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Find the following determinants

  1. Apr 25, 2005 #1
    I know how to find the determinant in general but these two problems here are tough for me:

    1. Find the determinant of:
    3 -1 0 0 0
    -1 3 -1 0 0
    0 -1 3 -1 0
    0 0 -1 3 -1
    0 0 0 -1 3

    2. Find the determinant of:
    0 2 2 2 2
    2 0 2 2 2
    2 2 0 2 2
    2 2 2 0 2
    2 2 2 2 0

    Now, I know that I need to select a row or column that contains all zeros except for one number. The row or column that the one non-zero number is on would be selected as well. Then you would use the formula (for selecting row i and column j):

    det A = (-1)^(i + j) * det A (without row i and column j)

    But I can't figure out how to create a row or column full of zeros (except one element) for both of these problems. Thanks for any help.
     
  2. jcsd
  3. Apr 25, 2005 #2

    OlderDan

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    Science Advisor
    Homework Helper

    Haven't done one of these for an age. How about this one. Subtract the row below from each row

    -2 2 0 0 0
    0 -2 2 0 0
    0 0 -2 2 0
    0 0 0 -2 2
    2 2 2 2 0

    Add the first 4, then the next 3 then the next 2

    -2 0 0 0 2
    0 -2 0 0 2
    0 0 -2 0 2
    0 0 0 -2 2
    2 2 2 2 0

    Add the first 4 to the last

    -2 0 0 0 2
    0 -2 0 0 2
    0 0 -2 0 2
    0 0 0 -2 2
    0 0 0 0 8

    Subtrace 1/4 of last from each of the others

    -2 0 0 0 0
    0 -2 0 0 0
    0 0 -2 0 0
    0 0 0 -2 0
    0 0 0 0 8

    Det = 128
     
  4. Apr 25, 2005 #3
    you can use cofactor formula or just do elimination like what OlderDan did. i prefer doing elimination and multiply diagonals to get determinant. i think thats how computer does it also. however on a test, you might have to write down the cofactors explicitly. i don't really remember all the details, but i believe the general idea is to reduce the problem from finding det of large matrix to a small one.
    I'll illustrate w/ a smaller matrix
    a b c
    d e f
    g h i

    if you expand along the first row, it's
    a * det(e f; h i) - b * det(d f; g i) + c * det(d e; g h)
    so you can see, the general pattern is
    (entry j from row 1) * (det of matrix w/ row 1, column j erased) * (-1)^(row+col)
    i think the sign is built into cofactors, but from here i hope you can see its like reduction formula. i mean you can reduce it to finding det of 1x1 matrix.
    det(e f; h i) = e * det(i) - f * det(h)
    again the sign comes from (-1)^(row+col)
     
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