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Inverses of matrices

  1. Aug 25, 2008 #1
    1. The problem statement, all variables and given/known data
    hi, i have 2 nxn matrices A and B i found that (AB)^-1=A^-1xB^-1 but how do i do
    (A^2B^2)^-1=????

    plz i really need some help here im kinda lost i have
    AA^-1=I
    AI=A
    (AB)(A^-1xB^-1)=I

    anyone have any ideas?
     
  2. jcsd
  3. Aug 25, 2008 #2

    HallsofIvy

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    Staff Emeritus
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    Well, it's really unfortunate that you "found" that because it is not true!

    No! you can't cancel BA-1! That's why (AB)-1 is NOT A-1B-1.

    In order to be able to "cancel" A with A-1 or B with B-1, because multiplication of matrices is not commutative, you have to reverse the order of multiplication. (AB)(B-1A-1)= A(BB-1)A-1= A(I)A-1= AA-1= I. (AB)-1= B-1A-1.

    One thing you could do is think of (A2B2)-1 as (UV)-1= V-1U-1 where U= A2 and V= B2- that is (A2B2)-1= (B2)-1(A-1)2= (A-1)2(B-1)2.

    Similarly, (AABB)(B-1B-1A-1A-1)= AAB(BB-1)B-1A-1A-1= AA(B-1B-1)A-1A-1= A(AA-1)A-1= AA-1= I


     
    Last edited: Aug 25, 2008
  4. Aug 25, 2008 #3
    oh yes... sorry that was a typo it should of been B^-1A^-1 because of socks and shoes formula..... sorry, ok yes i understand what u mean, yer i kept thinking of it as A^2 and B^2 instead of AA and BB just got myself confused... thankyou heaps!
     
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