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Matrix - solving linear system

  1. Sep 14, 2009 #1
    linear system:
    x + y +z = 2
    2x + 3y+ 2z = 3
    2x + 3y+ (a2 - 2)z = a+1


    when reducing it to row echelon form, the last step looks like the following (if my calculations are right)

    1 1 1 2
    0 0 1 1
    0, 1, a2-5, a-4


    the question is to determine all values of a for which he system has
    (a)no solution, (b) infinitely many solutions, (c) only one solution

    for (a) to happen a2-5 = 0 and a-4 not=0

    but I can't seem to factorize a2-5=0 or are my calculations just wrong?


    thanks Amy-Lee
     
  2. jcsd
  3. Sep 14, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Amy-Lee! Welcome to PF! :smile:
    I haven't checked how you got there, but a2 - 5 = 0 is just (a + √5)(a - √5) = 0, or a = ±√5 :wink:
     
  4. Sep 15, 2009 #3
    Thank you Tiny Tim!!!:approve:
     
  5. Sep 15, 2009 #4

    HallsofIvy

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    Did it occur to you that [itex]a^2- 5= 0[/itex] is the same as [itex]a^2= 5[/itex] and so [itex]a= \pm\sqrt{5}[/itex]?
     
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