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Matrices simple question

  1. Jun 5, 2012 #1
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    For part b:
    Could anyone why it is + or - 3? I really don't understand why there would be two solutions as det|P| as it would just be the absolute value of P, meaning just +ve?
     
  2. jcsd
  3. Jun 5, 2012 #2

    SammyS

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    It's "±" because your equation has the absolute value of the determinant of p .
     
  4. Jun 5, 2012 #3

    Mark44

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    You seem to be confusing P (the matrix) and p (a number). Uppercase and lowercase are significant here, and you seem to be ignoring the difference.

    The notation |P| doesn't mean "absolute value" of P; it is the determinant of P, also written as det(P).

    What you show as |det p| makes no sense, because you're not taking the deteriminant of the number p - you want the determinant of the matrix P.
     
  5. Jun 5, 2012 #4
    Why?


    What is shown in the OP is what the book has shown, not me.

    Thanks for your input so far.
     
  6. Jun 5, 2012 #5

    Mark44

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    OK, then that's an error in the book. Apparently the author got confused between P and p.

    What is written as |det p| should be written as |det P| or |det(P)|.
    Since |det (P)| = 3, then det(P) = ± 3.
     
  7. Jun 5, 2012 #6

    SammyS

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    The solution to

    4|x|=12

    is

    x = ±3 .

    Do you agree?
     
  8. Jun 5, 2012 #7
    No i've never learnt this before, could you explain it please?
     
  9. Jun 5, 2012 #8
    the solution to 4|x|=12 is x=±3 BECAUSE if you plug x back in it works for both 3 and -3 ie. if you only list one of these as a solution then you have not completely solved the equation.

    Therefore if you have 4|det P|=12 then |det P| = 3 meaning that det P can be either 3 or -3

    make sense?
     
  10. Jun 5, 2012 #9

    Mark44

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    The solutions to |y| = k, where k > 0 are y = k or y = -k.
     
  11. Jun 6, 2012 #10
    yes, thank you and to everyone else.
     
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