Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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
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
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 $$U B D^{-1} = \left(A - B D^{-1} C\right)^{-1} B D^{-1} = \left( D B^{-1} \left(A - B D^{-1} C\right)\right)^{-1} = -V $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.