Re: Partitioned matrix operations

*To*: mathgroup at smc.vnet.net*Subject*: [mg100664] Re: Partitioned matrix operations*From*: "Steve Luttrell" <steve at _removemefirst_luttrell.org.uk>*Date*: Wed, 10 Jun 2009 17:11:21 -0400 (EDT)*References*: <h0l467$oof$1@smc.vnet.net> <h0nug5$b6e$1@smc.vnet.net>

Your proposed use of Inverse works only if A, B and C commute (where appropriate). You can verify that your proposed inverse is not correct by premultiplying it by the original matrix thus: {{A, B}, {C, 0}} . {{0, C^(-1)}, {B^(-1), -A (B C)^(-1)}} When you expand this out the upper right element is A C^(-1) - B A (B C)^(-1) which does not simplify to 0 (as you would have liked) when A, B and C don't commute (where appropriate). Anyway, back to the original problem. Since it is only a 2x2 block matrix you can solve it manually thus: {{A,B},{C,0}} . {{U,V},{W,X}} = {{1,0},{0,1}} where you have to solve for the matrices U, V, W, X, and the 1 and 0 symbols on the r.h.s. stand for the appropriate matrices. I won't say any more in case this is a homework problem. -- Stephen Luttrell West Malvern, UK "dh" <dh at metrohm.com> wrote in message news:h0nug5$b6e$1 at smc.vnet.net... > > > Hi, > > you may symbolically invert your matrix: > > Inverse[M] > > giving: > > {{0, 1/C}, {1/B, -(A/(B C))}} > > Now it is up to you to ensure that C,B and B C are invertible. > > Daniel > > > > > > Joe Hays wrote: > >> Hello, > >> Here's a Mathematica newbie question. Say I have a matrix, M, defined as, > >> > >> M = {{A, B}, {C, 0}} > >> > >> where A is nxn, B is nxm, C is mxn, and the zero sum matrix is mxm. I >> would > >> like to perform an operation on the matrix M without fully defining A, B, > >> and C and get a result in terms of A, B, and C. For example, if I wanted >> to > >> determine the matrix inverse of M in terms of A, B, and C. > >> > >> Is this possible in Mathematica? I've unfortunately not found anything in > >> the docs that indicates that this is possible. > >> > >> Thx. > >> > >> > > >