Reproduce a solution set to a linear system with 2 equations and 8 variables

I was mistakenly trying to express x1 and x4 as linear combinations of the remaining variables, but of course it should have been the other way around. Thanks again for your help.In summary, the given solution set of [x1,x2,x3,x4,x5,x6,x7,x8]^T = [1,-1,0,0,0,0,0,0]^T, [0,-1,1,0,0,0,0,0]^T,[0,0,0,1,-1,0,0,0]^T,[0,0,0,-1,0,1,0,0]^T,[0,0,0,0,-1,
  • #1
Alex1812
7
0

Homework Statement


Show how the following solution set: [x1,x2,x3,x4,x5,x6,x7,x8]^T =[1,-1,0,0,0,0,0,0]^T, [0,-1,1,0,0,0,0,0]^T,[0,0,0,1,-1,0,0,0]^T,[0,0,0,-1,0,1,0,0]^T,[0,0,0,0,-1,0,1,0]^T,[0,0,0,0,0,-1,0,1] is obtained from the two linear equations x1+x2+x3-x4-x5-x6-x7-x8=0 and x4+x5+x6+x7+x8=0.


Homework Equations


none given other than the two linear equations.


The Attempt at a Solution


The two equations can be put into a matrix A= [1,1,1,-1,-1,-1,-1,-1,]; [0,0,0,1,1,1,1,1]. Then row 2 is added to row 1 and the following solution is obtained (where v1, v2, .. v6 are arbitrary values) x1=-v1-v2; x2= v1, x3=v2, x4= -v3 -v4 -v5 -v6; x5=v3; x6=v4; x7=v5; x8=v6. But this corresponds to a solution set different than the one provided in the question: [x1,x2,x3,x4,x5,x6,x7,x8]^T = [-1,1,0,0,0,0,0,0]^T, [-1,0,1,0,0,0,0,0]^T, [0,0,0,-1,1,0,0,0]^T, [0,0,0,-1,0,1,0,0]^T, [0,0,0,-1,0,0,1,0]^T,[0,0,0,-1,0,0,0,1]^T.

So the problem is that I don't know how to reproduce the given solution set to the given linear equations. We were also told verbally that Excel can be used to help us with this problem set, but I don't see how that can help if I can't even manually reproduce this solution.
 
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  • #2
Thanks Karen. I verified that the provided solution set is valid in that it satisfies the given linear equations. But I have no idea how to obtain this solution set if it were not already given. Is there something I'm missing?
 
  • #3
Put the coefficient matrix in row reduction form:

[tex]\left(\begin{array}{cccccccc}
1&1&1&-1&-1&-1&-1&-1\\
0&0&0&1&1&1&1&1
\end{array}\right)[/tex]
and subtract the second row from the first to row reduce it:

[tex]\left(\begin{array}{cccccccc}
1&1&1&0&0&0&0&0\\
0&0&0&1&1&1&1&1
\end{array}\right)[/tex]

This tells you you can solve for x1 in terms of x2 and x3, and x4 in terms of x5 through x8. Write out the solution vector (x1,x2,...,x8) in terms of the free variables than express it as a sum of vectors each multiplied by one of the free variables. Your solution will pop out.
 
  • #4
I tried writing out the solution in terms of the six free variables, but my solution set is different thant the one given (sorry for poor pensmanship). I don't understand how the given solution set uses no more than two free varaibles for each of (x1, ..., x8) when x4 depends on 4 other variables.
 

Attachments

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  • #5
LCKurtz said:
Put the coefficient matrix in row reduction form:

[tex]\left(\begin{array}{cccccccc}
1&1&1&-1&-1&-1&-1&-1\\
0&0&0&1&1&1&1&1
\end{array}\right)[/tex]
and subtract the second row from the first to row reduce it:

[tex]\left(\begin{array}{cccccccc}
1&1&1&0&0&0&0&0\\
0&0&0&1&1&1&1&1
\end{array}\right)[/tex]

This tells you you can solve for x1 in terms of x2 and x3, and x4 in terms of x5 through x8. Write out the solution vector (x1,x2,...,x8) in terms of the free variables than express it as a sum of vectors each multiplied by one of the free variables. Your solution will pop out.

Alex1812 said:
I tried writing out the solution in terms of the six free variables, but my solution set is different thant the one given (sorry for poor pensmanship). I don't understand how the given solution set uses no more than two free varaibles for each of (x1, ..., x8) when x4 depends on 4 other variables.

The two rows in the array represent the equations:

x1+x2+x3=0
x4+x5+x6+x7+x8=0

Solving these for x1 and x4 gives:

x1=-x2-x3
x4=-x5-x6-x7-x8

so

(x1,x2,x3,x4,x5,x6,x7,x8)
=(-x2-x3,x2,x3,-x5-x6-x7-x8,x5,x6,x7,x8)

Now separate that last vector into a sum with each vector having it's own xi factored out.

[Edit] I can't read your attachment at the moment; I hope I have addressed your question.
 
  • #6
Thanks for your help. I got it now.
 

Related to Reproduce a solution set to a linear system with 2 equations and 8 variables

1. How many solutions are possible for a linear system with 2 equations and 8 variables?

There can be infinitely many solutions or no solution at all, depending on the coefficients and constants in the equations.

2. How do you represent a solution set to a linear system with 2 equations and 8 variables?

The solution set can be represented using an ordered pair or vector notation, where each variable is assigned a value that satisfies both equations in the system.

3. What is the process for reproducing a solution set to a linear system with 2 equations and 8 variables?

To reproduce a solution set, the equations must first be simplified and rearranged to the form y = mx + b, where m and b represent the slope and y-intercept, respectively. Then, the equations can be graphed and the point of intersection will represent the solution set.

4. Can a linear system with 2 equations and 8 variables have no solution?

Yes, it is possible for a linear system to have no solution if the two equations are parallel and never intersect, or if they represent the same line and have infinitely many points of intersection.

5. How can you check if a set of values is a solution to a linear system with 2 equations and 8 variables?

To check if a set of values is a solution, simply substitute the values into both equations and see if they satisfy both equations. If they do, then the values are a solution to the linear system.

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