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Range of a Matrix Transformation linear algebra

  1. Jul 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Given,
    A =
    [ 1 -3 4;
    -3 2 6;
    5 -1 -8]

    b =
    [b_1;
    b_2;
    b_3]
    Show that there does not exist a solution to Ax = b for every b in R3 and describe the set of all {b1,b2,b3} for which Ax = b does have a solution.


    2. Relevant equations
    row reduction


    3. The attempt at a solution
    I row reduced until I got the following augmented matrix:
    [ 1 -3 4 | b_1;
    0 1 (-18/7)| (b_2 + 3*b_1)/7;
    0 0 8 | b_3-5*b_1 - 14*b_2]
    I'm confused about this because I was lead to believe that since there is a pivot for every value then there should exist a solution for every b in R3. And There is in fact a pivot in every row. Can someone explain to me if there is a solution or not and why please. I'm just not seeing why there wouldn't be one. Thank you
     
  2. jcsd
  3. Jul 10, 2010 #2

    vela

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    Are you sure you wrote down the matrix correctly? In particular, is the 4 in the first row supposed to be a -4?
     
  4. Jul 10, 2010 #3
    oops, wow I feel dumb. I even double checked it. So, I had the right idea at least right?
     
  5. Jul 10, 2010 #4
    ok so with that -4 in place I got the matrix to be:

    [ 1 -3 -4 | b_1;
    0 1 (6/7)| (b_2 + 3*b_1)/7;
    0 0 0 | b_3-5*b_1 - 14*b_2]

    from this i can see that there isn't an answer for all b in R3. The 2nd part of the question asks "describe the set of all {b1,b2,b3} for which Ax = b does have a solution"

    this would be where b_3-5*b_1 - 14*b_2 is equal to zero. Otherwise there would be no solution. Would I just write this or do i need to say something about b1 and b2? Would i need to say that b1 and b2 could be anything? Or how would I go about saying that?
     
  6. Jul 10, 2010 #5

    vela

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    It depends on what "describe" means here. You might be expected to give a geometric interpretation of the solution set.
     
  7. Jul 11, 2010 #6

    HallsofIvy

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    Good: [itex]b_3- 5b_1- 14b_2= 0[/itex] and you can write that as [itex]b_3= 5b_1+ 14b_2[/itex] so that
    [tex]\begin{pmatrix}b_1 \\ b_2 \\ b_3\end{pmatrix}= \begin{pmatrix} b_1 \\ b_2 \\ 5b_1+ 14b_2 \end{pmatrix}= b_1\begin{pmatrix}1 \\ 0 \\ 5\end{pmatrix}+ b_2\begin{pmatrix}0 \\ 1 \\ 14\end{pmatrix}[/tex].

    Geometrically, what is that?
     
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