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Math Help - Need Help with Elem. Linear Algebra Problem

  1. #1
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    Need Help with Elem. Linear Algebra Problem

    Express the invertible matrix

    \left(\begin{array}{ccc}1&2&1\\1&0&1\\1&1&2\end{ar  ray}\right)

    as a product of elementary matrices.
    --------------------------
    The answer in the back of the book has a product of 6 matrices. I think I'm supposed to use the identity matrix and perform the same row operations on it as I would to obtain the above matrix... or something like that. Any help with this would be appreciated!
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  2. #2
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    Quote Originally Posted by paupsers View Post
    Express the invertible matrix

    \left(\begin{array}{ccc}1&2&1\\1&0&1\\1&1&2\end{ar  ray}\right)

    as a product of elementary matrices.
    --------------------------
    The answer in the back of the book has a product of 6 matrices. I think I'm supposed to use the identity matrix and perform the same row operations on it as I would to obtain the above matrix... or something like that. Any help with this would be appreciated!
    Yes, that's right. A "row operation" is one of three kinds:
    1) swap two rows
    2) multiply every member of a row by the same constant.
    3) replace every member of a row by itself plus a constant time the corresponding member of a second row.

    An "elementary matrix" is a matrix constructed from the identity matrix by a single row operation.

    What you want to do is reduce the given matrix to the identity matrix (which is possible because it is invertible) by row operations, writing down the elementary matrix corresponding to each row operation.

    For example, starting from
    \left(\begin{array}{ccc}1&2&1\\1&0&1\\1&1&2\end{ar  ray}\right)
    My first steps would be to reduce the first column to the column \left(\begin{array}{c}1 \\ 0 \\ 0\end{array}\right).
    The number in the upper left is already 1 so I don't need to change that. I can get 0 in the next row by subtracting the first row from the second row. The elementary matrix corresponding to that is
    \left(\begin{array}{ccc}1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)
    To get 0 in the third row, I need to subtract the first row from the third row. That gives the elementary matrix \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1\end{array}\right)

    and they convert the matrix to
    \left(\begin{array}{ccc}1 & 2 & 1 \\ 0 & -2 & 0 \\ 0 & -1 & 1\end{array}\right)
    Now you need to convert the second column properly and then the third. Can you continue?

    After you have converted the matrix to the identity matrix and written down all the corresponding elementary matrices (in the proper order) so they multiply to give the original matrix.

    Normally it would take 9 row operations to reduce a 3 by 3 matrix to the identity matrix and so the matrix would be represented as the product of 9 elementary matrices. Notice that the upper left number was already 1 so we didn't need a row operation for that. Also that 0 now in the second row, third column will stay there we won't need a row operation to change that. I suspect that sort of thing, a 0 or 1 already in the correct position will happen once more to give 6 rather than 9 elementary matrices.
    Last edited by HallsofIvy; July 18th 2009 at 06:05 AM.
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  3. #3
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    This is exactly what I tried doing, but it's not the answer the book gives (although it is similar). Is the solution to this problem unique?
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  4. #4
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    No, it is not because you can follow different "paths" to row reduce to the identity matrix. What I showed is the most commonly used "path".
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