Solving a system on linear equations

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Homework Help Overview

The discussion revolves around solving a system of linear equations with more variables than equations, specifically involving six variables and four equations. The original poster presents their attempts at manipulating the equations using row operations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of having more variables than equations, questioning the validity of the problem statement. Some suggest moving certain variables to the right-hand side to simplify the system.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the problem. Guidance has been offered regarding the nature of the solutions, including the possibility of infinitely many solutions due to the excess of variables.

Contextual Notes

There is a noted discrepancy between the number of variables and equations, which raises questions about the problem's setup and the nature of the solutions. The original poster confirms the problem is as stated in their textbook.

noamo48
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Homework Statement


Row A) x1 + 2x2 - x3 + x4 + 2x5 = 1
Row B) 2x1 + x2 + x3 - x4 + 2x6 = -1
Row C) x1 + 4x2 - 2x3 + x4 - x5 = 0
Row D) x1 + x2 + 3x3 + x4 + x6 = 2


Homework Equations


Ri <--> Rj
Ri --> cRj, c not equal to 0
Ri --> Ri + cRj, c [tex]\neq[/tex] 0, i [tex]\neq[/tex] j


The Attempt at a Solution


1) Swap rows B and C

2a) Add (-1) x Row A to Row B
2b) Add (-2) x Row A to Row C
2c) Add (-1) x Row A to Row D

3a) Multiply Row B by (1/2)
3b) Add (-2) x Row B to Row A
3c) Add (3) x Row B to Row C
3d) Add Row B to Row D

4a) Multiply Row C by (2/3)
4b) Add (1/2) x Row C to Row B
4c) Add (-7/2) x Row C to Row D

5a) Multiply Row D by (1/7)
5b) Add (-1) x Row D to Row A
5c) Add Row D to Row B
5d) Add (2) x Row D to Row C

I wind up with a very funky solution set full of large fractions as coefficients...can someone let me know what they get please so I can compare...thanks!
 
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The first thing you need to recognize is that there are more variables (x1-x6) than there are equations. This means there are either no solutions or an infinite number of solutions. Are you sure the problem statement is correct?
 
I am positive that is the problem in my book...
 
I don't know how your instructor would want you to approach this problem but I'd solve it by moving x5 and x6 to the right hand side and solve the resulting 4 x 4 problem recognizing that x5 and x6 can take on arbitrary values (meaning there are infinitely many solutions).
 

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