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Determine T-inverse (1,11)

  1. Jun 25, 2011 #1
    I thought that I could find the inverse of the coefficient matrix, but it's originally 2x3, so I redacted the linearly dependent row and found the 2x2 A inverse. I'm not sure what to do after that.

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110625_165538.jpg [Broken]

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110625_170257.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jun 25, 2011 #2

    micromass

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    Hi Shackleford! :smile:

    The problem is not asking you to determine the inverse of T. The problem is asking you to calculate [itex]T^{-1}(1,11)[/itex] which is the set of all triples (a,b,c) such that

    [tex]T(a,b,c)=(1,11)[/tex]

    There will NOT be a unique (a,b,c) that satisfies this (in general). We will expect a set of triples as answer.

    The equation brings us to a system of equations that you need to solve:

    [tex]\left\{\begin{array}{c} a+b=1\\ 2a-c=11\\ \end{array}\right.[/tex]
     
  4. Jun 26, 2011 #3
    Oh. Well, I quickly misread that problem. It's no problem now. I'll do it in the morning. Thanks!
     
  5. Jun 26, 2011 #4
    The strategy is to find the set of solutions to the homogeneous equation and then find a particular solution.

    a + b = 1
    2a - c = 11

    a + b = 0
    2a - c = 0

    Implies a = a, b = -a, c = 2a.

    KH = a(1, -1, 2)

    Solving the system yields

    a -(1/2)c = 11/2
    b -(1/2)c = -9/2

    The book's answer sets c = 0 and gives the particular solution as (11/2, -9/2, 0). Not too difficult a problem. It's a good problem to test your fundamental understanding of the theory and technique.
     
  6. Jul 10, 2011 #5
    my daughter is studying in Class XI in a AP board school with BiPC.She is struggling to find the value of Tan inverse 4. Pl help.
     
  7. Jul 10, 2011 #6

    Ray Vickson

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    Do you mean arctan(4) or tan(1/4)? I can read your question either way. Also: what angular units are used (degrees? radians?).

    RGV
     
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