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Elementary Matrices problem

  1. Sep 23, 2007 #1
    Hello all, i'm taking my first year in linear algebra and i'm having some issues understanding how to deal with some problems involving elementary matrices.

    First off, i have a set of problems that ask to find the elementary matrix E such that AE=B, and secondly i have a set of problems asking to find the elementary matrix E such that EA=B. I've reread this section in the book a couple of times and there isn't much about matrix algebra involving elementary matrices, other than the fact that they do row operations on matrices. How exactly am i supposed the row operations in these sets of problems?

    For example, one problem is

    Find an elementary matrix E such that EA=B

    [tex]A=\left(\begin{array}{ccc}2&1&3\\-2&4&5\\3&1&4\end{array}\right), B=\left(\begin{array}{ccc}2&1&3\\3&1&4\\-2&4&5\end{array}\right)[/tex]

    it's obvious to me that row's 2 and 3 are switched in A to make B, but how do i know what elementary matrix does that? The back of the book says that


    and after performing the matrix multiplication i get it, but there has to be a better way to learn how to do it (especially since i don't know how to do the other problems in the set without looking in the back of the book).


    Find an elementary matrix E such that AE = B

    [tex]A=\left(\begin{array}{ccc}4&-2&3\\-2&4&2\\6&1&-2\end{array}\right), B=\left(\begin{array}{ccc}2&-2&3\\-1&4&2\\3&1&-2\end{array}\right)[/tex]

    the back of the book states that


    column 1 is halved in the transformation from A to B, so that makes sense, however there is another problem (from the same AE = B set)

    [tex]A=\left(\begin{array}{cc}2&4\\1&6\end{array}\right), B=\left(\begin{array}{cc}2&-2\\1&3\end{array}\right)[/tex]

    here column 2 is halved and negative, so i'd assume the elementary matrix to be similar to the one in the first problem, but the back of the book says it is


    can anyone point me in the right direction here? even a link to a site that explains it well would be helpful. thank you for your time.
    Last edited: Sep 23, 2007
  2. jcsd
  3. Sep 23, 2007 #2
  4. Sep 23, 2007 #3


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    Inverses are overkill. The exercise wants him to recognize elementary row & column operations and relate them to the elementary matrices.
  5. Sep 23, 2007 #4


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    No it's not. The negation of half of 6 is -3, not 3.
  6. Sep 23, 2007 #5
    i know how to do matrix inverses... how are they related to elementary matrices, though?

    if i have EA=B is there a way to solve for E using inverses?
  7. Sep 23, 2007 #6
    [tex]EA=B \implies E=BA^{-1}[/tex]
  8. Sep 23, 2007 #7


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    Are you sure you know WHAT an "elementary matrix" is. It is a matrix derived by applying a particular row or column operation to the identity matrix. In your last problem you go from A to B by subracting twice the first column from the second column. If you do that to the identity matrix, you get the corresponding row operation.
  9. Feb 8, 2009 #8
    Edit: sorry for the needless bump.... I only just realized how old this topic was.

    I am currently taking Linear Algebra as well, and this one was rather easy for me to figure out. First, I assumed [tex]E=\left(\begin{array}{cc}a&b\\c&d\end{array}\right)[/tex] and then did some basic algebra....



    So [tex]E=\left(\begin{array}{cc}1&-3\\0&1\end{array}\right)[/tex]
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