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Matrix algebra question

  1. Oct 21, 2009 #1
    1. Show that if a square matrix A satisfies A2 - 3A + I = 0, then A-1 = 3I - A



    2. A-1A = I and A-1A = I and more that I can't think of



    3. 3A = A2 + I

    A = (A2 + I)/3

    ???


    This question is weird :eek:
    Anyone know how to do it?
     
  2. jcsd
  3. Oct 21, 2009 #2
    I'm so stupid I got the template wrong. I think I figured it out:

    Show that if a square matrix A satisfies A2 - 3A + I = 0, then A-1 = 3I - A

    A2 - 3A + I = 0
    A2 - 3A = -I
    A(A - 3I) = -I
    A - 3I = -I/A

    I/A = A-1, therefore:

    A-1 = 3I - A
     
  4. Oct 21, 2009 #3

    lanedance

    User Avatar
    Homework Helper

    i would take out the divide step, its not really defined for matricies, there is only multiplication by the inverse

    so before that step, just muliply through by -1 and you're finished
     
  5. Oct 22, 2009 #4
    Oh you're right! Matrix multiplication doesn't actually exist >.>

    Thanks for the input!
     
  6. Oct 22, 2009 #5

    Mark44

    Staff: Mentor

    Matrix multiplication actually DOES exist. It's matrix division that doesn't exist.
     
  7. Oct 22, 2009 #6

    Mark44

    Staff: Mentor

    I buy it up to the step where you have A(A - 3I) = -I. After that, I'm not buying. First off, and as already noted, there is no matrix division, so -I/A doesn't make sense.

    Second, you say that I/A = A-1. What makes you think that A has an inverse? All you are given is that A is a square matrix. There is nothing said about A being invertible (i.e., having an inverse).

    So how can you justify your last statement, that A-1 = 3I - A?

    Hint: A(A - 3I) = -I, or equivalently, A(3I - A) = I.
     
  8. Oct 22, 2009 #7
    Isn't it said that a matrix is invertible if the product is an identity matrix, whereas (3I - A) is the inverse of A, so you know it's invertible. Thus instead of dividing, you multiply both sides by the inverse of A which would make 3I - A = A-1

    Or wait is it AB = BA = I, so it has to commute? I'm not too sure how to prove that A is invertible in this case, if this isn't right. :(
     
  9. Oct 22, 2009 #8

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What are A(3I- A) and (3I- A)A ?
     
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