Linear Algebra question, finding Elementary Matrix

Discussion in 'Undergraduate Math' started by bummed, Feb 15, 2004.

  1. bummed

    bummed Guest

    Hello folks:

    Was wondering if anyone could shed some insight. I'm taking linear
    algebra and studying for a test. This particular problem was not really
    covered in class and my textbook is not doing a good job in explaining
    this. In-fact, it does not explain this at all.

    Typical question is this:

    Find the elementary matrix E such that EA = B.

    For this problem:
    A = 2 3 B = 2 3
    -1 4 2 -8

    I can simply (maybe overkill) solve for 4 unknowns like so:

    E = a b A = 2 3
    c d -1 4

    and do the following
    2a - b = 2
    3a + 4b = 3
    2c - d = 2
    3c + 4d = -8

    and I get a = 1, b = 0, c = 0, d = -2 such that
    1 0
    0 -2

    is my elementary matrix. In-fact, I can use this approach for any cases
    where all matrixes involved are equivelant sized M x N square matrices.


    Now, here's where I'm thrown at a loss:

    Here's a problem that says:
    A = 1 2 and B = 5 6
    3 4 3 4
    5 6 1 2

    Find elementary matrix E such that EA = B

    In this case, I cannot do

    a b c 1 2
    d e f x 3 4
    g h i 5 6

    and solve for my 9 unknowns specifically due to the fact that there will
    be free variables involved... Can anyone show me an easy way to derive
    the elementary matrix above without having to simply guess and perform
    arbitrary elementary row operations until I get the right answer?

    The book says the answer is:

    0 0 1
    0 1 0
    1 0 0

    I suspect there's a "formulatic" way to do this but my textbook (as well
    as my cliff notes) don't show me this. If guessing by trial and error
    is the only way to do this, then that's ok too. But if there's a
    specific way to do a formula, I'd be much graeful for any help.

    thanks!!!
     
    bummed, Feb 15, 2004
    #1
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  2. bummed

    Paul Sperry Guest

    The elementary matrix you're looking for is the identity matrix
    modified in the same way A is modified to get B.

    For your first example the second row of A is multiplied by -2.
    Multiply the second row of I by -2 to get [[1, 0], [0, -2]].

    For your second example, the first and third rows of A are swapped so
    swap the first and third rows of I to get your book's answer.

    As a third example take your second A and let
    C = [[1, 2], [3, 4], [0, -4]] which is A with the first row multiplied
    by 5 and subtracted from the third row. Do the same thing to I to get
    E = [[1, 0, 0], [0, 1, 0], [-5, 0, 1]] then EA = C.
     
    Paul Sperry, Feb 15, 2004
    #2
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  3. bummed

    Chergarj Guest

    Typical question is this:
    You're starting to drift off-path. You might need to look for a square matrix,
    the inverse of A. You might find a clearer description of the theory and the
    method in a college algebra book. It works like hooking together two square
    matrices, one of them the identity matrix, and the other being the matrix for
    which you want the inverse of this. Be careful, though because I stopped
    studying this about 3 months ago, and I'm a bit mixed up. Check into an old
    college algebra book and you'll likely find what you need.

    G C
     
    Chergarj, Feb 15, 2004
    #3
  4. bummed

    Virgil Guest

    A superficial look at A and C shows that they only differ in the order
    of their rows with rows 1 and 3 being interchanged.

    Since the problem itself refers to *elementary* matrices one supposes
    that you are supposed to know about them. There are 3 types, one of
    which, when multiplied onto the left of anonther matrix, swaps the
    positions of two rows in that other matrix.

    What is the (elementary) 3x3 matrix which, as a premultiplier, will
    interchange rows 1 and 3 of any 3xn matrix?

     
    Virgil, Feb 15, 2004
    #4
  5. bummed

    Daniel Guest

    Hi there,

    There is a very easy way to find these elementary matrices.
    If you can find out what kind of 'row-operation' you
    must perform to get from matrix A to matrix B, then the
    elementary matrix is simply that row-operation performed on
    the identity matrix (think about why this is true).
    Ex. For the first problem you had:
    A = 2 3 B = 2 3
    -1 4 2 -8

    You can see B can be obtained from A by multiplying the
    second row by -2.
    Therefore the elementary matrix will be the identity matrix with
    the bottom row multplied by -2:
    E=1 0
    0 -2

    For the second problem you have:
    A = 1 2 B=5 6
    3 4 3 4
    5 6 1 2

    B can be obtained from A by switching row 1 and row 3.
    Now you can easily write down the elemntary matrix that performes
    this row-operation. Simply perform it on the identity matrix.
     
    Daniel, Feb 15, 2004
    #5
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