Explanation Given: S = { A 1 , A 2 , A 3 , A 4 , A 5 } , where A 1 = [ 1 − 2 1 − 1 ] A 2 = [ 2 − 3 4 − 3 ] A 3 = [ − 1 1 − 3 2 ] A 4 = [ 1 − 1 4 0 ] , and A 5 = [ 12 − 17 30 − 11 ] Approach: Let, S be a vector space, and suppose B = { A 1 , A 2 , A 3 , ⋅ ⋅ ⋅ , A n } be a spanning set for S . If B is linearly independent, then B is a basis for S . Calculation: Suppose, a 1 A 1 + a 2 A 2 + a 3 A 3 + a 4 A 4 + a 5 A 5 = 0 … ( 1 ) a 1 [ 1 − 2 1 − 1 ] + a 2 [ 2 − 3 4 − 3 ] + a 3 [ − 1 1 − 3 2 ] + a 4 [ 1 − 1 4 0 ] + a 5 [ 12 − 17 30 − 11 ] = [ 0 0 0 0 ] [ a 1 + 2 a 2 − a 3 + a 4 + 12 a 5 − 2 a 1 − 3 a 2 + a 3 − a 4 − 17 a 5 a 1 + 4 a 2 − 3 a 3 + 4 a 4 + 30 a 5 − a 1 − 3 a 2 + 2 a 3 − 11 a 5 ] = [ 0 0 0 0 ] Compare both sides and form the augmented matrix. a 1 + 2 a 2 − a 3 + a 4 + 12 a 5 = 0 − 2 a 1 − 3 a 2 + a 3 − a 4 − 17 a 5 = 0 a 1 + 4 a 2 − 3 a 3 + 4 a 4 + 30 a 5 = 0 − a 1 − 3 a 2 + 2 a 3 − 11 a 5 = 0 The augmented matrix is, [ A | θ ] = [ 1 2 − 1 1 12 − 2 − 3 1 − 1 − 17 1 4 − 3 4 30 − 1 − 3 2 0 − 11 | 0 0 0 0 ] Reduce the above matrix to row echelon form. [ 1 2 − 1 1 12 0 1 − 1 1 7 0 0 0 1 4 0 0 0 0 0 | 0 0 0 0 ] The system of equations from above matrix is, a 1 + 2 a 2 − a 3 + a 4 + 12 a 5 = 0 a 2 − a 3 + a 4 + 7 a 5 = 0 a 4 + 4 a 5 = 0 Further, a 1 = − 2 a 2 + a 3 − a 4 − 12 a 5 a 2 = a 3 − a 4 − 7 a 5 a 4 = − 4 a 5 That is, a 1 = − a 3 − 2 a 5 … ( 2 ) a 2 = a 3 − 3 a 5 … ( 3 ) a 4 = − 4 a 5 … ( 4 ) Substitute, 1 for a 3 and 0 for a 5 to find a 1 , a 2 and a 4 . a 1 = − 1 a 2 = 1 a 4 = 0 Substitute, − 1 for a 1 ; 1 for a 2 ; 1 for a 3 ; 0 for a 4 ; 0 for a 5 in equation ( 1 ) To determine (b) To find: A subset of S that is a basis for Sp ( S ) . Express each element of S that does not appear in the basis as a linear combination of the basis vectors. To determine (c) To find: A subset of S that is a basis for Sp ( S ) . Express each element of S that does not appear in the basis as a linear combination of the basis vectors.

Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)

5th Edition

ISBN: 9780134689531

Author: Lee Johnson, Dean Riess, Jimmy Arnold

Publisher: PEARSON

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Chapter 5.SE, Problem 6SE

To determine

(a)

To find:

A subset of S that is a basis for Sp(S). Express each element of S that does not appear in the basis as a linear combination of the basis vectors.

To determine

(b)

To find:

A subset of S that is a basis for Sp(S). Express each element of S that does not appear in the basis as a linear combination of the basis vectors.

To determine

(c)

To find:

A subset of S that is a basis for Sp(S). Express each element of S that does not appear in the basis as a linear combination of the basis vectors.

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Chapter 5.SE, Problem 5SE

Chapter 5.SE, Problem 7SE

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Chapter 5 Solutions

Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)

Ch. 5.2 - For u,v and w given in given Exercises 13,...Ch. 5.2 - For u,v and w given in given Exercises 13,...Ch. 5.2 - For u,v and w given in given Exercises 13,...Ch. 5.2 - Prob. 4E Ch. 5.2 - Prob. 5E Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...Ch. 5.2 - Prob. 7E Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...

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(a) To find: A subset of S that is a basis for Sp ( S ) . Express each element of S that does not appear in the basis as a linear combination of the basis vectors .

Solution Summary: The author analyzes the subset of S that is a basis for Sp(S).

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(a)

To find:

A subset of S that is a basis for Sp(S). Express each element of S that does not appear in the basis as a linear combination of the basis vectors.

(b)

To find:

A subset of S that is a basis for Sp(S). Express each element of S that does not appear in the basis as a linear combination of the basis vectors.

(c)

To find:

A subset of S that is a basis for Sp(S). Express each element of S that does not appear in the basis as a linear combination of the basis vectors.

Want to see the full answer?

Chapter 5 Solutions

(a) To find: A subset of S that is a basis for Sp ( S ) . Express each element of S that does not appear in the basis as a linear combination of the basis vectors .

Solution Summary: The author analyzes the subset of S that is a basis for Sp(S).

Videos

(a) To find: A subset of S that is a basis for Sp(S). Express each element of S that does not appear in the basis as a linear combination of the basis vectors.

(b) To find: A subset of S that is a basis for Sp(S). Express each element of S that does not appear in the basis as a linear combination of the basis vectors.

(c) To find: A subset of S that is a basis for Sp(S). Express each element of S that does not appear in the basis as a linear combination of the basis vectors.

Want to see the full answer?

Chapter 5 Solutions

(a)

To find:

A subset of S that is a basis for Sp(S). Express each element of S that does not appear in the basis as a linear combination of the basis vectors.

(b)

To find:

A subset of S that is a basis for Sp(S). Express each element of S that does not appear in the basis as a linear combination of the basis vectors.

(c)

To find:

A subset of S that is a basis for Sp(S). Express each element of S that does not appear in the basis as a linear combination of the basis vectors.