k3 b0
k1 b1 b2 b0
k1 b0 b1 b2
k2 b1
k3 b2

b ( x) (b0 b1 b2 ) b1  2
p b2  3
p
The vectors span the vector space V

15. 
( d ) q1 
2, q2 x 
1, q 3 x2 x 1; v p2

Let b ( x ) b0 b1 x b2 x 2
 
b ( x ) k1 q1 
k 2 q2 
k3 q3
(b0 b1 x b2 x 2 ) 
k1 q1 
k 2 q2 
k3 q3
k1 (2) k2 ( x 1) k3 ( x 2 x 1)
2k1 k2 x k2 k3 x 2 k3 x k3
(2 k1 k2 k3 ) (k2 x k3 x ) ( k3 x 2 )
b0 2k1 k2 k3 .....................................[1]
b1 k2 k3 ..............................................[2]
b2 k3 .....................................................[3]
[3] in [2] : b1 k2 b2
k2 b1 b2 ............................... .............[4]
[3],[4] in [1] : b0 2 k1 (b1 b2 ) b2
b0 2k1 b1 2b2
b0 b1 2b2
k1
2
b0 b1
k1 b2
2
 b0 b1   
b ( x) ( b2 ) q1 (b1 b2 ) q2 b2 q3
2
The vectors span the vector space V

16. 0 2 1

(a) u 1 
, v 0 
, w 1 ;V R3
3 -3 0

Let k1u 
k2 v 
k3 w 
o
0 2 1 0
k1 1 k2 0 k3 1 0
3 3 0 0
k1 2k 2 k3 0
k1 k3 0
3k1 3k 2 0
0 2 1
Let A 1 0 1
3 3 0
A 0(0 3) 2(0 3) 1(3 0)
9
A 0, A has invertible.
The vector are linearly dependent.

17.  0 0  1 0
(b) v1 , v2 ;V M 22
1 1 1 0

Let k1v1 
k 2 v2 
o
0 0 1 0 0 0
k1 k2
1 1 1 0 0 0
k2 0 0 0
k1 k2 k1 0 0
k2 0
k1 0 k1 0
k1 k2 0
The system has trivial solution, k1 k2 0
The vector arelinearlyindependent

18. (c )  3 x 2 x
p 
1, q x 1; v p2
k1  k2 q 0
p 
k1 (3 x 2 x 1) k2 ( x 1) 0x2 0x 0
3k1 x 2 k1 x k1 k2 x k2 0x2 0x 0
3k1 0 k1 0
k1 k2 0 k2 0
k1 k2 0
The system has trivialsolution, k1 k2 0 k2
The system has trivial solution, k 1 0
The vectors span the vector space V
The vector arelinearlyindependent

19. 
(d ) x1  
(0, 3,1,1) , x2 (5, 5,1, 3) , x 3 ( 1, 0, 5,1); V R4
0   
k1 x1 k2 x2 k3 x 3
k1 (0, 3,1,1) k2 (5, 5,1, 3) k3 ( 1, 0, 5,1)
(5k2 k3 , 3k1 5k2 , k1 k2 k3 , k1 k2 5k3 , k1 3k 2 k3 )
0 5k 2 k3
0 3k1 5k2
0 k1 k2 k3
0 k1 3k2 k3
5 1 0 0 1 3 1 0
3 5 0 0 r4 r1 0 8 15 0
3 r3 r2
1 1 1 0 1 1 15 0
1 3 1 0 0 5 1 0

20. 1 3 1 0 1 3 1 0
r3
r1 r3 0 8 15 0 2
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
5 r3 r4 0 8 15 0 r2 r3 0 1 2 0
r4
0 1 2 0 11
0 8 15 0
0 0 11 0 0 0 1 0
1 3 1 0 1 3 1 0
r3
8 r2 r3 0 1 2 0 2
0 1 0 0
2 r4 r2
0 0 31 0 0 0 1 0
0 0 1 0 0 0 1 0
1 3 1 0 1 0 0 0
r3
31
0 1 0 0 r3 3 r2 r1
0 1 0 0
2 r4 r2
0 0 1 0 0 0 1 0
0 0 0 0 0 0 0 0

21. 
(5) v1 
( 1, 3) , v2 
(2,1) R 2 are w1 ( 1, 2)
  
and w2 ( 3, 5) element span ( v1 , v2 )

w1 ( 1, 2) .

