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Homework Help: Linear Algebra Solution Set Question

  1. Sep 25, 2008 #1
    1. The problem statement, all variables and given/known data
    Find all solutions of a for which the resulting linear system has (a) no solution, (b) a unique solution, (c) infinetly many solutions.

    x+z=4
    2x+y+3z=5
    -3x-3y+(a^2-5a)=a-8


    2. Relevant equations

    ...Row reduction

    3. The attempt at a solution

    I solved the augmented matrix and got z=a-5/[(a-2)(a-3)], then got x and y in terms of z. I solved the homogeneous and only got the trivial solution regardless of a's value. I'm thinking that the system has no solution when a=2 or a=3, and has a unique solution when a equals any other real number. Is this correct?
     
  2. jcsd
  3. Sep 25, 2008 #2

    HallsofIvy

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    Science Advisor

    Yes, that is correct. The only way you could have "infinitely many solutions" would be if you arrived at something like "z= (a-2)/[(a-2)(a-3)]". In that case, any value of z would satisfy (2-2)(2-3)z= (2-2) so there would be infinitely many solutions for a= 2.. (I haven't checked your calculations. I am assuming you did not cancel something like (a-b)/(a-b) in your solution.)
     
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