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Matrix determinants

  1. Apr 4, 2016 #1
    1. The problem statement, all variables and given/known data
    abe7cd6b9a.png

    2. Relevant equations


    3. The attempt at a solution
    6407b645fb.jpg

    The answer in the solutions is given as : (2x+1)(x-1)(1-x), they did their matrix differently so thats how they got that answer. I used wolfram alpha to factorise my quadratic on the last line and it gave me alternative forms which had 2 factors from the solutions provided to me, so my answer is correct. My issue is, how do I sort of convert my answer that I got into the ones that the solutions had as a lot of our exams are MCQ and I probably wouldn't have realized that they were the same answers.
     
  2. jcsd
  3. Apr 4, 2016 #2

    LCKurtz

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    Your mistake was using Wolfram for such a simple problem. You have ##(-2x^2+x+1)## that you want to factor. Standard would be to factor out the leading minus sign giving ##-(2x^2-x-1)## which you should easily be able to factor yourself into ##-(2x+1)(x-1)##.
     
  4. Apr 4, 2016 #3
    I never used wolfram while i was doing it. I used the x = -b formula to get the roots which gave me x=-1/2 and x=1
     
  5. Apr 4, 2016 #4

    LCKurtz

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    I'm not familiar with the "x=-b" formula for getting the roots. But after you said that I noticed that your two answers are in fact not the same. You don't want the roots, you want the factors, and they aren't the same. If you multiply out the last three factors in your image, you don't get the expression you have.
     
  6. Apr 4, 2016 #5

    SteamKing

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    I think the OP is referring to the Quadratic formula here.

    ##x_{1,2} = \frac{-b±\sqrt{b^2-4ac}}{2a}##
     
  7. Apr 4, 2016 #6
    I used that formula above, to get the roots and then i used the factor theorem to get the facctors.
     
  8. Apr 4, 2016 #7

    LCKurtz

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    But your last expression is not the correct factorization of your original polynomial. It is off by a factor of 2. The problem is$$
    -2x^2+x+1 = -(2x^2-x-1) = -(2x+1)(x-1)\ne (x-1)(x+\frac 1 2)$$You have to use the factor theorem correctly. Note that both ##2x+1## and ##x+\frac 1 2## have a root of ##-\frac 1 2##. All this confusion would have been avoided if you had just factored the expression in the first place instead of looking for roots.
     
  9. Apr 4, 2016 #8

    mfb

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    The roots are correct, but if you just look at the roots you miss the prefactor of 2.

    2x and x both have 0 as root, but they are not the same. Similar with (x+1/2) and (2x+1).
    Just multiply your factors together again and see if you get the right result.
     
  10. Apr 4, 2016 #9
    Yeah. I've gotten the right answer now, I made it complicated for no reason.
     
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