Explanation Let A , B and C be three players and x , y and z be starting dollars respectively. In first round, Player A loses first, then the players B and C are winners. Each player has the following amount. Player A has an amount x - y - z . Player B has an amount 2 y . Player C has an amount 2 z . In second round, Player B loses, then the players A and C are winners. Now, each player has the following amount. Player A has an amount 2 x - 2 y - 2 z . Player B has an amount 2 y - x - y - z . Player C has an amount 4 z . In third round, Player C loses, then the players A and B are winners. Now, each player has the following amount. Player A has an amount 4 x - 4 y - 4 z . Player B has an amount - 2 x + 6 y - 2 z . Player C has an amount - x - y + 7 z . It is given that after three rounds each player has $ 24 . 4 x - 4 y - 4 z = 24 - 2 x + 6 y - 2 z = 24 - x - y + 7 z = 24 The augmented matrix for a system of linear equations is, Let A = 4 - 4 - 4 24 - 2 6 - 2 24 - 1 - 1 7 24 Condition for reduced echelon form. 1. First nonzero entry in any row is the only nonzero entry in its column. The given matrix is not in reduced echelon form. To transform to reduced echelon matrix elementary row operations are applied. The elementary row operation R 1 × 1 4 has to be done in R 1 . A = 4 × 1 4 - 4 × 1 4 - 4 × 1 4 24 × 1 4 - 2 6 - 2 24 - 1 - 1 7 24 Simplifying, A = 1 - 1 - 1 6 - 2 6 - 2 24 - 1 - 1 7 24 The elementary row operation R 2 + 2 R 1 has to be done in R 2 A = 1 - 1 - 1 6 - 2 + 2 · 1 6 + 2 · - 1 - 2 + 2 · - 1 24 + 2 · 6 - 1 - 1 7 24 Simplifying, A = 1 - 1 - 1 6 0 4 - 4 36 - 1 - 1 7 24 The elementary row operation R 3 + R 1 has to be done in R 3 . A = 1 - 1 - 1 6 0 4 - 4 36 - 1 + 1 - 1 + - 1 7 + - 1 24 + 6 Simplifying, A = 1 - 1 - 1 6 0 4 - 4 36 0 - 2 6 30 The elementary row operation R 2 × 1 4 has to be done in R 2 . A = 1 - 1 - 1 6 0 × 1 4 4 × 1 4 - 4 × 1 4 36 × 1 4 0 - 2 6 30 Simplifying, A = 1 - 1 - 1 6 0 1 - 1 9 0 - 2 6 30 The elementary row operation R 3 + 2 R 2 has to be done in R 3

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 1.2, Problem 51E

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The money each have on the beginning of the game.

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Chapter 1.2, Problem 50E

Chapter 1.2, Problem 52E

Chapter 1 Solutions

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

Ch. 1.1 - Which of the equations in Exercises 1-6 are...Ch. 1.1 - Which of the equations in Exercises 1-6 are...Ch. 1.1 - Which of the equations in Exercises 1-6 are...Ch. 1.1 - Which of the equations in Exercises 1-6 are...Ch. 1.1 - Which of the equations in Exercises 1-6 are...Ch. 1.1 - Which of the equations in Exercises 1-6 are...Ch. 1.1 - In Exercises 7-10, coefficients are given for a...Ch. 1.1 - In Exercises 7-10, coefficients are given for a...Ch. 1.1 - In Exercises 7-10, coefficients are given for a...Ch. 1.1 - In Exercises 7-10, coefficients are given for a...

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The money each have on the beginning of the game.

Solution Summary: The author explains the augmented matrix for a system of linear equations: Let A, B and C be three players with x, y, and z being starting dollars.

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To find: The money each have on the beginning of the game.

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To find:

The money each have on the beginning of the game.