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Help on simple linear algebra problem

  1. Apr 28, 2006 #1
    Let A and I be as follows.

    A = [1 d]
    [c b]
    I=[1 0]
    [0 1]

    Prove that if b - cd != 0, then A is row equivalent to I

    I'm CLUELESS as to WHERE TO START. Please help me

    I tried simplifying to the matrix

    [1 d]
    [0 b - cd]

    And have no clue what to do next.
     
  2. jcsd
  3. Apr 28, 2006 #2
    What is the criteria for row equivalent matrices? How would you perform the allowable operations to get to I?
     
  4. Apr 28, 2006 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    To get [itex]\left(\begin{array}{cc}1 & 0\\0 & 1\end{array}\right)[/itex]
    you will need a 1 in place of that b-cd. What row operation will give you that?
     
  5. May 18, 2006 #4
    It has been a while since I took linear algebra so I forget the terms for these things, but I know what you're getting at. If A is row equivalent to I, that means that elementary row operations can reduce it to such. That can only be done if the determinant is not zero. (Then we say A is either singular or not singular, don’t remember which) The determinant of A is b - cd. So in a sense, you're done, unless you actually need to prove what I just said.

    In that case, argue by contradiction. Show that if b - cd = 0, The reduced row echelon form of A is not the identity matrix.

    Hope that helps.
     
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