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Matrix/Determinant/Inverse Q's

  1. Jun 5, 2004 #1

    dcl

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    [tex]
    c = \left[ {\begin{array}{*{20}c}
    {2 - x} & 5 & 1 \\
    { - 3} & 0 & x \\
    { - 2} & 1 & 2 \\
    \end{array}} \right][/tex]

    a) Calculate det(C).
    My answer was [tex]x^2 - 12x + 27[/tex].

    b) Calculate det(2C).
    Umm, would this just be 2*det(C)?
    Couldn't find anything more helpful in my notes.

    c) State the values for 'x' for which C is not invertible.
    I believe the value for 'x' that would make this non invertable would be the solution that det(c) = 0. (A matrix has no inverse when the determinant = 0 yeh?)
    which would be x = 9 or 3
    is this correct?
     
  2. jcsd
  3. Jun 5, 2004 #2

    Janitor

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    Science Advisor

    I get the same answer as you on (a) and (c). I'm thinking that det(aC)=a^N det(c) where a is a constant and N is the "dimension" (wrong terminology? NxN matrix). If I am right about that, then for a 3x3 matrix, det(2C) = 8 det(C).

    You may be thinking about the trace of a matrix: Tr(2C) = 2 Tr(C).
     
    Last edited: Jun 5, 2004
  4. Jun 5, 2004 #3

    dcl

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    Hmm, yeh, what you're saying would make more sense.
    Thanks for confirming the other answers.
     
  5. Jun 5, 2004 #4
    Janitor is correct about (b). To see it, just consider what happens if C is the identity: 2I has 2 along the diagonal and 0 everywhere else, so the determinant is det(2I) = 8 = 23det(I).
     
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