Solving a system on linear equations

AI Thread Summary
The discussion revolves around solving a system of linear equations with more variables than equations, leading to either no solutions or infinitely many solutions. The user attempts various row operations to simplify the equations but ends up with complex fractions in the solution set. Another participant points out that since there are more variables (x1-x6) than equations, the problem likely has infinitely many solutions. They suggest moving x5 and x6 to the right side to focus on a 4x4 system, acknowledging the arbitrary nature of x5 and x6. This approach emphasizes the need to recognize the implications of having more variables than equations in linear systems.
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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 \neq 0, i \neq 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).
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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