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I am asked to prove that $$\begin{pmatrix} \\ A & B\\ C &D\end{pmatrix}^{-1}=\begin{pmatrix} M & -MBD^{-1} \\ -D^{-1}CM & D^{-1}+D^{-1}CMBD^{-1} \end{pmatrix}$$ Where $M=(A-BD^{-1}C)^{-1}$.

Unfortunately, I have no idea what to do about it.

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    Try to multiply the matrix on the right-hand side with $\begin{pmatrix} \\ A & B\\ C &D\end{pmatrix}$ and check if the result is an identity matrix.2012-10-07
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    What have you tried? The first thing to try would be to simply multiply these two matrices, wouldn't it?2012-10-07
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    It is probably more interesting to ask: how is the inverse of the matrix obtained in the first place:)2012-10-07
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    @PatrickLi: OK, I will. Thanks. I forgot about inverse matrices and the way we used to solve such problems at high school.2012-10-07
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    @Shiyu: For someone like you who have spent their whole life solving mathematical problems it might be easy, but for the ones who have not and are not interested in maths, it doesn't look as easy at first glance. I wanted a hint to start solving it and got it, you are not forced to leave comments to offend other users with your comments.2012-10-07
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    I think you take offence too easily. @Shiyu did not say it would be easy, only that it is more interesting. The Riemann hypothesis is more interesting still, by far, but nobody claims it's easy.2012-10-07
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    I never intended to mock or offend you. I think that would be more interesting. I just tried to support your question to get a more beneficial answer, which I think Sasha has given. @HaraldHanche-Olsen: Thanks.2012-10-08

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I guess rather than verifying the inverse stated in the assignment, you should derive it. Let $$ \begin{pmatrix} A & B \cr C & D \end{pmatrix}^{-1} = \begin{pmatrix} U & V \cr W & X \end{pmatrix} $$ We have: $$ \begin{pmatrix} 1 & 0 \cr 0 & 1 \end{pmatrix} = \begin{pmatrix} A & B \cr C & D \end{pmatrix} \cdot \begin{pmatrix} U & V \cr W & X \end{pmatrix} = \begin{pmatrix} A U + B W & AV + B X \cr CU + D W & CV + DX \end{pmatrix} $$ Thus $AV = -BX$ and $CU = -DW$, giving $X = -B^{-1} A V$ and $W=-D^{-1} C U$. Substituting back into block-diagonals: $$ A U - B D^{-1} C U = \left(A - B D^{-1} C U\right) U= 1 \qquad C V-D B^{-1} A V = \left(C - D B^{-1} A \right) V= 1 $$ Hence $$ U = \left(A - B D^{-1} C\right)^{-1} \qquad V = \left(C - D B^{-1} A\right)^{-1} $$ Now $$U B D^{-1} = \left(A - B D^{-1} C\right)^{-1} B D^{-1} = \left( D B^{-1} \left(A - B D^{-1} C\right)\right)^{-1} = -V $$