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Matrix Proof help please

  1. Nov 20, 2005 #1
    "Show that if a square matrix C satisfies
    C^3 + C^2 + C + I = 0
    then the inverse C^-1 exists and
    C^-1 = -(C^2 + C + I) :surprised
  2. jcsd
  3. Nov 20, 2005 #2

    all you need to remember is that
    c^2 = cc
    c^3 = ccc
    and the associative property of matrix multiplication
    P.S. c^(-1) is taken here as the left inverse of c
  4. Nov 20, 2005 #3
    Ok, i didn't realise i could write matrix's like that and deal with them just as variables...
    OK so this is what i did...

    CCC + CC + C + I = 0 (re arrange)
    C = - (CCC + CC + I) (I = C^(1) C)
    C = - (CCC + CC + (C^(-1)C)) (GET C^(-1)C on other side)
    C^(-1)C = - (CCC + CC + C) (DIVIDE BY C)
    C^(-1) = - (CC + C + 1) (1 = Identity matrix)
    C^(-1) = - (CC + C + I)

    DONE !? :!!) :uhh:
  5. Nov 20, 2005 #4
    you are okay up to your third equation

    "Show that if a square matrix C satisfies
    C^3 + C^2 + C + I = 0
    then the inverse C^-1 exists and
    C^-1 = -(C^2 + C + I)

    Lets do it simpler....

    Just multiple your left hand side (every term on it !) by C^(-1) and the right hand side by the same.

    C^(-1) ( C^3 + C^2+ C +I) = C^(-1) 0

    What do you get from here is the result that you're looking for
  6. Nov 20, 2005 #5
    Because well, you cannot exactly treat matrices as variables..

    The Division (Divide by C) is not defined, just the multiplication.
    and the multiplication of a matrix by its (left) inverse has the property that


    And because the matrix multiplication is associative, if

    C^2=CC then C^(-1)C^2= C^(-1)CC = (C^(-1) C) C = I C = C

    something similar for C^3 or an arbitrary power of C
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