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Use of guassian elimination to find the determinate of a matrix

  1. Nov 21, 2013 #1
    1. The problem statement, all variables and given/known data

    I have to find that the characteristic polynomial for a 3x3 matrix A, which is effectively the same as finding the det(A - I*lamda)

    matrix A is: attachment.php?attachmentid=254147&d=1385085115.jpg

    the problem is that when I've written out 'A - I*lamda' in matrix form, the three values at the bottom left of the matrix (where I've circled in blue in the pic above)

    I know that guassian elimination involves subtracting one row (or a multiple of a row) from another, so that I end up changing those three values to zero. But I don't get how this method would work when for one of the rows, theres more than one value which I need to change to zero.
     
  2. jcsd
  3. Nov 21, 2013 #2

    LCKurtz

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    The three values what? You didn't finish the sentence.

    Why don't you just write down ##A -\lambda I## for your original matrix and calculate its determinant?
     
  4. Nov 21, 2013 #3

    Ray Vickson

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    If you have two things to 'zero out' in column 1, you need to do two row-subtractions---one for row 2 and another for row 3. I am very, very surprised that you have not seen this before!

    However, I would urge you to NOT start in row 1, because you will obtain rational (not polynomial) values in columns 2 and 3. (Can you see why?) Better: interchange rows 1 and 2 (say), then start with the new row 1---that is, you will be adding or subtracting multiples of 3 to make zeros.
     
  5. Nov 21, 2013 #4

    SteamKing

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    The usual method to find the characteristic polynomial of a matrix is to first form the matrix (A - Iλ)

    Because of the variable λ in this matrix, Gaussian elimination (a numerical technique) is of limited value in calculating this determinant. However, since you are dealing with a 3x3 matrix, there is an algorithm (the Rule of Sarrus) which allows one to calculate the determinant with a fair amount of algebra involved.

    http://en.wikipedia.org/wiki/Rule_of_Sarrus

    http://en.wikipedia.org/wiki/Characteristic_polynomial
     
  6. Nov 21, 2013 #5

    The three values I was referring to, are the ones in the matrix: A - I*lamda

    they're located in the same positions are the three values I've circled for matrix: A

    I know I have to get them to zero, right?
     
  7. Nov 21, 2013 #6
    but theres two values in row 3 that I need to get to zero

    I can't make a subtraction that would get them both to zero, because they're not the same values


    sorry I can't see why
     
  8. Nov 21, 2013 #7

    LCKurtz

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    Like I asked before, why don't you just write down ##A-\lambda I## and take its determinant?

    Also, I don't see any ##\lambda##'s in your matrix, so it isn't ##A-\lambda I## in the first place.
     
  9. Nov 21, 2013 #8
    seems like a decent method, never come across it before though

    is there any significance to the values next to each of 'a' in the matrix shown in your first link?
     
  10. Nov 21, 2013 #9
    the matrix I posted in the pic inb my 1st post was for matrix A, not (A - I*lamda)

    I've done what you've suggested, by writing down A - I*lamda, its basically the same as matrix A, except for the three values in the central diagonal, they each have '- lamda'

    I'm trying to calculate the determinate of A - I*lamda by using guassian elimination
     
  11. Nov 21, 2013 #10

    LCKurtz

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    OK then. I thought from your OP you were asking about reducing ##A##, which seemed irrelevant. Both methods should work on ##A-\lambda I## if your arithmetic skills are good.
     
  12. Nov 21, 2013 #11

    SteamKing

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    The values next to the 'a' entries represent the row and column indices of that matrix element.
    For instance, a21 tells you this represents the element at row 2, column 1 of the matrix.

    This is basic stuff. How much do you know about matrix algebra?
     
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