1812Alex Posted June 21, 2011 Share Posted June 21, 2011 I'm stuck on a homework question: 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. 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. I've tested the given solution by plugging it into the equations and it seems to work, but I have no clue how to reproduce it. Link to comment Share on other sites More sharing options...
DrRocket Posted June 21, 2011 Share Posted June 21, 2011 (edited) I'm stuck on a homework question: 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. 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. I've tested the given solution by plugging it into the equations and it seems to work, but I have no clue how to reproduce it. You have two linear equations, x1+x2+x3-x4-x5-x6-x7-x8=0 and x4+x5+x6+x7+x8=0, in eight variables, an undertermined set of equations. The term "solution set" normally refers to the set of all valid solutions to the given equations. One would suppose then that the given task is to show that the solutions that you list belong to that set. Perhaps what is expected is that you will characterize the complete solution set and then show that the given solutions lie within it. Have you tried to do this, by analyzing the two given equations and maybe combinations of them ? Edited June 22, 2011 by DrRocket Link to comment Share on other sites More sharing options...
1812Alex Posted June 21, 2011 Author Share Posted June 21, 2011 Thanks, you're right. I've looked at the problem again and it seems the provided "solution set" is actually a solution basis obtained be solving the linear system for x2 and x8 and then looking at pairs of variables in the solution for x4 + x5 + x6 + x7 +x8 = 0 by assuming x6=x7=x8, then x4+x5= 0 and so on for all the pairs of that solution. Link to comment Share on other sites More sharing options...
DrRocket Posted June 22, 2011 Share Posted June 22, 2011 Thanks, you're right. I've looked at the problem again and it seems the provided "solution set" is actually a solution basis obtained be solving the linear system for x2 and x8 and then looking at pairs of variables in the solution for x4 + x5 + x6 + x7 +x8 = 0 by assuming x6=x7=x8, then x4+x5= 0 and so on for all the pairs of that solution. The statement of the problem is a bit odd. But what is going on is this: The second equation constrains x4--x8 in a way that can be used directly in the first equation to find a relation among x1--x3. Those two relations let you find lots of simultaneous solutions of the two equations. Link to comment Share on other sites More sharing options...
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