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Linear algebra -- compute the following without finding invA

  1. Mar 6, 2015 #1
    1. The problem statement, all variables and given/known data

    upload_2015-3-6_12-4-18.png
    2. Relevant equations
    A=LU, U^-1 * L^-1= A^-1 , U^-1 * L^-1 * U^-1 * L^-1 = A^-2,

    3. The attempt at a solution
    I used MATLAB and the relations:
    U^-1 * L^-1= A^-1 , U^-1 * L^-1 * U^-1 * L^-1 = A^-2,
    to find a solution

    I found U^-1*L^-1 , let =B

    Then, found B^2 and took the inverse of B to get A^-2.

    B=A^-1

    C=inv(B^2)

    so we have B*x+C*y= [2;5;10]

    My question is: is there an easier way to do this? We are supposed to do this problem by hand. This would have taken a good amount fo effort to find so many inverse matrices.


    Any input is appreciated!

    Thank you!
     
  2. jcsd
  3. Mar 6, 2015 #2

    Mark44

    Staff: Mentor

    L and U are triangular matrices, so I don't think it would have been all that hard to compute their inverses by hand, especially since they're only 3 x 3. Also, once you have A-1, just multiply it by itself to get (A-1)2. You don't have to do the product of four matrices you show.
     
  4. Mar 6, 2015 #3

    SammyS

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    When you found matrices, B and C, that violated your instruction which was that you should not form A-1 and A-2 explicitly.

    Simply find L-1 and U-1 and use those. They are easy to find by hand.
     
  5. Mar 6, 2015 #4

    HallsofIvy

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    As you say, since A= LU, [itex]A^{-1}= L^{-1}U^{-1}[/itex] and [itex]A^{-2}= L^{-1}U^{-1}L^{-1}U^{-1}[/itex] so the first thing I would do is find [itex]L^{-1}[/itex] and [itex]U^{-1}[/itex]. Since L and U are triangular, that should be easy to do "by hand".

    (Well, that was three answer in quick succession- and we are all saying the same thing.)
     
  6. Mar 7, 2015 #5

    Ray Vickson

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    You can do it by solving equations with ##A## on the left, which is doable easily from the LU decomposition. You want to compute ##v+w##, where ##v = A^{-1} x## and ##w = A^{-2} y##. We can get ##v## as the solution of the system ##A v = x## and we can get ##w## as the solution of ##A^2 w = y##, and this, in turn, can be obtained from the systems ##A w_0 = y##, ##A w = w_0##.

    I don't know if this is what the questioner really wants.
     
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