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Matrix related to inverse

  1. May 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Let B = [tex]\left(\begin{array}{cc}2&4\\1&2\end{array}\right)[/tex] and A be a matrix satisfying A2 = 0, AB = 0

    1. Calculate (I+A) (I-A), I = identity

    2. Calculate (I+A)-1 (I+2A)-1B

    2. Relevant equations



    3. The attempt at a solution
    1. Ans = I

    2.
    In the manual, it's written : (I+2A)-1 = 1/4 (I-2A)
    I don't understand how 1/4 pops up...

    Thanks
     
  2. jcsd
  3. May 26, 2010 #2

    Hurkyl

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    What is the definition of the inverse of a matrix? What happens if you tried plugging (I-2A) into the definition?


    Incidentally, it's worth making an analogy with the power series for 1/(1+2a).
     
  4. May 26, 2010 #3
    The inverse of a matrix A is a matrix A-1 such that AA-1=I

    (I-2A) (I-2A)-1 = I...I still don't get it

    power series of 1/(1+2a) = (1+2a)-1 = 1 - 2a - 8a2 - ...

    What is the relation to the inverse?


    Thanks
     
  5. May 26, 2010 #4

    Mark44

    Staff: Mentor

    Should be an alternating series, and the coefficient on a2 is wrong.
    1/(1 + 2a) can be written as (1 + 2a)-1.

    (I + 2A)-1 has a similar expansion. That's the connection.
    I also don't get what your solution manual shows for (I + 2A)-1; namely, the 1/4 factor. I think it's a typo.
     
  6. May 26, 2010 #5

    Hurkyl

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    (I've changed the variable in the above to reduce confusion)

    Sorry, I meant plugging in (I-2A) in for the "inverse of I+2A" -- i.e. B=I+2A and B-1=I-2A.

    My mistake, though -- when you make the substitution, things do work out. The equation
    (I+2A)-1 = (1/4) (I-2A)​
    is, indeed, in error.
     
  7. May 26, 2010 #6
    yeah, it should be 4

    a is variable and A is matrix. Can matrix also be expanded?

    So, (I + 2A)-1 should be equal to (I-2A). Am I right?

    Thanks
     
  8. May 26, 2010 #7

    Mark44

    Staff: Mentor

    Yes, since (I + 2A)(I - 2A) = I. This means that I + 2A is the inverse of I - 2A.
     
  9. May 26, 2010 #8
    Thanks a lot Mark and Hurkyl :smile:
     
  10. May 26, 2010 #9

    Hurkyl

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    Yes; you can do calculus with matrices! It's a little trickier, but useful. However, you don't have to know matrix analysis to use it as inspiration -- e.g. armed with the knowledge of geometric series, you can guess at a formula for the inverse of (I - A) is for any nilpotent matrix A... and once you have made the guess, it's easy to show directly that the formula does give the inverse.
     
  11. May 26, 2010 #10
    hm...haven't studied about it yet but I read about it a little on wiki. Thanks a lot hurkyl
     
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