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

[Alex1812]

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.

 

[Alex1812]

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?

 

[LCKurtz]

Put the coefficient matrix in row reduction form:

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

$$\left(\begin{array}{cccccccc} 1&1&1&0&0&0&0&0\\ 0&0&0&1&1&1&1&1 \end{array}\right)$$

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]

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.

 

[LCKurtz]

Put the coefficient matrix in row reduction form:

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

$$\left(\begin{array}{cccccccc} 1&1&1&0&0&0&0&0\\ 0&0&0&1&1&1&1&1 \end{array}\right)$$

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.

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 x_i factored out.

[Edit] I can't read your attachment at the moment; I hope I have addressed your question.

 

[Alex1812]

Thanks for your help. I got it now.