Solving a system on linear equations

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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).