# Homework Help: What coefficents make these L.I. Vectors 0? I row reduced, and

1. Nov 26, 2005

### mr_coffee

Hello everyone, its me.
I was suppose to determine if these vectors, A, B, and C, and D Linear Independant, Well A, B, C are linear independant, but D is not, I got

D =
3/5
-4/5
-1/5
0

A =
1
0
0
0

B =
0
1
0
0

C =
0
0
1
0

I row reduced and got:
1 0 0
0 1 0
0 0 1
If they are linearly dependent, determine a non-trivial linear relation - (a non-trivial relation is three numbers which are not all three zero.) Otherwise, if the vectors are linearly independent, enter 0's for the coefficients, since that relationship always holds.
A + B + C + D = 0.

So they want me to find Coefficents that make A + B +C +D= 0, and i'm confused on how i'm suppose to do that! Thanks!

Last edited: Nov 26, 2005
2. Nov 26, 2005

### shmoe

You found them to be linearly independant, they tell you what to pick as coefficients:

"...if the vectors are linearly independent, enter 0's for the coefficients, since that relationship always holds."

Will any other choice of coefficients work? (Remember the definition of linear independance?)

3. Nov 26, 2005

### mr_coffee

Shmoe, sorry u must have gotten my orginal post, after posting it i realized that hah, but i came across a simllair problem, in which A, B, and C, are L.I. But D isn't. And its throwing me off! thanks!

4. Nov 26, 2005

### HallsofIvy

Write it out in detail:
$$\alpha A+ \beta B+ \gamma C+ \delta D= 0$$
is
$$(\alpha, 0, 0, 0)+ (0, \beta, 0, 0)+ (0, 0,\gamma, 0)+ (\frac{3}{5}\delta, -\frac{4}{5}\delta,-\frac{1}{5}\delta, 0)$$
$$= (\alpha+ \frac{3}{5}\delta,\beta-\frac{4}{5}\delta,\gamma-\frac{1}{5}\delta, 0)= (0,0,0,0)$$
So we must have
$$\alpha+ \frac{3}{5}\delta= 0$$
$$\beta-\frac{4}{5}\delta= 0$$
$$\gamma-\frac{1}{5}\delta= 0$$
That gives 3 equations for the 4 unknown numbers $\alpha, \beta, \gamma, \delta$. Of course, there are an infinite number of solutions. I suggest taking $\delta$ equal to some easy (non-zero) number and solving for the others.

Last edited by a moderator: Nov 26, 2005