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Math Help - Calculate the inverse of the matrix using power series.

  1. #1
    Senior Member sfspitfire23's Avatar
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    Calculate the inverse of the matrix using power series.

    Calculate the inverse of [1, i/2 ; -1/3, 1] = I - [0, -i/2 ; 1/3, 0] using a power series.


    I know that (I - A)^-1 = I + A + A^2 +A^3 but am unsure of how to calculate this without using a computer.

    Thanks
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  2. #2
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    Re: Calculate the inverse of the matrix using power series.

    There are a few ways to do this. Typically you want to diagonalize any matrix that you're going to raising to powers, but here you can avoid that.

    \text{Let } A = \left(\begin{matrix}0 & -i/2 \\ 1/3 & 0 \end{matrix}\right). \text{ Then } A^2 = cI, \text{ where } c = \frac{-i}{6}.

    \text{Thus }\left(\begin{matrix}1 & i/2 \\ -1/3 & 1 \end{matrix}\right)^{-1}

    = (I - A )^{-1}

    = I + A + A^2 + A^3 +...

    Spoiler:


    = I + A + cI + cA + c^2I + c^2A + c^3I + c^3A + ...

    = (1+c+c^2+...)I + (1+c+c^2+...)A

    = (1+c+c^2+...)(I + A)

    = \frac{1}{1-c}\left(\begin{matrix}1 & -i/2 \\ 1/3 & 1 \end{matrix}\right).

    \text{Thus }\left(\begin{matrix}1 & i/2 \\ -1/3 & 1 \end{matrix}\right)^{-1} = \frac{1}{1-\frac{-i}{6}}\left(\begin{matrix}1 & -i/2 \\ 1/3 & 1 \end{matrix}\right) = \frac{1}{6+i}\left(\begin{matrix}6 & -3i \\ 2 & 6 \end{matrix}\right) = \frac{6-i}{37}\left(\begin{matrix}6 & -3i \\ 2 & 6 \end{matrix}\right).

    Last edited by johnsomeone; October 26th 2012 at 03:50 PM.
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  3. #3
    Senior Member sfspitfire23's Avatar
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    Re: Calculate the inverse of the matrix using power series.

    Thanks. How did you calculate c?
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  4. #4
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    Re: Calculate the inverse of the matrix using power series.

    I just did the matrix multiplication (they're 2x2 matricies with half zeros, so that's about than 10 second of work). I didn't know ahead of time it would work out this way. If was only after I did the multiplication that I saw that the power series would thereby algebraically "collapse" the way it did.

    As an aside, that's a kinda universal approach I have to math problems. When I'm given something general or abstract, I try to do (or apply it to) a few specific, more concrete examples - or in this case compute a term or two of a series. That often helps me "see" what's going on. Sometimes, like with this problem, I'll find something unexpected that will give me the key to how to solve the problem.
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