Are Vectors a, b, and c Linearly Independent?

  • Thread starter Thread starter halfoflessthan5
  • Start date Start date
  • Tags Tags
    Linearly Vectors
halfoflessthan5
Messages
15
Reaction score
0
just a quick one:

Homework Statement


Show that the vectors a=2i -2j, b=3j - k and c = i + 2j +k are linearly independant


Homework Equations





The Attempt at a Solution



What does 'linearly independent' mean and what's the test for it? Its from a really old exam paper so i might just know this theory under a different name.

thankyou :smile:
 
Physics news on Phys.org
It means that the equation x*a+y*b+z*c=0 (where x,y,z are scalars and a,b,c are your vectors) has only the trivial solution x=y=z=0.
 
Is this right then:

(2i -2j)x + (3j - k)y + (i + 2j +k)z = 0

multiply out and rearrange

(2x + z)i + (-2x + 3y +2 z)j + (z - y)k = 0

comparing is js and ks on each side

2x + z = 0
-2x + 3y + 2z = 0
z - y = 0

as matrices

[2 0 1] [x] = [0]
[-2 3 2] [y] = [0]
[0 -1 1] [z] = [0]

(like in the eigenvalue problem) there is a non-trivial solution only if determinent of the co-efficients is zero

detM= 12

=/= 0

=> vectors a,b,c where linearly independant
 
Looks right.
 
thankyouuu
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top