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Vectors Linear Independent

by CSNabeel
Tags: independent, linear, vectors
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CSNabeel
#1
Aug25-09, 11:21 AM
P: 12
1. The problem statement, all variables and given/known data
Considering the following vectors R[tex]^{4}[/tex]:

v1 = (1,2,0,2) v2 = (2,3,1,4) v3 = (0,1,-1,0)

Determine if these vectors are linearly independent. Let S be the linear span of the three vectors. Define a basis and the dimensions of S. Express the vector v=(3,5,1,6) as a linear combination of the three vectors. Can this be achieved in a unique way? Justify your answer?

2. Relevant equations
I tried to put it into matrix form and reduce via row echolon but I'm not if this is the correct or proper way


3. The attempt at a solution

[ 1 2 0 2
2 3 1 4
0 1 -1 0
3 5 1 6]

[ 1 2 0 2
0 -1 1 0
0 1 -1 0
0 0 0 0 ]

x +2y = 2
y - z = 0
-y + 2 = 0
therefore
y=z making it linearly independent
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Дьявол
#2
Aug25-09, 11:51 AM
P: 365
You need to prove that p=q=r=0, for v1,v2,v3 to be linear independent:

[tex]pv_1 + qv_2 +rv_3=0[/tex]

[tex]p(1,2,0,2)+q(2,3,1,4)+r(0,1,-1,0)=0[/tex]

You should express the vector v in same manner as linear combination of v1,v2,v3: i.e pv1+qv2+rv3=v

p,q,r are random scalars.

Regards.
CSNabeel
#3
Aug25-09, 12:02 PM
P: 12
so with that being said which of the two do I follow from below to work out the answer?

a)

1p + 2q = 0
2p +3q +r = 0
q - r = 0
2p + 4q = 0

b)

1p + 2q = 3
2p +3q +r = 5
q - r = 1
2p + 4q = 6

and if I follow b I'm I right to think that p = 1 q =2 and r = 0

Дьявол
#4
Aug25-09, 12:23 PM
P: 365
Vectors Linear Independent

Quote Quote by CSNabeel View Post
so with that being said which of the two do I follow from below to work out the answer?

a)

1p + 2q = 0
2p +3q +r = 0
q - r = 0
2p + 4q = 0

b)

1p + 2q = 3
2p +3q +r = 5
q - r = 1
2p + 4q = 6

and if I follow b I'm I right to think that p = 1 q =2 and r = 0
Ok, your task have two parts,

a) to check the linear independence of the vectors v1,v2 and v3

b)to find out if the vector v can be represented as linear combination of the vectors v1,v2 and v3.

So you need to solve both a) and b).

Regards.
CSNabeel
#5
Aug25-09, 12:36 PM
P: 12
a)

1p + 2q = 0 (1)
2p +3q +r = 0 (2)
q - r = 0 (3)
2p + 4q = 0 (4)

(3) q = r
(1) p = -2q
put (3)and(1) into (2) 2(-2q) + 3(q) +q = -4q +3q + q = 0

p=-2
q = 1
r = 1

vectors are dependent


b)

1p + 2q = 3 (1)
2p +3q +r = 5 (2)
q - r = 1 (3)
2p + 4q = 6 (4)

(3) q - 1 = r
(3) into (1) 2p + 3q + (q-1) = 5 ; 2p +4q = 6 (same as 4)
(4) can be divide by 2 to equal (1) answer therefore is

p = 1
q = 1
r = 0

so it that then correct?

Thank you by the way your really helpful
Дьявол
#6
Aug25-09, 12:56 PM
P: 365
I am glad that I helped you.

Just a little correction:
a)
r=q
p=-2q
q any number in R, you chose q=1

The vectors are linear dependent

b)
r=q-1
p=3-2q
q any number in R, you chose it q=1

Regards.


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