Testing linear dependence of three vectors

[doctordiddy]

Homework Statement

http://imgur.com/P9udvTs

Homework Equations

The Attempt at a Solution

So I set the scalar multiples of a, b, and c as x,y,z

so i had 3 equations

4x-4y+4z=0
-4x+4y-4z=0
-2x-4y-5z=0

i tried solving it numerous times each time trying to use a different x value, but i cannot get it. Did I even do it correctly up to the point where i get my 3 equations?

Thanks for any help

 

[Rikardus]

The equations are correct. Notice the first and second equations are equivalent so you'll only really use the second and third equation and will get an undetermined solution.
Write x and y as functions of z for, example. You will hopefully arrive at a solution where (3/2)*a + (1/2)*b - 1*c = 0.

 

[HallsofIvy]

Homework Statement

http://imgur.com/P9udvTs

Homework Equations

The Attempt at a Solution

So I set the scalar multiples of a, b, and c as x,y,z

so i had 3 equations

4x-4y+4z=0
-4x+4y-4z=0

The second equation is just -1 times the first so you really have just two equations.

-2x-4y-5z=0

i tried solving it numerous times each time trying to use a different x value, but i cannot get it. Did I even do it correctly up to the point where i get my 3 equations?

Thanks for any help

"get it"? What are you trying to get? What do you mean by "use a different x value"?

As said, you really just have two equations: 4x- 4y+ 4z= 0 and -2x- 4y- 5z= 0.
If you subtract the second equation from the first, you eliminate y leaving 6x+ 9z= 0 so that 9z= -6z and then z= -(2/3)x. Putting that back into the first equation 4x- 4y+ 4(-(2/3)x)= 4x- 4y- (8/3)x= (4/3)x- 4y= 0 so that 4y= (4/3)x and then y= (1/3)x.

That is, for any value of x, y= (1/3)x and z= -(2/3)x is a solution. x= 3, y= 1, z= -2 is one solution, x= 6, y= 2, z= -4 is another. The set of all solutions to this system of equations forms a one-dimensional vector space.