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Basis/Solution set/linear algebra

  1. Apr 15, 2008 #1

    fsm

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    1. The problem statement, all variables and given/known data
    Find the basis of the solution space of the homogeneous system of linear equations.
    x-2y+3z=0
    -3x+6y-9z=0


    2. Relevant equations
    Ax=0


    3. The attempt at a solution
    I first set up my equation
    [tex]\left[ \begin{array}{cccc} 1 & 2 & -3 \\ -3 & 6 & 9 \end{array} \right][/tex]*[tex]\left[ \begin{array}{cccc} x \\ y \\ z \end{array} \right][/tex]=0

    Then I put A in rref to get:
    [tex]\left[ \begin{array}{cccc} 1 & -2 & 0 \\ 0 & 0 & 1 \end{array} \right][/tex]

    So I get the system:
    x-2y=0
    z=0

    From here I'm lost. I don't know if I should paramterize or what. I did try setting z=t x=s, but the vectors I got were not correct. Any help would be great.
     
  2. jcsd
  3. Apr 15, 2008 #2
    you made a mistake


    you should get x - 2y + 3z = 0


    Look at your matrix, where you have a 2 you should have -2, and where you have a -3 you should have a 3
     
  4. Apr 15, 2008 #3

    fsm

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    Thanks for your help. I incorrectly labeled those values. Actually on my paper have those values correct. What you did make me do was double check my work and I did 9 instead of -9. The answer checks.
     
  5. Apr 16, 2008 #4

    HallsofIvy

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    Staff Emeritus
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    In my (not so humble) opinion, too many students get the idea that "linear algebra" is all about "matrices". To combat that, I prefer to avoid using them for problems like this.
    The first thing I would do is try to "solve" those equations in the "usual" way: seeing "x" in one and "-3x" in the other I multiply the first equation by 3 and add to the second equation. Much to my surprise, everything cancels! That tell me that the two equations are not independent and there really is just one equation there: x- 2y+ 3z= 0.

    Okay, that means every <x, y, z> in the solution space must satisfy x- 2y+ 3z= 0. I can pick two of the variables to be anything I want, and solve for the third. That means the solution space is 2 dimensional and I need 2 basis vectors. As I said, I could let any two variables be any two numbers I want. Since it is easy to solve for x: x= 2y- 3z, I can choose y and z to be whatever I want. I prefer to use 1 and 0. (I have a preference for really easy numbers!)

    If y= 1 and z= 0, then x= 2(1)-3(0)= 2. <2, 1, 0> is in the solution space. If y= 0 and z= 1, then x= 2(0)- 3(1)= -3. <-3, 0, 1> is also in the solution space. It is easy to see those are independent (it is a result of choosing "0, 1" and "1, 0") so a basis for the solution space is {<2, 1, 0>, <-3, 0, 1>}.

    You say that the vectors you got were "not correct". How did you determine that? You should realize that there are an infinite number of correct answers so your (correct) answer might well be different from that got by someone else or given in your text.

     
    Last edited: Apr 16, 2008
  6. Feb 4, 2010 #5
    HallsofIvy,

    I just wanted to say thank you for your wonderful explanation. Google brought me to this page and your explanation of the problem truly helped my understanding. I registered an account just now just so I could thank you for the great work you have done.

    All the best.
     
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