Let w1 
k1 v1 
k 2 v2
( 1, 2) k1 ( 1, 3) k 2 ( 2,1)
( k1 2k 2 , 3k1 k2 )
5
k1 2k 2 1 k1
7
1
3k1 k2 2 k2
7
 5 1 
w1 v1 v2
7 7
  
w1 element is in span ( v1 , v2 )

22.    
w2 ( 3,5) . Let w2 k1 v1 k2 v2
( 3,5) k1 ( 1,3) k2 ( 2,1)
( k1 2k2 ,3k1 k2 )
13
k1 2k2 3 k1
7
4
3k1 k2 5 k2
7
 13 v 4 v
w2  
1 2
7 7
  
w2 element isin span ( v1 , v2 )

23. 1 2

a ) x1 
, x2 12 b) 
y1 , 2
y
3 2

k1 x1 
k 2 x2 0 k1 
y1 k 
2 y 2 0
1 2 k1 0
k1 0
12
k2 0 2 k2 0
3
1 2
Let A = 12 , Let A = ,
3
2
Linearly dependent: A 0 Linearly dependent: A 0
2 1 2
4 0
0
4
1 1 ( 2)( 2) 0
( )( ) 0
2 2 2
1
2

24. Assume 
k1 x1 
k2 x2  
k3 x3 = z
1 1 k1 0
Question 7
1
1
1 k
1 k2
3
0
0
Ak 0
• Congratulation!! 1
1
1 1
1 1
•A
1 1 1 1
This is your second assignment
( 1) 3
( 1)
• Pleasedependent Atutorial class next week
3
Linearly
submit in 0
2 3
3
2 3 0
3
3 2 0
2, 1

k1 x1 
k 2 x2 
0
k1 0, k 2 0 linearly dependent
k1 k2 k3 0 linearly independent
A 0 linearly independent

25. second method
1 1

a ) x1 
1 , x2 
, x3 1
1 1

k1 x1 
k 2 x2 
k3 x3 0
1 1 k1 0
1 1 k2 0
1 1 k3 0
1 1
Let A = 1 1 ,
1 1
A 0
2
( 1) ( 1)( 1) ( 1)(1 )
3
( 1) ( 1 )
3
1 1 )
3
3 2
1, 2

26.   
(8) v1 , v2 is linearly independent v3 span  
v1 , v2 ,
  
then, v1 , v2 , v3 also linearly independent.

v3 span  
v1 , v2

v3  
v1
v2 ................................[1]
1 2
 
Assume that v1 , v2 , v3are linearly independent.
B ,B ,B  
IR : B v , B v , B v  0
1 2 3 1 1 2 2 3 3
B3 0,

B1 v1 
B2 v2 0

B3 v3 
B1 v1 
B2 v2
 B1  B2 
v3 v1 v2
B3 B3
contradiction with equation1
  
Hence v , v ,v is linearly independent
1 2 3

27. (9) Prove theorem 1.13 in chapter 1
Theorem 1.1.3 (Cauchy-Schwarz Inequality)
 
If u=(u1 ,u 2 ,u 3 ............u n ) and v=(v1 ,v 2 ,v3...........v n )
are vectors in R n

then, u.v u v
1 1
2 2 2 2 2 2 2 2 2 2
u1v1 u2 v2 ......... un vn u 1 u2 u3 .... un v
1 v2 v3 .... v3

28. Reference from Wikipedia(universal reference) ;p

29. “Belajar dari kesalahan
membuatmu dewasa dan
belajar dari pengalaman
orang lain membuatmu
bijaksana…”
-dr Radz