Proving Zero Vectors for A, B, and C

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Homework Help Overview

The discussion revolves around proving the linear independence of the vectors A, B, and C, defined as A= [1 1 -1], B=[0 1 2], and C=[3 0 1]. The original poster attempts to show that the only solution to the equation rA + sB + tC = 0 is r=s=t=0.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants question the original poster's understanding of the problem, suggesting that it is about vectors rather than matrices. Others inquire about how to demonstrate that the equation has only the trivial solution.

Discussion Status

Participants are exploring different interpretations of the problem and discussing methods to prove linear independence. Guidance has been offered regarding the use of matrix representation and row reduction to analyze the system of equations.

Contextual Notes

There is some confusion regarding terminology, specifically the use of "matrice" and the distinction between vectors and matrices. The original poster's approach is questioned, and there is a need for clarity on the definitions and methods involved in proving linear independence.

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Homework Statement


A= [1 1 -1], B=[0 1 2], C=[3 0 1]

Show that r=s=t=0



Homework Equations





The Attempt at a Solution


I said r=s=t=0
so

0[ 1 1 -1] + 0[0 1 2] + 0[3 0 1] so
0 + 0 + 0= 0
so rA + sB + tC = 0 and r=s=t=0

Is that right way to do it?
 
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Is that the right way to do what? You title is misleading; it is apparently not a matrix (there is no such word as "matrice" in English) problem, but a problem about vectors.

Are you trying to show that the vectors are linearly independent? If so, you have to show that the equation rA + sB + tC = 0 has only one solution for the constants r, s, and t. The equation rA + sB + tC = 0 always has what is sometimes called the trivial solution (namely r = s = t = 0). Your job is to show that there are no other solutions for these constants.
 
Could you please help me on how to start the problem because I don't know how to prove it only has one solution etc..
 
Write the vectors as columns in a matrix, and then row-reduce the matrix. If you end up with three nonzero rows (three rows, each with a nonzero leading entry), then that means that the only solution is r = 0, s = 0, t = 0, and there are no other solutions.

If you end up with one or more rows that have all zeros, then there are multiple solutions.

To help you understand what is going on here, think back to we're trying to do, namely find all solutions of the equation rA + sB + tC = 0. Try to picture this equation with the vectors A, B, and C written in vertical form.

The preceding equation can be written as matrix equation that looks like this:
[A B C][r s t]^T = [0] (Note: [r s t]^T is a column vector)

This matrix equation can be written as an augmented matrix like so:
[1 0 3 | 0]
[1 1 0 | 0]
[-1 2 1 | 0]

Row reduce this matrix as described at the beginning of this post.

Is that enough for you to start in on?
 
Mark44 said:
Is that the right way to do what? You title is misleading; it is apparently not a matrix (there is no such word as "matrice" in English) problem, but a problem about vectors.

Err. . . vectors are an 1 x n (row) or n x 1 (column) matrix. . .
 
descendency said:
Err. . . vectors are an 1 x n (row) or n x 1 (column) matrix. . .
True enough.
 

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