Final answer:
Three linearly dependent vectors in R₃ that are not multiples of each other can be found by creating a set where no single vector is an exact scalar multiple of another yet their linear combination can result in the zero vector, such as {(1, 0, -1), (0, 1, -1), (1, 1, -2)}.
Explanation:
Finding three linearly dependent vectors in R₃ (three-dimensional space) such that none of them is a multiple of the other requires understanding linear dependence and the structure of vectors in three dimensions. A set of vectors is linearly dependent if there are scalars, not all zero, such that a linear combination of these vectors gives the zero vector. In simpler terms, one of the vectors can be written as a combination of the others.
An example of such a set of vectors is {(1, 2, 3), (2, 4, 6), (3, 6, 9)}. At first glance, it might seem that each vector is a multiple of the others, but let's consider a slightly altered set: {(1, 0, -1), (0, 1, -1), (1, 1, -2)}. In this set, no single vector is an exact scalar multiple of another. However, they are linearly dependent because they satisfy the condition that a linear combination of these vectors equals the zero vector, explicitly: 1*(1, 0, -1) + 1*(0, 1, -1) - 1*(1, 1, -2) = (0, 0, 0).
This example demonstrates that it is possible to have linearly dependent vectors in R₃ that are not simple multiples of each other, but together, they fail to span the three-dimensional space entirely, indicating their linear dependence.
