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Homework Help: Linear algebra - set of solutions

  1. Nov 11, 2012 #1

    Is it true that the following system of linear equations would always have a single solution (i.e. would never have an infinite number of solutions nor none) for any value of λ?

    λx + 3y -z = 1
    x + 2y -z = 2
    -λx + y + 2z = -1

    May someone kindly confirm?
  2. jcsd
  3. Nov 11, 2012 #2
    How did you conclude that? What did you try?
  4. Nov 11, 2012 #3
    Via a series of elementary operations I got:
    lambda*x +7y = 1
    (1/2)x + y -(1/2)z = 1
    4y + z = 0

    So, rank of the matrix is equal to the number of columns, 3, is it not? Wouldn't that always have a solution, for any lambda that is?
  5. Nov 11, 2012 #4
    Alas, I am still experiencing difficulties trying to find the values of lambda for which number of solutions of the system of equations is: (i) 1; (ii) infinite; (iii) 0
    Having substituted the equations into a matrix and performed a series of elementary operations on the rows, I have found out that there is a single solution for lambda different than 3 and 7/6. However, when lambda is 3 the system still has a single solution, hence a contradiction!
    What am I doing wrong? I have tried various row operations several times, and each time upon substitution of those values for lambda (for which it couldn't possibly have a single solution) the system still has a single solution!
    Would anyone please help with this?
  6. Nov 11, 2012 #5

    Ray Vickson

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    For some values of λ the determinant of the coefficient matrix vanishes. What does that tell you?

  7. Nov 11, 2012 #6
    We haven't dealt with determinants so I wouldn't know, and even if I did I am not permitted to use employ any knowledge thereof. Could you please just tell me what it is I might be doing wrong? For hours I have tried reducing the matrix and each time the lambda's I got for which the number of solutions would be one, ended up, upon substitution, yielding a solution! So I am pretty stuck and would appreciate some more practical guidance, please.
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