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Given A^-1 find A

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Given

    [tex]A^{-1} =
    \left[\begin{array} {cc}
    2&-1\\
    3&5
    \end{array}\right]
    [/tex]

    find A.

    The only thing that we have learned regarding inverses so far is that for a 2x2 matrix, the inverse is given by

    [tex]A^{-1} = \frac{1}{ad-bc}
    \left[\begin{array} {cc}
    d&-b\\
    -c&a
    \end{array}\right]
    [/tex]

    So I set corresponding entries equal to each other and got the following 4 EQs:

    [tex]a = \frac12-\frac{bc}{d}[/tex]

    [tex]c = \frac{ad}{b}-1[/tex]

    [tex]b = \frac13+\frac{ad}{b}[/tex]

    [tex]d = \frac15 +\frac{ad}{c}[/tex]


    So my question is, what is the best way to solve these? I have been trying substitution, but I guess I really don't know how to use that method with so many EQs?

    I feel like I am just running in circles here! I took the 1st EQ and plugged the 2nd in for 'c' then I plugged the 3rd in for 'b' and the 4th for 'd.'

    But that did not eliminate anything. I still have all 4 vars in the final EQ.

    I need something more systematic. I am not sure which moves are legal here. How can eliminate something here?
     
  2. jcsd
  3. Oct 11, 2009 #2

    Pengwuino

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    Gold Member

    No no, for 2x2, a,b,c,d are the entries in the actual matrix A. a = 2, b = -1, c = 3, d = 5.
     
  4. Oct 11, 2009 #3
    Sorry. I don't follow. I know that a,b,c,d are the entries in A. But I am given the entries in A-1. So I need to work backwards, right?

    Thanks.

    EDIT: I think that I see what you mean. I did that; I am just showing the final forms of my EQS.

    For example the 1st EQ comes from d/(ad-bc) = 2
     
  5. Oct 11, 2009 #4

    Pengwuino

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    Gold Member

    Oh wait! I didnt see you were given the inverse and need to find A. Hmm let me think about this.
     
  6. Oct 11, 2009 #5

    Dick

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    Uh, (A^(-1))^(-1)=A, right? Just invert A^(-1) to find A.
     
  7. Oct 11, 2009 #6
    Oh. I guess that would be a lot easier!

    But while you are here: do you see what I mean about the endless loop of substitutions if I were to use that method?

    Is there a better way to solve this system? I thought about using Gaussian Elim, but for some reason my gut is against it.
     
  8. Oct 11, 2009 #7

    Dick

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    No, I don't see any problems with endless loops of substitution. You seem to be using the same a,b,c,d notation for the entries of A as for A^(-1), if A^(-1)_11 is 'a' then A_11 probably shouldn't also be denoted by 'a'. Just invert A^(-1) to get A. Period. That's an order. In the world of invertible matrices there is no distinction between "matrices" and "inverse matrices". They all use the same formulas.
     
    Last edited: Oct 11, 2009
  9. Oct 11, 2009 #8
    Okay!! I will! BUT ( I don't expect you to read this, 'cause I know how that goes) I don't think that I abused any notation here.

    If

    [tex]A =
    \left[\begin{array} {cc}
    a&b\\
    c&d
    \end{array}\right]
    [/tex]

    then,

    [tex]
    A^{-1} = \frac{1}{ad-bc}
    \left[\begin{array} {cc}
    d&-b\\
    -c&a
    \end{array}\right]
    =\left[\begin{array} {cc}
    2&-1\\
    3&5
    \end{array}\right][/tex]

    Set corresponding entries equal and you get for example

    [tex]\frac{1}{ad-bc}*d=2\Rightarrow a = \frac12-\frac{bc}{d}[/tex]

    which is what I got in the first post so and so forth.

    So I believe that my EQs are correct and therefore I do see a problem with nasty substitutions.

    Maybe you are a substitution wizard. I am not.
     
  10. Oct 11, 2009 #9

    Hurkyl

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    The problem is that you never eliminated any of the variables. You eliminated the 'c' from the first equation by using the second equation to make a substitution. But because you did not eliminate 'c' from the third and fourth equations, it reappeared later in your calculation.
     
  11. Oct 11, 2009 #10
    Exactly! SO can I use Gaussian elimination? I just feel weird about it since I have a's times b's and so and so forth.

