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Matrix Algebra Inverse Matrix Question

  1. Sep 8, 2010 #1
    1. The problem statement, all variables and given/known data

    If A is an nxn matrix such that A^3 = 0 (the zero matrix) then (I-A)^-1 = ...?

    A. not invertible
    B. I+A^2
    C. I-A
    D. I+A
    E. I+A+A^2

    2. Relevant equations

    3. The attempt at a solution

    I just don't know how to work out what the inverse of (I-A) is if I know A^3... how is this even helpful?? It's a multichoice question and apparently the answer is 'E'
  2. jcsd
  3. Sep 8, 2010 #2


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    You don't need to work it out; you just need to check each one!
  4. Sep 8, 2010 #3
    If I do that, I get to option C That is (I-A)(I-A)^-1=(I-A)/(I-A)=I which seems correct, but the answer is apparently E. I+A+A^2

    *EDIT: Meant I+A+A^2
    Last edited: Sep 8, 2010
  5. Sep 8, 2010 #4
    By definition (I – A) (I – A)^-1 = I, so let C = (I – A)^-1

    So C(I – A) = I thus C – CA = I so (C – CA)A^2 = A^2 thus CA^2 – CA^3 = A^2 and since we are given A^3 = 0, we get CA^2 = A^2 and since identity is unique for matrixes, C = I. so C = (I – A)^-1 = I. So (I – A)I = I. So I – A = I. So A = 0.

    I could be off on this somewhere, but I’m pretty sure A=0 and your inverse is equal to I. So if this is true B, C,D and E should be solutions. Whereas E is trivial by plugging it in.
  6. Sep 8, 2010 #5
    Never mind! Click! Just figured out I should have been going (I-A)(I-A)=...

    Thanks for the help
  7. Sep 8, 2010 #6
    Here you seem to imply that (I-A)^-1 = 1/(I-A). What exactly does it mean to divide the number 1 by the difference of the identity matrix and matrix A? Can you divide a number by a matrix?

    This illustrates how important it is to live by your definitions in math. How was A^-1 defined in your class, where A is a matrix? I bet the concept of division was never used.
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