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Elementary Row Operations on Matrices

  1. Nov 22, 2003 #1

    cepheid

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    I understand why two of the three row operations do not change the solution set of a system:

    1. Interchange two rows. (Doesn't make much difference in what order one decides to write down the linear equations does it?)

    2. Multiply a row by a scalar. (This step doesn't change the solution set because e.g. writing 2x = 6 instead of x = 3 doesn't change the geometric situation at all)

    It's the third one that's giving me trouble:

    3. Replace one row with the sum of it and a multiple of another row.

    Even though this is the key method of solving simultaneous equations, I cannot justify to myself that it doesn't change the solution set of a system. Can anyone shed light on this?
     
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  3. Nov 23, 2003 #2

    Hurkyl

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    All 3 follow from a general principle:

    Applying an invertible transformation to both sides of an equation does not change the solution set.


    The third operation:

    Replace row i with the sum of row i and r times row j (where r is nonzero)

    can be inverted with the operation

    Replace row i with the sum of row i and -r times row j


    This row operation is invertible, and thus does not change the solution set.
     
  4. Dec 1, 2003 #3

    HallsofIvy

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    Another way of thinking about it is that "row operations" are precisely the way we solve systems of equations. Multiplying a row by a number is the same as multiplying the entire equation by that number. Swapping two rows is the same as switching the position of two equations in the system. Adding a multiple of one row to another is precisely what you do when you multiply one equation by a number in order that the coefficient of, say, x is the negative of the coefficient in another equation and then add the two equations to eliminate x.

    By the way, the general statement "row operations do not change the solution to a system of equations" is not true!

    The row operation "multiply a row by 0" is not invertible and does change the solution set! (Which is why some texts specifically exclude that from the definition of "row operation".)
     
    Last edited: Dec 1, 2003
  5. Dec 1, 2003 #4

    HallsofIvy

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    Yet another way of looking at it (perhaps the most basic way):

    A "fundamental" matrix is a matrix derived by applying a row operation to the identity matrix.

    For example, the fundamental 3 by 3 matrix corresponding to "add two times the first row to the third row" is

    [1 0 0]
    [0 1 0]
    [2 0 1].

    It's easy to show that multiplying a matrix by a fundamental matrix is exactly the same as applying the corresponding row operation to that matrix. It's also easy to see that any fundamental matrix (except that corresponding to "multiply a row by 0") is invertible.

    Applying a row operation to the "augmented matrix" of a matrix equation is the same as multiplying both sides of the matrix equation by the corresponding fundamental matrix.
     
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