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Math Help - Inverse matrix problem

  1. #1
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    Inverse matrix problem




    I don't really know how to go about solving this problem, since it's a partitioned matrix. If I write it out in its complete 4 * 4 form, it will take a long time to reduce it, and I won't be able to get an answer in the form required by the question. So what method of inverse should I use?
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  2. #2
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    For simple 2 x 2 matrices, the inverse of
    a b
    c d

    is
    d -b
    -c a

    divided by the determinant: ad-bc.

    Why don't you treat your 2x2 partitioned matrix as though it were a simple 2x2? You will be surprised how easily the inverse of A appears.
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  3. #3
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    Quote Originally Posted by temaire View Post



    I don't really know how to go about solving this problem, since it's a partitioned matrix. If I write it out in its complete 4 * 4 form, it will take a long time to reduce it, and I won't be able to get an answer in the form required by the question. So what method of inverse should I use?
    The inverse of the 2 by 2 matrix \begin{bmatrix}a & b \\ c & d\end{bmatrix} is \frac{1}{ad- bc}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}, as long as ad- bc\ne 0 of course.

    The whole point of "block matrices" is that the same thing is true of them: the inverse of \begin{bmatrix}A & B \\ C & D\end{bmatrix} is (AD- BC)^{-1}\begin{bmatrix}D & -B \\ -C & A\end{bmatrix}

    Here, A= B (sorry about that), B= 2I, C= I, and D= 0. AD- BC= 0- 2I so (AD- BC)^{-1}= -\frac{1}{2}I.

    The inverse of this matrix is \begin{bmatrix}-\frac{1}{2} & 0 & 0 & 0 \\ 0 & -\frac{1}{2} & 0 & 0\\ 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & -\frac{1}{2}\end{bmatrix}\begin{bmatrix}0 & -2I \\-I & B\end{bmatrix}= \begin{bmatrix}-\frac{1}{2} & 0 & 0 & 0 \\ 0 & -\frac{1}{2} & 0 & 0\\ 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & -\frac{1}{2}\end{bmatrix}\begin{bmatrix}0 & 0 & -2 & 0 \\ 0 & 0 & 0 & -2 \\ -2 & 0 & 1 & 2 \\  0 & -2 & 3 & 0\end{bmatrix} .
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  4. #4
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    This really helped guys. But I can't seem to get \begin{bmatrix}-\frac{1}{2} & 0 & 0 & 0 \\ 0 & -\frac{1}{2} & 0 & 0\\ 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & -\frac{1}{2}\end{bmatrix}\begin{bmatrix}0 & 0 & -2 & 0 \\ 0 & 0 & 0 & -2 \\ -2 & 0 & 1 & 2 \\ 0 & -2 & 3 & 0\end{bmatrix} in terms of B, B^{-1}, 0, I, as the question requires. Any ideas?

    Thanks for the help by the way.
    Last edited by temaire; March 4th 2010 at 02:39 PM.
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