    That's like having x1*x2 in GE and I have yet to come upon that. Is GE still a valid solution for problems involving variables with other variables as coefficients?

    Thanks :smile:
     
  12. Oct 11, 2009 #11

    Hurkyl

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    It's not Gaussian elimination.

    But yes, you can solve this system of equations by systematically eliminating variables through substitution.

    This is a slow and painful way of finding the inverse of A-1, but you can do it.
     
  13. Oct 11, 2009 #12
    Painful. Yes, I have noticed. So would it be safe to say that if I choose to start with the 1st EQ:

    [tex]a = \frac12-\frac{bc}{d}[/tex]

    and then start substituting EQS 2,3 & 4, I can use 2,3, & 4 as many times as needed to eliminate all of the variables?

    I just cannot reuse EQ 1 again.

    Or is that not a safe assumption?

    (I know that Dick's way is the "best" way. But sometimes I like to do it the 'hard' way. I learn more that way)
     
    Last edited: Oct 11, 2009
  14. Oct 11, 2009 #13

    Hurkyl

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    Well, I suspect you have the right idea -- but can't be sure. (it's easy enough for you to test yourself :wink:)

    The "can't reuse" bit isn't that it's mathematically unsound to do so -- it's that it merely acts contrary to the way you're simplifying the system of equations.
     
  15. Oct 11, 2009 #14

    Hurkyl

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    Incidentally, Gaussian elimination can be used to find the inverse of a matrix T. Presumably you've seen the algorithm in your book? It involves creating the augmented matrix
    [T | I]​
    and then row reducing.
     
  16. Oct 11, 2009 #15
    easy yes. but still painfully time consuming.

    Yeah. I just know that in order to eliminate a variable, the equations have to be linearly independent of one another.

    So if the whole reusing of EQs gets really confusing for me.

    My original approach was the following:

    1. Use EQ 1 which is solved for 'a'
    2. plug EQ 2 into 1 where ever I see 'c'
    3. plug EQ 3 into the resultant of the above where ever I see 'b'
    4. plug EQ 4 into the resultant of the above where ever I see 'd'

    rinse and repeat steps 2 - 4 until all variables are gone except for 'a.'

    It just got to the point, though, that I wasn't sure if that would actually ever happen!

    Will it? Or could it be that there is a unique sequence in which I would have to substitute the EQs in order to successfully solve for 'a' alone?



    Not there yet. But I peaked ahead a little and that does look familiar.
    Thanks!!!
     
  17. Oct 11, 2009 #16

    Hurkyl

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    And the problem you saw is that when you eliminated 'b' from equation 1, you used a substitution that introduced 'c', which undid your earlier work to eliminate 'b' from equation 1.

    What you need, then, is a substitution that eliminates 'b', but doesn't have any occurrences of 'c'.
     
  18. Oct 12, 2009 #17

    Mark44

    Staff: Mentor

    Re Gaussian elimination...
    C'mon, this is just a 2 x 2 matrix! How painfully time consuming can that be?
     
  19. Oct 12, 2009 #18
    This comment was regarding using substitution
    method using the 4 EQs in post #1, not GE.

    So if you would really like to know, give it a try :wink:

    As an aside, even if GE were employed, it still wouldn't be
    a 2x2 matrix as you say. I think maybe you misunderstood
    the thread.
     
  20. Oct 12, 2009 #19

    Mark44

    Staff: Mentor

    My comment pertained to the 2x2 matix in post #1, and using GE to solve the 2x4 matrix system [A-1|I] to get [I|A].
     
  21. Oct 12, 2009 #20
    I see. The way that you snipped my quote though made it seem like when I said it was "painfully time consuming" I was speaking solving [A|I]. But I was not.


    In post #12 I decided that I would like to further prod the idea of using substitution with the 4 EQs in post #1.

    It gets pretty nasty as pointed out by Hurkyl. I am still curious to know if there is only one exact sequence that the EQs must be substituted in in order to reduce the EQs to 1 in 1 unknown. Though I doubt there is only one, I have yet to one. But that is because I went to bed. But now I am up at at 'em :smile:

    Hopefully I won't spend all afternoon on this silly idea and will actually make some progress in the text :rofl:

    But I like exploring.... so if I do, that's cool too.
     
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