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Should be easy question on column space of a matrix

  1. Aug 18, 2012 #1
    1. The problem statement, all variables and given/known data
    So the actual problem "Find the value of a for which the following system of linear equations has a solution"

    2x + 4y + z = a
    -4x -7y + 0 = 1
    0 -1y -2z = 1


    2. Relevant equations



    3. The attempt at a solution
    I thought one approach was to find a basis for the corresponding matrix and see what value of a would make that vector in the space formed by the basis. That is, see what value of a would make (a, 1, 1) in the range of the matrix.

    But when I row reduce I get

    2 4 1
    0 1 2
    0 0 0

    So I want to say the 1st and 2cnd columns form a basis -> (2, -4, 0) and (4, -7, -1). But then some linear combination of these should equal (1, 0, -2). However, the first components of the basis appear to never be able to combine linearly to 1. That is there are no integers x,y such that 2x + 4y = 1 since -> x + 2y = 1/2.....? I feel like I must be making a really trivial mistake?!?!
     
  2. jcsd
  3. Aug 18, 2012 #2

    gabbagabbahey

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    Maybe you should show how you are row reducing. But first, ask yourself if there is any reason to think that there are any values of [itex]a[/itex] that would make it so there are no solutions.
     
  4. Aug 19, 2012 #3

    HallsofIvy

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    Row reduce the augmented matrix
    [tex]\begin{bmatrix}2 & 4 & 1 & a \\ -4 & -7 & 0 & 1 \\ 0 & -1 & -2 & 1\end{bmatrix}[/tex]
    instead. That will tell you immediately what a should be.

    So you want x and y such that 2x+ 4y= 1, -4x- 7y= 0, and -y= -2? From the last equation, y= 2 so the first two equations becom 2x+ 8= 1 and -4x- 14= 0. Both of thos equations reduce to 2x= -7.
     
  5. Aug 19, 2012 #4
    Hmm, I was being really dumb and thinking that a scalar multiple had to be an integer, since as I was saying in the OP there are no integer solutions to 2x = -7. Thanks!
     
  6. Aug 22, 2012 #5

    uart

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    Hi Fractal. Did you yet solve this problem? x = -7/2 is not a solution, so it doesn't look like you have really understood this one yet.

    Row reduce the augmented matrix that Halls' suggested.

    - For all values of "a" except one, there are no solutions.

    - For this one particular value of "a" there are infinite solutions, which you should express parametrically.
     
  7. Aug 22, 2012 #6
    Thanks for the look over! So the x = -7/2 bit was just in reference to expressing the 3rd column as a linear combination of the other two (-7/2 times the first plus 2 times the 2cnd gives the 3rd). The motivation for my original post was motivated by getting weirded out by thinking that my row reduction wasn't giving a basis for the range. I think I was able to get it from there, but I don't have my work anymore. Thanks again!
     
  8. Aug 22, 2012 #7

    uart

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    Ok, I agree that "-7/2 times the first plus 2 times the 2nd gives the 3rd". But how exactly did that allow you to solve for "a"?

    I think the easy way to solve it is to just row reduce the augmented matrix as suggested above. You should end up with something like,

    \begin{bmatrix}1 & 0 & -7/2 & -7a/2 - 2 \\ 0 & 1 & 2 & 2a+1 \\ 0 & 0 & 0 & 2a+2\end{bmatrix}
    from which you can conclude there are no solutions unless [itex]2a + 2 = 0[/itex].

    Then you can subst in this value of "a" to express x,y and z parametrically. Like,

    [itex]x = 3/2 + 7t/2[/itex]
    [itex]y = -1 -2t[/itex]
    [itex]z = t[/itex]
     
    Last edited: Aug 22, 2012
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