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Determinants By Row Reduction/Row Echelon Form

  1. Jul 15, 2010 #1
    Hello all,

    I have been studying some linear algebra, and I recently came upon the method of finding determinants by row reduction (to row echelon form). But isn't it true than a matrix can have any row echelon form? If so, this would mean different determinants, right?

    I am studying from "Elementary Linear Algebra With Applications" (ninth edition) by Howard Anton & Chris Rorres. In the first chapter it says "a row echelon form of a matrix is not unique", which only adds to my confusion.

    Hopefully someone can shed some light on this issue with examples.
  2. jcsd
  3. Jul 17, 2010 #2


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    I don't know what you mean by this. No, it is NOT true that a matrix can have "any" row echelon form. For example, a non-invertible matrix must have at least one row all zeros in any "row echelon form". It is true that the "row-echelon form" is not unique- but not that it can by "any" form.

    Yes, it is true that you can row-reduce a matrix to different row-echelon forms having different numbers on the main diagonal.

    You cannot just "get" the determinant of a matrix from its row-echelon form- you get the determinant from the way you row reduce it:
    1) If you swap two rows, you multiply the determinant by -1.
    2) If you add a multiple of one row to another, you don't change the determinant.
    3) If you multiply a row by a number, you multiply the determinant by that number.

    So, for example, suppose your matrix is
    [tex]\begin{bmatrix}2 & 1 & 2 \\ 1 & 1 & 1 \\ 2 & 2 & 5\end{bmatrix}[/tex]

    You can reduce this to row-echelon form by, say, subtracting 1/2 the first row from the second, and subtracting the first row from the third to get
    [tex]\begin{bmatrix}2 & 1 & 2 \\ 0 & 1/2 & 0 \\ 0 & 1 & 3\end{bmatrix}[/tex]
    then subtracting twice the second row from the third to get
    [tex]\begin{bmatrix}2 & 1 & 2 \\ 0 & 1/2 & 0 \\ 0 & 0 & 3\end{bmatrix}[/tex]

    Now this last matrix is in row-echlon form, an "upper triangular matrix", so its determinant is just the product of the numbers on its main diagonal: 2(1/2)(3)= 3.
    Further, since I only used row-operations of type (2), above, I haven't changed the determinant- the determinant of the original matrix is also 3.

    But seeing that "1 1 1" in the second row, it might have occured to me that it would be simpler to swap the first two rows, then subtract twice the (new) first row from the other rows to get
    [tex]\begin{bmatrix}1 & 1 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & 3\end{bmatrix}[/tex]

    Now, that matrix has determinant -3. But since I swapped two rows, it no longer has the same determinant as the original matrix- I need to multiply by -1 for that swap to get the determinant of the original matrix: -1(-3)= 3 again.

    Some times we might prefer to always get a "1" in the diagonal position- that is often done in computer programs to row-red
    uce because that way you can easily see what to multiply by and subtract from other rows. Here, I might start by dividing the first row by 2, then subtracting that (new) first row from the second and twice the (new) first row from the third to get
    [tex]\begin{bmatrix}1 & 1/2 & 1 \\ 0 & 1/2 & 0 \\ 0 & 1 & 3\end{bmatrix}[/tex]

    Now, multiply the second row by 2 and subtract it from the third row to get
    [tex]\begin{bmatrix}1 & 1/2 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 3\end{bmatrix}[/tex]. Now this new matrix has determinant 3 (which happens to be correct) but I need to think "I multiplied a row by 1/2 and I multiplied a row by 2 so I multiplied the entire determinant by 2(1/2) which happens to equal 1. If the product of all the numbers I multiplied rows by were some other number, I would have to divide by it to get the original determinant.
  4. Jul 17, 2010 #3
    ^Very nice explanation HallsofIvy.
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