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Matrices question

  1. Apr 11, 2009 #1
    I have been given the following matrix and i am told there is a quick trick to getting the determinant:

    0 1 2 3 4
    0 0 2 3 4
    0 0 0 3 4
    0 0 0 0 4
    5 4 3 2 1

    I used my calculator and found out the answer is 120 which is 5!, but i am not sure what rule or reason the answer is 5!

    Can ne one help?
     
  2. jcsd
  3. Apr 11, 2009 #2
    Hint: Use a co-factor expansion down the first column and use the knowledge that the determinant of an upper-triangular matrix is the product of the entries on its diagonal.
     
  4. Apr 11, 2009 #3
    I am not familiar with this can you help a little more?

    My teacher never covered this principle of co-factor expansion
     
  5. Apr 11, 2009 #4
    This is what i did can u confirm:

    5*( 1 2 3 4) inside parenthesis is the 4x4 matrix
    ( 0 2 3 4)
    ( 0 0 3 4)
    ( 0 0 0 4)

    which is 5*4*3*2*1 = 120
     
  6. Apr 12, 2009 #5
    Let i be some row and let j be some column of a matrix A. To find the determinant of A, you can use a co-factor expansion across a row, and you would use the following formula for a fixed i and varying j. Or, you can use co-factor expansion down a column, where you fix a j and sum over all i's.

    det(A) = [tex]\sum[/tex]Cij = [tex]\sum[/tex](-1)[tex]^{i+j}[/tex]*(i,j)*det(Aij), where Aij denotes the matrix formed by removing the ith row and jth column from A; and, (i,j) denotes the entry in the ith row and jth column.

    So, in this case, you would fix j at 1 and sum from i = 1 to i = 5.

    det(A) = 0+0+0+0+5*det(A5,1) = 5*(1*2*3*4) = 120.

    Is this clear?
     
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