That solved the problem, thank you very much!
Afterall, it ocurred that I don't need this procedure since I will only use 
diagonal matrices and inverse of such are trivial.

It will come in handy someday though, I am sure.
Thanks again.

--- 30.3.11 (?r), Hong Zhang <hzhang at mcs.anl.gov> napisa?(a):

Od: Hong Zhang <hzhang at mcs.anl.gov>
Temat: Re: [petsc-users] Matrix inverse and LU decomposition
Do: "PETSc users list" <petsc-users at mcs.anl.gov>
Data: 30 Marzec 2011 (?roda), 15:07

You need call
MatSetType(I,MATDENSE);MatSetType(Dinv,MATDENSE);
See?petsc-dev/src/mat/examples/tests/ex1.c (attached) on how to 
useMatMatSolve().Can you send me your complete code so I can check why Dinv is 
empty?

Hong

On Wed, Mar 30, 2011 at 8:31 AM, Bart?omiej W <bartlomiej.wach at yahoo.pl> 
wrote:

Hello, 

I want to obtain the inverse of a matrix, even though it's not advised by the 
FAQ ;-) and I have a problem that earlier responses didn't solve for me. Simply 
speaking : the inverse matrix that is a result of? DI=Dinv as told by the FAQ, 
is empty for me. Here is my code(the significant part) :


? ierr = MatCreate(PETSC_COMM_WORLD,&D);CHKERRQ(ierr);
? ierr = MatSetSizes(D,PETSC_DECIDE,PETSC_DECIDE,n,n);CHKERRQ(ierr);
? ierr = MatSetFromOptions(D);CHKERRQ(ierr);

? ierr = MatCreate(PETSC_COMM_WORLD,&Dinv);CHKERRQ(ierr);

? ierr = MatSetSizes(Dinv,PETSC_DECIDE,PETSC_DECIDE,n,n);CHKERRQ(ierr);
? ierr = MatSetFromOptions(Dinv);CHKERRQ(ierr);

? ierr = MatCreate(PETSC_COMM_WORLD,&I);CHKERRQ(ierr);
? ierr =
 MatSetSizes(I,PETSC_DECIDE,PETSC_DECIDE,n,n);CHKERRQ(ierr);
? ierr = MatSetFromOptions(I);CHKERRQ(ierr);

(...)

? ierr = MatAssemblyEnd(D,MAT_FINAL_ASSEMBLY);CHKERRQ(ierr);
? ierr = MatAssemblyEnd(I,MAT_FINAL_ASSEMBLY);CHKERRQ(ierr);

? ierr = MatAssemblyEnd(Dinv,MAT_FINAL_ASSEMBLY);CHKERRQ(ierr);

? MatView(D,PETSC_VIEWER_STDOUT_SELF);
??? 
? ierr = MatGetOrdering(D,? MATORDERING_NATURAL,? &perm,? &iperm); 
CHKERRQ(ierr); ??? 
? ierr = MatFactorInfoInitialize(&info); CHKERRQ(ierr);

? ierr = MatLUFactor(D, perm, iperm, &info); CHKERRQ(ierr);

? MatView(D,PETSC_VIEWER_STDOUT_SELF);
? MatView(I,PETSC_VIEWER_STDOUT_SELF);
??? 
? ierr = MatMatSolve(D,I,Dinv); CHKERRQ(ierr);

? MatView(Dinv,PETSC_VIEWER_STDOUT_SELF);


// Results in such output:

//====
 MATRIX D
row 0: (0, 2) 
row 1: (1, 1) 
row 2: (2, 1) 
row 3: (3, 1) 

//==== MATRIX D after LU decomposition
row 0: (0, 0.5) 
row 1: (1, 1) 
row 2: (2, 1) 
row 3: (3, 1) 

//==== Identity matrix

row 0: (0, 1) 
row 1: (1, 1) 
row 2: (2, 1) 
row 3: (3, 1) 

//==== inverse of D? (result of MatMatSolve(D,I,Dinv))
row 0:
row 1:
row 2:
row 3:


The approach with :

??? ierr = MatGetOrdering(D,? MATORDERING_NATURAL,? &perm,? &iperm); 
CHKERRQ(ierr); ??? 

??? ierr = MatGetFactor(D, MAT_SOLVER_PETSC, MAT_FACTOR_LU,&F);
??? ierr = MatFactorInfoInitialize(&info); CHKERRQ(ierr);
??? ierr = MatLUFactorSymbolic(F,D,perm,iperm,&info);CHKERRQ(ierr);
??? ierr = MatLUFactorNumeric(F,D,&info);CHKERRQ(ierr);

??? ierr = MatLUFactor(D,
 perm, iperm, &info); CHKERRQ(ierr);

Provides same output

Thank You for Your help.



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