2
$\begingroup$

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.

  • 0
    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

1 Answers 1

7

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 $