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Simple Linear Algebra matrix problem

  1. Jun 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Let a1 = [1 4 -2] supposed to be a column
    Let a2 = [-2 -3 7] supposed to be a column
    Let b = [4 1 h] supposed to be a column

    For what value(s) of h is b in the plane spanned by a1 and a2?

    3. The attempt at a solution

    Well, the first thing I did as in 99% of all Linear Alg problems was to put it into an augmented matrix:

    [1 -2 l 4]
    [4 -3 l 1]
    [-2 7 l h]

    which became from -4R1 +R2 -> R2:

    [1 -2 l 4]
    [0 5 l -15]
    [-2 7 l h]

    and I'm stuck. I could do 2R1+ R3 -> R3, but this doesn't seem to render a correct answer of h= 7/2.
     
  2. jcsd
  3. Jun 25, 2012 #2
    I think the simpler solution is to construct the whole 3x3 matrix and take the determinant. If b lies in the plane of a1, a2, then the three of them put together no longer span all of 3D space, and the determinant of the matrix will be zero.
     
  4. Jun 25, 2012 #3

    I like Serena

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    Hi anniecvc! :smile:

    No need to be stuck.
    You can divide row 2 by 5: R2/5 -> R2

    Then you can use row 2 to simplify row 1.

    From rows 1 and 2 you get the solution for the linear combination.

    Substitute those in the equation represented by row 3 and you get h.
     
  5. Jun 25, 2012 #4

    HallsofIvy

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    Personally, I prefer to do problems like this right from the definitions.
    b is in the space spanned by a_1 and a_2, then it is a linear combination of them. That is, there must exist numbers x and y such that
    [tex]x\begin{bmatrix}1 \\ 4\\ -2\end{bmatrix}+ y\begin{bmatrix}-2 \\ -3 \\ 7\end{bmatrix}= \begin{bmatrix}4 \\ 1\\ h\end{bmatrix}[/tex]
    which is the same as
    [tex]\begin{bmatrix}x- 2y \\ 4x- 3y \\ -2x+ 7y\end{bmatrix}= \begin{bmatrix}4 \\ 1 \\ h\end{bmatrix}[/tex]

    which, in turn, is equivalent to the three equations x- 2y= 4, 4x- 3y= 1, -2x+7y= h.

    Solve the first two equations for x and y and put those values into the third to find h.
     
  6. Jun 25, 2012 #5
    If the required answer is 7/2, it is wrong, as can be checked easily. What you described initially does work.
     
  7. Jun 26, 2012 #6
    Ougoah, you're right, the answer is h=-17, which using Serena's help comes out nicely. Thank you!!! Ivy, that does work too but I think defeats the purpose of being in a linear algebra class rather than an Alg 2 class. Still, thanks for the insight!
     
  8. Jun 26, 2012 #7

    I like Serena

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    You're welcome! :smile:
     
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