Explanation Given: The given system of linear equations is, { 2 x + y + 2 z = 4 2 x + 2 y = 5 2 x − y + 6 z = 2 Approach: A matrix derived from a system of linear equations (each written in standard form with the constant term on the right) is the augmented matrix. Thus, obtain augmented matrix based on the given system of equations. Now, to apply Gaussian elimination with back substitution, the augmented matrix has to be converted to the row-echelon form by the elementary row transformation. Calculation: Consider the given system of equations. { 2 x + y + 2 z = 4 2 x + 2 y = 5 2 x − y + 6 z = 2 Here, the variables are already aligned. Now, using the coefficients and constant terms as the matrix entries, the augmented matrix is, [ 2 1 2 ⋮ 4 2 2 0 ⋮ 5 2 − 1 6 ⋮ 2 ] Solve the system by the Gaussian elimination method. Convert the augmented matrix in row echelon form. Apply row operation R 2 → R 2 − R 1 and R 3 → R 3 − R 1 . [ 2 1 2 ⋮ 4 2 2 0 ⋮ 5 2 − 1 6 ⋮ 2 ] → R 3 → R 3 − R 1 R 2 → R 2 − R 1 [ 2 1 2 ⋮ 4 0 1 − 2 ⋮ 1 0 − 2 4 ⋮ − 2 ] Apply row operation R 1 → R 1 − R 2 and R 3 → R 3 + 2 R 2 . [ 2 1 2 ⋮ 4 0 1 − 2 ⋮ 1 0 − 2 4 ⋮ − 2 ] → R 3 → R 3 + 2 R 2 R 1 → R 1 − R 2 [ 2 0 4 ⋮ 3 0 1 − 2 ⋮ 1 0 0 0 ⋮ 0 ] Apply row operation R 1 → 1 2 R 1

Bundle: Precalculus, Loose-leaf Version, 10th + WebAssign Printed Access Card for Larson's Precalculus, 10th Edition, Single-Term

10th Edition

ISBN: 9781337605090

Author: Ron Larson

Publisher: Cengage Learning

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Chapter 8.CR, Problem 23CR

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The solution for the following system of linear equations based on Gaussian elimination with back-substitution method.

{2x+y+2z=42x+2y =52x−y+6z=2

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Chapter 8.CR, Problem 22CR

Chapter 8.CR, Problem 24CR

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Ch. 8.1 - Prob. 1CP Ch. 8.1 - Prob. 2CP Ch. 8.1 - Prob. 3CP Ch. 8.1 - Prob. 4CP Ch. 8.1 - Prob. 5CP Ch. 8.1 - Prob. 6CP Ch. 8.1 - Prob. 7CP Ch. 8.1 - Prob. 8CP Ch. 8.1 - Prob. 9CP Ch. 8.1 - Prob. 1E

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The solution for the following system of linear equations based on Gaussian elimination with back-substitution method. { 2 x + y + 2 z = 4 2 x + 2 y = 5 2 x − y + 6 z = 2

Solution Summary: The author explains how to solve the given system of linear equations based on Gaussian elimination with back-substitution method.

Videos

To find:

The solution for the following system of linear equations based on Gaussian elimination with back-substitution method.

{2x+y+2z=42x+2y =52x−y+6z=2

Want to see the full answer?

Chapter 8 Solutions

The solution for the following system of linear equations based on Gaussian elimination with back-substitution method. { 2 x + y + 2 z = 4 2 x + 2 y = 5 2 x − y + 6 z = 2

Solution Summary: The author explains how to solve the given system of linear equations based on Gaussian elimination with back-substitution method.

Videos

To find: The solution for the following system of linear equations based on Gaussian elimination with back-substitution method. {2x+y+2z=42x+2y =52x−y+6z=2

Want to see the full answer?

Chapter 8 Solutions

To find:

The solution for the following system of linear equations based on Gaussian elimination with back-substitution method.

{2x+y+2z=42x+2y =52x−y+6z=2