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Simple Ax = B

  1. Jul 9, 2011 #1
    Hello all, I'm really sorry that this may come across as a ridiculously simple question, but I'm taking a linear algebra class online with a teacher who is no help at all. I was hoping somebody could guide me through the process to solve this problem. Please be as thorough as possible for me. Thank you in Advance.

    Let A = matrix whose columns are u; v:
    2 -4
    A = 3 2
    4 1

    Solve the equation Ax = b:

    (and I think it's giving me the answer here, but I'm not sure cause I can't even begin to solve the problem)

    [ -4
    8 ]


    Please help if you can. Thank you.
     
  2. jcsd
  3. Jul 9, 2011 #2

    micromass

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    Hi Emily A! :smile:

    What is b??
     
  4. Jul 9, 2011 #3
    Unfortunately, that's all I'm given for the problem. Is that why I can't seem to understand this problem? Because I don't even have all of a problem? It is possible that where I think it's giving me the answer, maybe that is the "B", but I'm not sure, the book is poorly worded/organized, and as I said, the teacher is of no use to any of the students.
     
  5. Jul 9, 2011 #4
    Hi Emily_A
    what does it mean "2 -4" in the second line in your problem?Please identify your problem.It must not be so complicated :)
     
  6. Jul 10, 2011 #5

    HallsofIvy

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    Then please state the problem exactly as it is given.
     
  7. Jul 10, 2011 #6
    I have stated the problem exactly as given, I've typed out everything the book says word for word. @shstar_2008, I just saw that the forum didn't format as I thought it would, but either way, the origional matrix A is:
    2 -4
    3 2
    4 1

    Sorry about that formatting error.

    again, the book states exactly this:

    Let A = matrix whose columns are u; v:
    A =
    2 -4
    3 2
    4 1

    Solve the equation Ax = b:

    -4
    8

    (these last 2 numbers are in a matrix stacked vertically, and I don't know if they are the answer, or B or what?)

    That is ALL the book gives me to go on.
     
  8. Jul 10, 2011 #7

    I like Serena

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    Hi Emily_A! :smile:

    You are dealing with a matrix and 2 vectors.
    The vector b has 3 elements, and the vector x has 2 elements.
    So I'm going to assume you've been given the vector x, and the question is to calculate vector b.

    Do you know how to calculate the following?
    [tex]
    A \cdot x =
    \begin{pmatrix}
    2 & -4 \\
    3 & 2 \\
    4 & 1
    \end{pmatrix}
    \cdot \begin{pmatrix}
    -4 \\
    8
    \end{pmatrix}
    [/tex]
     
  9. Jul 11, 2011 #8
    Hi Emily_A
    As "I like Serena" has written,you have two vectors,x and b. and you have a vector with 2 elements. Due to dimension given for A,I think this given vector would be x. so you must only multiply A by x and calculate b.
    In another idea, this vector can be b. So in the equation you must have A'x=b to have an equation with right dimensions(A' is the transpose of A).
    In my book,the problem may be is in this way.
    Be successful
     
  10. Jul 13, 2011 #9
    It must be the case that x has dimensions 2*1, because if it were b that had dimensions 2*1 then the
    matrix would have to have two rows (but it has three).

    We are solving for b, we are given A*x.
    Take your left hand pointer finger and put it on the first row, with the right hand finger put it on the first element of the vector x.
    Move over right one element at a time with the matrix, as you simultaneously move down on vector x.
    Multiply terms the fingers meet, and sum all such terms when you get to the end of a row of a matrix.
    This sum is the first element of the new vector b.
    Repeat for the 2nd row of A starting again at the first element of x; this is element 2 of b.
    Again for last row, this is the final (third) row (element) of b.
    (2*-4 + (-4)(8) ) = (-8 -32) = (-40)
    (3*(-4) + (2)(8) ) = (-12 + 16)=( +4)
    (4*-4 + (1*8) ) = (-16 + 8) = (-8)
     
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