Explanation Approach: If I denote the ( n × n ) identity matrix, and E be the matrix that results when an elementary column operation is applied to the identity matrix I , then the matrix E is called elementary matrix . Calculation: Reduce the matrix A to a lower triangular matrix L by the following sequence of column operations as follows: A = [ 0 − 1 2 1 3 − 1 1 2 1 ] → C 1 ↔ C 2 [ − 1 0 2 3 1 − 1 2 1 1 ] → C 3 + 2 C 1 [ − 1 0 0 3 1 5 2 1 5 ] → C 3 − 5 C 2 [ − 1 0 0 3 1 0 2 1 0 ] Therefore, A E 1 E 2 E 3 = L . Here, E 1 = [ 0 1 0 1 0 0 0 0 1 ] , E 2 = [ 1 0 2 0 1 0 0 0 1 ] , E 3 = [ 1 0 0 0 1 − 5 0 0 1 ] , and L = [ − 1 0 0 3 1 0 2 1 0 ] . Evaluate the product Q = E 1 E 2 E 3 as, Q = [ 0 1 0 1 0 0 0 0 1 ] [ 1 0 2 0 1 0 0 0 1 ] [ 1 0 0 0 1 − 5 0 0 1 ] = [ 0 1 0 1 0 2 0 0 1 ] [ 1 0 0 0 1 − 5 0 0 1 ] = [ 0 1 − 5 1 0 2 0 0 1 ] Thus, Q = [ 0 1 − 5 1 0 2 0 0 1 ]

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 6.5, Problem 18E

To determine

To find:

The elementary matrices E1, E2, and E3 and the value of the product Q=E1E2E3 and verify that AQ=L and det(Q)=det(QT).

A=[0−1213−1121]

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Chapter 6.5, Problem 17E

Chapter 6.5, Problem 19E

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Chapter 6 Solutions

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

Ch. 6.2 - In Exercises 1-8, evaluate the determinant of the...Ch. 6.2 - Prob. 2E Ch. 6.2 - In Exercises 1-8, evaluate the determinant of the...Ch. 6.2 - Prob. 4E Ch. 6.2 - Prob. 5E Ch. 6.2 - In Exercises 1-8, evaluate the determinant of the...Ch. 6.2 - In Exercises 1-8, evaluate the determinant of the...Ch. 6.2 - Prob. 8E Ch. 6.2 - In Exercises 9-14, calculate the cofactors...Ch. 6.2 - Prob. 10E

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The elementary matrices E 1 , E 2 , and E 3 and the value of the product Q = E 1 E 2 E 3 and verify that A Q = L and det ( Q ) = det ( Q T ) . A = [ 0 − 1 2 1 3 − 1 1 2 1 ]

Solution Summary: The author explains the elementary matrices E_1,

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To find:

The elementary matrices E1, E2, and E3 and the value of the product Q=E1E2E3 and verify that AQ=L and det(Q)=det(QT).

A=[0−1213−1121]

Want to see the full answer?

Chapter 6 Solutions

The elementary matrices E 1 , E 2 , and E 3 and the value of the product Q = E 1 E 2 E 3 and verify that A Q = L and det ( Q ) = det ( Q T ) . A = [ 0 − 1 2 1 3 − 1 1 2 1 ]

Solution Summary: The author explains the elementary matrices E_1,

Videos

To find: The elementary matrices E1, E2, and E3 and the value of the product Q=E1E2E3 and verify that AQ=L and det(Q)=det(QT). A=[0−1213−1121]

Want to see the full answer?

Chapter 6 Solutions

To find:

The elementary matrices E1, E2, and E3 and the value of the product Q=E1E2E3 and verify that AQ=L and det(Q)=det(QT).

A=[0−1213−1121]