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Using eigenvalues to get determinant of an inverse matrix

  1. Nov 21, 2016 #1
    1. The problem statement, all variables and given/known data

    Screen Shot 2016-11-21 at 11.00.51 AM.png
    2. Relevant equations
    determinant is the product of the eigenvalues... so -1.1*2.3 = -2.53
    det(a−1) = 1 / det(A), = (1/-2.53) =-.3952

    3. The attempt at a solution
    If it's asking for a quality of its inverse, it must be invertible. I did what I showed above, but my answer was incorrect.
     
  2. jcsd
  3. Nov 21, 2016 #2

    PeroK

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    How many eigenvalues does a ##3 \times 3## matrix have?

    Didn't you have a question yesterday about the trace of a matrix?
     
  4. Nov 21, 2016 #3
    Yes I did. For the record, our instructor has been MIA for about 4 days and missed the last lecture, so we are afloat...

    From your question regarding determinants, I realize we are missing a det. I have looked through the entirety of our textbook. every single homework problem and do not see a single problem like this one. I am not sure how to find the determinant considering we are missing 4 numbers in our matrix. We weren't told it was symmetric, so no clues there.
     
  5. Nov 21, 2016 #4

    PeroK

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    This is the difference between "plug and chug" and real maths. With this problem, you have to think things through. You have to work out how to use what you already know to get the answer.

    My question about the trace, by the way, was a big hint. What do you know about the trace of a matrix?
     
  6. Nov 21, 2016 #5
    it is the sum of the eigenvalues
     
  7. Nov 21, 2016 #6

    Ray Vickson

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    You claimed that ##51 \times (-1.1) \times 2.3 = -2.53##. Do you see anything wrong with that?
     
  8. Nov 21, 2016 #7
    Ive been thinking about how this math is different from Calc2 and its predecessors, and you are right, please have patience with me as I embark on the process of undoing 14 years worth of being taught "plug and chug"
     
  9. Nov 21, 2016 #8
    Hi Ray, 51st number. The 51th out of a lovely 95 :) but I can totally see why that would make you wonder... cute dog by the way.
     
  10. Nov 21, 2016 #9

    Ray Vickson

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    OK, that makes sense.

    If the three eigenvalues are ##r_1 = -1.1##, ##r_2 = 2.3## and ##r_3## (unknown), can you figure out how the quantities ##r_1 + r_2 + r_3##, ##r_1 r_2 r_3## and ##r_1 r_2 + r_1 r_3 + r_2 r_3## are related to the coefficients of the characteristic polynomial of ##A##?
     
  11. Nov 21, 2016 #10

    PeroK

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    Exactly. And you can calculate the trace of your matrix?
     
  12. Nov 21, 2016 #11
    I got so excited. Though I had it.... But I didnt, it was wrong. I took the diagonal values and added them to get the trace. I then used that to create a ratio for the determinant. Since Im looking for det of an inverse, I then divided the answer by 1.
    So:
    -2.1 + 4.2 + 1.7 3.8
    (-1.1)(2.3)x = 3.8 => -2.53

    3.8/-2.53 = -1.5019
    1/-1.5019 = -.6657
     
  13. Nov 21, 2016 #12

    PeroK

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    The trace is the sum of the eigenvalues, sk you can get the third eigenvalues from that.

    Then the determinant is the product of all three eigenvalues.
     
  14. Nov 21, 2016 #13

    Ray Vickson

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    Does your book really not tell you how to compute ##3 \times 3## determinants? If it does tell you, just use that method to figure out ##P(x) = \det(x I - A)##, the characteristic polynomial. (That is, you want to compute the determinant of the ##3 \times 3## matrix ##B = x I - A##.) It will be a lengthy and somewhat messy calculation, and your ##P(x)## will have the parameters ##a,b,c,d## in it (as well as ##x##, or course). However, until you have actually done it, you cannot predict what will drop out.

    If your book does not tell you how to do ##3 \times 3## determinants, there are hundreds of web pages that do so.
     
  15. Nov 22, 2016 #14
    It does, as I'm sure you know. There is no need to be condescending, I'm here for help, not to be judged. It does not however show how to solve for a 3x3 matrix that has 4 unknown coefficient variables.

    I get that you are trying to help without flat out giving answers, but there are nicer ways of saying things, or... not saying certain things at all. I already get this attitude from the teacher, when I reach out for help, I don't need it from you too.

    I haven't been in a math class of any kind in many years, and I have struggled enormously. When I finally do reach out for help, you can be guaranteed I've already tried for an hour or more to figure it out myself. I'm 43 years old, and I know that I have to put in the time and the work. I'm not some kid who wants everyone to do everything for her.
     
  16. Nov 23, 2016 #15

    PeroK

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    Why not pick up from post #12? I thought you were nearly there.
     
  17. Nov 23, 2016 #16

    Ray Vickson

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    N0, I don't know what is the content of your unnamed book; I don't know it its treatment is superficial or detailed; I don't know if it is very elementary or moderately advanced. And I certainly do not know what it has presented about determinants. That is why I asked.

    If your book does do some examples of ##3 \times 3## determinants, that is really all you need: just repeat the steps, but now with some symbolic instead of numerical entries. (That is what I indicated later: you will get terms with x, a,b,c, and d in them, but the steps you take ought to be the same as what your book has done.) The business about not knowing what you will get until you have done it is 100% true: you have a problem with 4 parameters (a,b,c,d), but will get a polynomial with only three coefficients (excluding the leading ##x^3## term), so it looks at first that maybe you will not have enough information to finish the problem. It turns out that you will have all you need, but you can only see that by going through the steps.

    And, of course, if your book does not do a good job of illustrating determinant computations, you really will have no choice but to consult another source. I used to suggest that people go the the library, but that has not worked for some time, so now I suggest looking on-line instead. You would perhaps be surprised at the number of people who never think to do an on-line search, so suggesting that is not insulting or condescending; it is just a matter of not knowing what is your background or the extent of your independence, confidence or initiative. I do not see that as "judging" you.
     
    Last edited: Nov 23, 2016
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