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I am looking for some help on a little matrix/determinant problem please?

  1. May 26, 2008 #1
    i get the feeling that its gonna be hard to type this out but here goes. sorry about the dots, it was the only way.
    i gotta show that the determinant

    |..1.......1......-1...|
    |..a......-b.......c...| = (a+b)(a+c)(c-b).
    |a^2...b^2...-c^2.|

    however, i cannot quite get the answer to this. there is definitely one of these questions on my exam on friday and i just cannot ever reach the finish line with these ones.

    here is as far as i get with this particular one:

    |..1.......1.......-1..|
    |..a......-b........c..|
    |a^2...b^2...-c^2|

    =

    |..1-1............1...........1-1....|
    |..a+b...........-b..........a+c....| (took column 2 from column 1 - determinant unchanged)
    |a^2+b^2....b^2....a^2-c^2..|

    =

    |......0.............1.............0......|
    |.....a+b..........-b...........a+c....| (neatened things up a bit)
    |(a+b)(a-b).....b^2....(a+c)(a-c)|

    =

    |..0......1......0...|
    1/(a+b)(a+c)|..1......-b.....1...| (multiplied column 1 by 1/(a+b) & column 2 by 1/(a+c)
    |a-b...b^2...a-c.| - therefore multiplied determinant by same amount)

    = (b-c)/(a+b)(a+c) when that simplified determinant is calculated.

    ive tried messing around a few ways but this is the closest i have got to the answer, as in this attempt i at least got the (a+b) and (a+c) parts.

    can somebody show me the right way to do this please, and most importantly explain it?

    thanks.
     
  2. jcsd
  3. May 26, 2008 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Direct calculation of the determinant from the above gives the correct answer - since the first row is 0 1 0, the determinant is given by 2x2 determinant from rows 2 and 3 and columns 1 and 3, with a sign reversal.
     
  4. May 26, 2008 #3
    oh, thank you so much, i feel like an idiot, i have been on AGES trying to figure these out, i feel like an idiot lol
     
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