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Homework Help: Linear Algebra determine the conditions

  1. Aug 29, 2010 #1
    1. The problem statement, all variables and given/known data

    determine the conditions on a b c and d such that the matrix
    a b
    c d
    will be row equivalent to the given matrix:
    1 0
    0 1

    and

    1 0
    0 0

    2. Relevant equations


    3. The attempt at a solution

    I have no idea what its asking, I mean on no. 1:
    a = 1 b = 0 c = 0 and d =1
    but what does that mean? I tried taking the resultant

    for the first one would be 1(1) - 0(0) = 1

    which would be ad - bc = 1 ??
     
  2. jcsd
  3. Aug 29, 2010 #2

    Mark44

    Staff: Mentor

    This looks to me like two separate problems, or one problem with two parts.
    Determine the conditions on a b c and d such that the matrix
    a b
    c d
    will be row equivalent to the given matrix:
    a)
    1 0
    0 1

    b)
    1 0
    0 0

    By "resultant" I think you mean determinant, and you're on the right track with that approach.
     
  4. Aug 29, 2010 #3
    yes, they are 2 problems, and yes I meant the determinant :P
    still if im on the right approach. . .thats as far as I can go. . .is that the answer?
    a)
    ad - bc = 1


    b)
    ad - bc = 0
     
    Last edited: Aug 29, 2010
  5. Aug 29, 2010 #4

    Mark44

    Staff: Mentor

    You should start a new thread for the other problem.

    For the original problem, suppose the matrix is
    [2 1]
    [3 0]

    Will it be row-equivalent to the identity matrix or to the other one (the one that has all zeroes except for the upper left entry)?
     
  6. Aug 29, 2010 #5
    2(0) - 3(1) = -3 . . .Im guessing no? (its not zero nor 1)
    or if you do some opperations
    -r1 + r2 =
    [1 -1]
    [3 0]
    and r_3 - 3r1

    [1 -1]
    [0 3]
    divide by 3 on row 2
    [1 -1]
    [0 1]

    add row_1 + row_2
    [1 0]
    [0 1]
    what I dont understand is what
    row equivalency is? unless it means that its like "multiples" (or row-operation-wise) of each other?

    if so then what you mentioned is row equivalent to a) ??
    still how would I write the answer?
     
  7. Aug 30, 2010 #6

    HallsofIvy

    User Avatar
    Science Advisor

    Okay, now do the same that to the matrix given: actually row-reduce the given matrix!

    From
    [tex]\begin{bmatrix}a & b \\ c & d\end{bmatrix}[/tex]
    as long as [itex]a\ne 0[/itex] we can divide the first row by a, then subtract c times the first row from the second to get
    [tex]\begin{bmatrix}1 & \frac{b}{a} \\ 0 & d- \frac{bc}{a}\end{bmatrix}[/tex]

    If [itex]d- bc/a\ne 0[/itex] (which is the same as saying det= ad- bc= 0), we have the second form. If not, we can divide the second row by that and get the first form.

    If a= 0, we can swap the rows, getting
    [tex]\begin{bmatrix}c & d \\ a & b\end{bmatrix}[/tex]
    Now what?
     
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