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Homework Help: Express x and y as a linear combination of v's

  1. May 13, 2012 #1
    1. The problem statement, all variables and given/known data


    3. The attempt at a solution

    I can't figure out how to do this problem. If you set the matrix up like this

    2 -1 2 1
    3 1 1 1

    You can't solve it because there are 3 unknowns and two equations.

    I still don't really understand quite well when you set the matrix up vertically or horizontally but if you set it up like this

    2 3
    -1 1
    2 1

    there's no place to put the values for x and y to solve things.
  2. jcsd
  3. May 13, 2012 #2


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    Gold Member

    Reduce the matrix and you'll get two pivots: 2 and 5/2.
    Last edited: May 13, 2012
  4. May 13, 2012 #3
    But if I reduce it to

    x x x x
    0 0 x x

    where x is any number, I still won't be able to solve it because there will be two unknowns in the first row.
  5. May 13, 2012 #4


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    Gold Member

    You have only 3 v's. So, in column form, after reducing the matrix:
    \displaystyle\left[ {\begin{array}{*{20}{c}}
    2&-1&2 \\
    0&5/2&-2 \\
    \end{array}} \right][/tex]
    Hence, you can see that columns 1 and 2 are a basis.
    Last edited: May 13, 2012
  6. May 13, 2012 #5
    Well, if I reduce it to

    \displaystyle\left[ {\begin{array}{*{20}{c}}
    2&-1&2 \\
    0&5/2&-2 \\
    \end{array}} \right][/tex]

    Then I don't see what the next step should be.
  7. May 13, 2012 #6
    Hi Robert
    You have 3 2D vectors, v1, v2, v3. as you can see, none is perpendicular to any, but they can't form a basis since they are 3 and the dimensionality is 2.
    So any two of them will form a basis.
    So for instance, you could have a first basis (v1,v2) a second one (v2,v3) and a third one (v1,v3) and that would be the 3 ways in which you can later express x and y
    So for instance, in the first base, you would want to express x in (v1,v2)
    (and later y too, but first let's look at x)
    How do you express x in (v1, V2) ?
    x must be a linear combination of v1 and v2, so x=av1+bv2
    x being (1,1) that would mean
    av1+b2=(1,1), so a(2,3)+b(-1,1)=(1,1)
    therefore you would just have to solve (2a-b, 3a+b)=(1,1)
    I let you continue at this point

  8. May 13, 2012 #7
    here are the answers


    by using v1 and v2 and setting up the matrix as follows

    2 -1 1
    3 1 1

    I was able to get .4 -.2 which is the answer listed in 13

    Using that technique I was not able to get the other two answers for x which are

    -4/5 v + 7/5v + 2v

    I don't see how I could even get values for three different v's when I'm only using two equations.
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