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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. –  Patrick Li Oct 7 '12 at 12:40
    
What have you tried? The first thing to try would be to simply multiply these two matrices, wouldn't it? –  Hagen von Eitzen Oct 7 '12 at 12:41
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It is probably more interesting to ask: how is the inverse of the matrix obtained in the first place:) –  Shiyu Oct 7 '12 at 12:56
    
@PatrickLi: OK, I will. Thanks. I forgot about inverse matrices and the way we used to solve such problems at high school. –  Gigili Oct 7 '12 at 13:05
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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. –  Harald Hanche-Olsen Oct 7 '12 at 13:16
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up vote 7 down vote accepted

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 $$U B D^{-1} = \left(A - B D^{-1} C\right)^{-1} B D^{-1} = \left( D B^{-1} \left(A - B D^{-1} C\right)\right)^{-1} = -V $$

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