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Unable to calculate the Determinant of a large matrix 
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#37
May1809, 12:27 AM

P: 398

Here goes: There are 120 possible permutations of the five rows and they can be listed in seven different classes according to how many elements they move. I'm going to use the group theory notation where (123) means 1 goes to 2, 2 goes to 3, 3 goes back to 1, and 4 and 5 are unmoved: Class 1: Similar to (12345) where everything moves: 24 permutations Class 2: Similar to (1234) (5) where only four rows move: 30 permutations Class 3: Similar to (123) where only 3 rows move: 20 permutations Class 4. Similar to (123)(45) where all 5 rows move (with incomplete mixing!) 20 permutations Class 5: Similar to (12)(34) where 4 rows move: 15 permutations Class 6: similar to (12) where only 2 rows move: 10 permutations Class 7: the identity e, where nothing moves: 1 permuation A quick check shows I've accounted for all 120 possible permutations of 5 rows. Interesting that each class has a definite parity...members of Class 2, for instance, can all be generated by a composition of 3 simple swaps, hence Clas 2 has odd parity. Since we need to clear all the zeroes out of the diagonal, we need only be interested in those classes for which every row gets moved. So only Class 1 and Class 4 need be counted. The 24 permutations in Class 1 are even, so they give a net of +24; the 20 permutations in Class 4 are odd so they net 20. Total = 4, which is the determinant of the 5x5 matrix. 


#38
May1809, 08:04 AM

P: 2,158

If you first replace row i by row i minus row 1 for all i > 1, you get the matrix:
[tex]\begin{pmatrix} 0&1&1&1&1\\ 1&1&0&0&0\\ 1&0&1&0&0\\ 1&0&0&1&0\\ 1&0&0&0&1 \end{pmatrix}[/tex] And if you now evaluate the determinant by summing over permutations, it is even simpler. 


#39
May1809, 08:06 AM

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#40
May1809, 12:42 PM

P: 226

I now see what you mean. I get a lot of zeros for the top rows all matrices. It seems that the determinant is 4 if there are odd number of rows, while 4 if there are even number of rows in the column. 


#41
May1809, 01:55 PM

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#42
May1809, 02:01 PM

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I would have you start with the 1x1 matrix, find the det for that one, continue with the 2x2 matrix. Just use the simplest way of calculating the determinant (diagonal > right > down = + and diagonal > left >down = ). 


#43
May1809, 03:01 PM

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#44
May1809, 03:48 PM

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#45
May1909, 01:09 AM

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Here is another approach to the problem. Write the matrix as AI, where A is the matrix for which every entry is 1. If we could find the eigenvalues of A, it would be easy to calculate the determinant. Compute A^{2}; you should find that A^{2} = nA. What does that tell you about the allowed eigenvalues of A?



#46
May1909, 02:10 AM

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Ha! I like that method, but you missed a neat trick. We can just do matrix algebra now  we can find the characteristic polynomial for A, and thus for the original matrix.



#47
May1909, 02:44 AM

P: 226

for rows n >= 2 det(n) = (n  1) * (1)^(n+1) A few examples of determinants: n det 2 1 3 2 4 3 5 4 6 5 ... 777 776 The result should now be correct. 


#48
May1909, 07:51 AM

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#49
May1909, 08:10 AM

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#50
May1909, 08:18 AM

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#51
May1909, 04:45 PM

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I'm not sure what Hurkyl had in mind, but you can get the multiplicities by noting that Tr A = n. (I had this in my original post, but then edited it out so as not to be giving out too much info to the OP.)



#52
May1909, 05:00 PM

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#54
May1909, 10:36 PM

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Why? This ones been solved at least two ways. I'm sure there's a million other ways. I'm still kind of leaning to the row reduction as the simplest.



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