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Linear Algebra - Matrix Equations

  1. Feb 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Let A be a matrix of size 3x3. Let u and v be vectors in R3, such that Au = v. Prove that the matrix equation Ax=2v has a solution.

    2. Relevant equations



    3. The attempt at a solution

    Alright, so this is for an intro linear algebra course, so the solution shouldnt be too coomplicated. But what im confused about, is can i make A equal a matrix, and then solve this problem with actual numbers? Or to prove this, should i be using variables instead???
    cause what i did was made:
    A =
    1 2 3
    1 2 3
    1 2 3

    And then i mad u=
    3
    2
    1

    Then i multiplied A by u to get v, and got v=
    10
    10
    10

    Then i moved onto proving that Ax=2v has a solution.
    So i wrote:
    [1 2 3] [x1] = 20
    [1 2 3] [x2] = 20
    [1 2 3] [x3] = 20
    Then i wrote out the matrix:
    1 2 3 20
    1 2 3 20
    1 2 3 20

    Passing to RREF i of course got
    1 2 3 20
    0 0 0 0
    0 0 0 0

    Which would therefore indicate that with the A and u that i picked, and the other info given that Ax = 2v has a solution because my RREF has 1 basic variable and 2 free variables, with no rows being 0 0 0 ≠ 0.

    BUT i just don't know if i should be somehow proving this with variables instead of numbers, cause i see him saying, 'would this be true for a different A and u and v?'

    Any help would be appreciated. :) THANKS!
     
  2. jcsd
  3. Feb 12, 2012 #2

    Dick

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    You are way off down the wrong track. If A(u)=v, what's A(2u)? Just using the properties of linear transformations?
     
  4. Feb 12, 2012 #3
    ahhh! my midterm is this friday and im trying to do some practice now, and im not getting anything :(

    A(2u) = 2v right?
    So when a Ax = 2v, does that mean x is 2u?
     
  5. Feb 12, 2012 #4

    Dick

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    It means 2u is a solution to Ax=2v, sure. It might have more solutions depending on the matrix A, but you only need one.
     
  6. Feb 12, 2012 #5
    Ohhh k, that makes sense. But i don't get how to prove it though?
    i just say that since A(u) = v and A(2u) = 2v , Ax = 2v has a solution
     
  7. Feb 12, 2012 #6
    "for all y, there is a x=? such that f(x)=y" means f has a solution for every y.
    "given y, there is an x=? such that f(x)=y" means f has a solution for that y.

    you were only given one such y.
     
  8. Feb 12, 2012 #7
    what y was i given? the u? :/
     
  9. Feb 12, 2012 #8

    Dick

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    If A(2u)=2v, which it does, given A(u)=v, then A(x)=2v has the solution x=2u. It's that simple. You are done with this one. As you suggested, it's not complicated. If you have a midterm on Friday, I'd suggest moving on to the next one.
     
  10. Feb 12, 2012 #9
    Oh wow, i totally over thought this one. Thanks so much. He gives these practice questions, then puts up the answer keys with ow many marks they'd be out of, and usually his qustions are worth like 10-20 marks, so i assumed there'd be more work to it.

    I may as well ask another question here instead of making a new topic.

    For what values of h is v3 in Span(v1, v2) where
    v1 = [1 -5 3]
    v2 = [-2 10 6]
    v3 = [2 -9 h]
    (of course those are supposed to be written vertically, i just cant do that haha)

    K so i wrote out the matrix and began passing to REF.
    and i came to here:
    1 -2 2
    0 0 1
    0 12 -6+h

    The second row has the form 0 0 ≠ 0, so doesnt that mean this system isnt going to have a solution? Where did i go wrong?
     
  11. Feb 12, 2012 #10

    Dick

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    You are going wrong because you don't really understand the problem you are supposed to solve. If v3 is in the span of v1 and v2 then a*v1+b*v2=v3. If you split that into components then you have three linear equations for the three variables a, b and h. Solve them. Don't just blindly REF. You could also do this with a determinant, but it's better to understand the basics first.
     
  12. Feb 12, 2012 #11

    Oh.. k yeah we havent done the determinant yet so ill do the basic way.
    So the three linear equations with the three variables would be the 3 rows in my augmented matrix correct? well like those would be the coefficients of the vairables. When you say solve them, do you mean solve each row as its own equation? im a little confused, cause the only way i would think of solving equations is doing ref, thats why i always get confused with how to do questions cause i just assume i need to do ref :/
     
  13. Feb 12, 2012 #12

    Dick

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    Write down the equations you have to solve first.
     
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