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Row Reduction to solve for 6 Unknowns

  1. Apr 8, 2014 #1
    1. The problem statement, all variables and given/known data
    I am in a calculus class where we are learning the introduction to row reduction. I have done this before in other courses, so I am familiar with the process, but I am not sure about this one. We were given:

    x4 + 2x5- x6 = 2
    x1 + 2x2 + x5 -x6 = 0
    x1 + 2x2 + 2x3 - x5 + x6 = 2

    We were supposed to: solve for each unknown or tell how many solutions this has(usually something like 0 or infinity).

    2. Relevant equations

    3. The attempt at a solution

    We haven't learned it exactly yet, but there is no real row eschelon form for this, is there? I don't know what all I can do mathematically, it just seems to me that we only have 3 equations and 6 unknowns, so we cannot possibly solve for this, right? Am I missing something here? I wrote out the matrix exactly as it is written (if it didn't state a variable, i set it to 0. e.g. in equation 1, it does not mention x1 through x3 so I made those 0).

    No matter what you do, you have a 3x6 matrix with 3 equations and 6 unknowns.

    Am I missing something here? How could I mathematically answer this question and not just state what I have stated above?

    Any help would be appreciated, thanks.
  2. jcsd
  3. Apr 8, 2014 #2


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    Just go ahead and row reduce it as far as you can. You can get rid of the ##x_1## in the middle equation with the last one, which will also get rid of the ##x_2## but that's OK. You should be able to get three rows whose first element is ##1##. Solve for those variables in terms of the others.
  4. Apr 8, 2014 #3
    That's what I thought, thanks. So I have reduced it as far as I can, and it is obvious that you still cannot solve, so would an appropriate answer just be that there are No Solutions?
  5. Apr 8, 2014 #4


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    I'm thinking you can likely solve for three of the variables in terms of the others, right? The extra variables are called free variables and can be anything. You have lots of solutions. The case when there would be no solutions is of one of your rows is all zeros and a nonzero on the right side.
  6. Apr 9, 2014 #5


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    What do you mean by "cannot solve"? You cannot solve for a single specific value for each number but you didn't expect to, right? Part of the question was "how many solutions are there?"

    You have three equations in six unknown values so if all three equations are independent you would expect to be able to solve for 6- 3= 3 of the values in terms of the other three.
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