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Matrices question

  • Thread starter Derill03
  • Start date
  • #1
63
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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?
 

Answers and Replies

  • #2
56
0
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.
 
  • #3
63
0
I am not familiar with this can you help a little more?

My teacher never covered this principle of co-factor expansion
 
  • #4
63
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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
 
  • #5
56
0
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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