Show that the set of vectors {(−4, 1, 3), (5, 1, 6), (6, 0, 2)} does not span R3, but that it does span the subspace of R3 consisting of all vectors lying in the plane with equation x + 13y − 3z = 0.

Show that the set of vectors {(−4, 1, 3), (5, 1, 6), (6, 0, 2)} does not span R3, but that it does span the subspace of R3 consisting of all vectors lying in the plane with equation x + 13y − 3z = 0.

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN:9781305658004

Author:Ron Larson

Publisher:Ron Larson

Chapter5: Inner Product Spaces

Section5.CM: Cumulative Review

Problem 23CM

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Linear Algebra

In mathematics, problems can be solved by the arithmetic operators like addition, subtraction, multiplication, and division. In algebra, we represent the numbers by any letter called a variable. If these variables are of degree 1, then it is studied under linear algebra. The first-degree equations involving algebraic expressions are called linear equations. That is, if you represent it by drawing a graph you will get a straight line.

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Show that the set of vectors {(−4, 1, 3), (5, 1, 6), (6, 0, 2)} does not span R³, but that it does span the subspace of R3 consisting of all vectors lying in the plane with equation x + 13y − 3z = 0.

Quantities that have magnitude and direction but not position. Some examples of vectors are velocity, displacement, acceleration, and force. They are sometimes called Euclidean or spatial vectors.

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Given a set of vectors {(−4, 1, 3), (5, 1, 6), (6, 0, 2)}.

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Begin with writing the set of vectors in matrix form.

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The vectors in component form will be given as,

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Expressing the above system in matrix form,

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We reduce the matrix to row echelon form:

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