What coefficents make these L.I. Vectors 0? I row reduced, and

[mr_coffee]

Hello everyone, its me.
I had a question again:
I was suppose to determine if these vectors, A, B, and C, and D Linear independent, Well A, B, C are linear independent, 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!

 

[shmoe]

You found them to be linearly independent, 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 independence?)

 

[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!

 

[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.