Vectors Linear Independent - Are These Vectors Linearly Independent?

[CSNabeel]

Homework Statement

Considering the following vectors R^{4}:

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?

Homework 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

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

 

[Дьявол]

You need to prove that p=q=r=0, for v1,v2,v3 to be linear independent:

pv_1 + qv_2 +rv_3=0

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

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]

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]

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]

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 dependentb)

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]

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.