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Determinant of a matrix using reduced echelon form

  1. Mar 21, 2007 #1
    Problem statement: Find the determinant of the following matrix by row reduction to echelon form.
    |1 3 3 -4 |
    |0 1 2 -5 |
    |2 5 4 -3 |
    |-3 -7 -5 2 |

    I reduced this matrix to
    |1 3 3 -4 |
    |0 1 2 -5 |
    |0 -1 -2 5 |
    |0 2 4 -10 |

    If I reduce this further, the entries in row 3 and 4 become 0. Is this still considered triangular form, and therefore the determinant will be 0?
     
  2. jcsd
  3. Mar 21, 2007 #2

    radou

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    Correct. It was obvious from the beginning, since the third and second row are proportional (after your first "reduction").
     
    Last edited: Mar 21, 2007
  4. Mar 21, 2007 #3

    Dick

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    Yes and yes.
     
  5. Mar 21, 2007 #4
    Ok, I didn't reduce it by row-eschelon form, but you should get a determinant of zero.

    Why? The second row is a linear combination of the other three.

    Upper Triangular form just requires that the entries below the diagonal are all zero. If entries above (or on) the diagonal are zero, that's ok.

    ZM
     
  6. Mar 21, 2007 #5
    If this is true, then will the determinant of any matrix containing a row vector consisting of all zeros will be zero?
     
  7. Mar 21, 2007 #6

    radou

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    Yes, it will.
     
  8. Mar 21, 2007 #7
    Yes. Think of it this way:

    The determinant function can be viewed as a machine that takes n n-dimensional vectors and spits out a number. However, if any of these vectors are linear combinations of any of the other vectors, then the determinant will be zero. A zero vector is a linear combination of every vector.

    Another way to see the same thing is just to break the determinant into minors along the row (or column) of zeros. The determinant is then zero.

    ZM
     
  9. Mar 21, 2007 #8

    radou

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    The exact reason follows directly from the definition of the determinant.
     
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