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

In summary, the conversation discusses determining if four given vectors are linearly independent and finding coefficients that make their sum equal to 0. The vectors A, B, and C are linearly independent while D is not. The solution involves setting up equations based on the definition of linear independence and solving for the coefficients.
  • #1
mr_coffee
1,629
1
Hello everyone, its me.
I had a question again:
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!
 
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  • #2
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
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
Write it out in detail:
[tex]\alpha A+ \beta B+ \gamma C+ \delta D= 0[/tex]
is
[tex](\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)[/tex]
[tex]= (\alpha+ \frac{3}{5}\delta,\beta-\frac{4}{5}\delta,\gamma-\frac{1}{5}\delta, 0)= (0,0,0,0)[/tex]
So we must have
[tex]\alpha+ \frac{3}{5}\delta= 0[/tex]
[tex]\beta-\frac{4}{5}\delta= 0[/tex]
[tex]\gamma-\frac{1}{5}\delta= 0[/tex]
That gives 3 equations for the 4 unknown numbers [itex]\alpha, \beta, \gamma, \delta[/itex]. Of course, there are an infinite number of solutions. I suggest taking [itex]\delta[/itex] equal to some easy (non-zero) number and solving for the others.
 
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FAQ: What coefficents make these L.I. Vectors 0? I row reduced, and

What are coefficients and L.I. Vectors?

Coefficients are the numbers that multiply a variable in an algebraic expression. L.I. Vectors are linearly independent vectors, meaning they are not linear combinations of each other.

How do you determine if vectors are linearly independent?

To determine if vectors are linearly independent, you can row reduce the vectors and check if there are any rows of zeros. If there are no rows of zeros, the vectors are linearly independent.

What does it mean if the row reduced vectors have all zero rows?

If the row reduced vectors have all zero rows, it means that the vectors are linearly dependent, and one or more vectors can be written as a linear combination of the others.

Can the coefficients of L.I. Vectors ever be 0?

Yes, the coefficients of L.I. Vectors can be 0. This would mean that the corresponding variables do not affect the linear combination of the vectors. However, for the vectors to be linearly independent, at least one coefficient for each vector must be non-zero.

What is the significance of finding the coefficients that make the L.I. Vectors 0?

Finding the coefficients that make the L.I. Vectors 0 is important because it helps determine if the vectors are linearly independent or dependent. It also allows for the creation of a basis for the vector space, which is essential in many mathematical and scientific applications.

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