Linear Algebra and Its Applications, 4th Edition

Published by Brooks Cole
ISBN 10: 0030105676
ISBN 13: 978-0-03010-567-8

Chapter 1 - Section 1.3 - An Example of Gaussian Elimination - Problem Set - Page 16: 7

Answer

2; 0; (3, -1)

Work Step by Step

To break down elimination permanently, we must make the system of equations have no solutions. A way to do this is to make the first equation equal to the second equation by multiplication of a constant. If we make a=2, our system will be: 2x+3y=-3 4x+6y=6 Which can easily be seen as having no solutions after multiplying the first equation by 2. 4x+6y cannot equal 6 and -6 at the same time: 4x+6y=-6 4x+6y=6 To break down elimination temporarily, we must make the initial system impossible to eliminate while still having a solvable equation through other means (i.e. switching rows and back-substitution). If a=0, our new system of equations would be: 3y=-3 4x+6y=6 From this system, you cannot eliminate x from the second equation since there is no x variable in the first equation. You must solve the equation through other means. If we switch the position of the two equations, we will have a triangular system of equations and it will be possible to solve for x and y using back substitution. 4x+6y=6 3y=-3 3y=-3 y=-1 Substituting y=-1 into 4x+6y=6: 4x+6(-1)=6 4x-6=6 4x=12 x=3 (3, -1)
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