# simpleMatrix code

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 May 21, 2013, 05:23 simpleMatrix code #1 Senior Member   Dongyue Li Join Date: Jun 2012 Location: Torino, Italy Posts: 756 Rep Power: 10 Hi, In OpenFOAM's test---simpleMatrix.I have this code: Code: ```int main(int argc, char *argv[]) { simpleMatrix hmm(3); hmm[0][0] = -3.0; hmm[0][1] = 10.0; hmm[0][2] = -4.0; hmm[1][0] = 2.0; hmm[1][1] = 3.0; hmm[1][2] = 10.0; hmm[2][0] = 2.0; hmm[2][1] = 6.0; hmm[2][2] = 1.0; hmm.source()[0] = vector(2.0, 1.0, 3.0); hmm.source()[1] = vector(1.0, 4.0, 3.0); hmm.source()[2] = vector(0.0, 5.0, 2.0); Info<< hmm << endl; //Info<< hmm.solve() << endl; //Info<< hmm << endl; //Info<< hmm.LUsolve() << endl; //Info<< hmm << endl; //Info<< "End\n" << endl; return 0; }``` What does the codes in red mean?Thanks in advance. This is the result in terminal: Code: ```3 3((-3 10 -4)(2 3 10)(2 6 1)) 3((2 1 3) (1 4 3) (0 5 2))```

 May 21, 2013, 10:18 #2 New Member   Pierre HORGUE Join Date: May 2009 Posts: 29 Rep Power: 10 Your system "hmm" is a linear system of vectors So you have A x = b with : - A the matrix - "x" the solution and "b" the source term (both are vector of vectors). The code in red fill the source vector "b" (composed by vector because your system is a simpleMatrix of vectors) Code: ``` hmm.source()[0] = vector(2.0, 1.0, 3.0); hmm.source()[1] = vector(1.0, 4.0, 3.0); hmm.source()[2] = vector(0.0, 5.0, 2.0);``` Pierre sharonyue likes this.

August 21, 2014, 23:46
#3
Senior Member

MultiComb
Join Date: May 2014
Posts: 204
Rep Power: 6
Quote:
 Originally Posted by Pedro24 Your system "hmm" is a linear system of vectors So you have A x = b with : - A the matrix - "x" the solution and "b" the source term (both are vector of vectors). The code in red fill the source vector "b" (composed by vector because your system is a simpleMatrix of vectors) Code: ``` hmm.source()[0] = vector(2.0, 1.0, 3.0); hmm.source()[1] = vector(1.0, 4.0, 3.0); hmm.source()[2] = vector(0.0, 5.0, 2.0);``` Pierre
hi，Pedro24。but A=3 3((-3 10 -4)(2 3 10)(2 6 1)) multiply with x1=(-0.452599 1.12232 -0.149847) (the result got by the test code) do not equal to b1=(2 1 3). why? i have misunderstand this? please help me. Thank you!!!

September 17, 2017, 02:10
#4
New Member

Big Orange
Join Date: Mar 2016
Posts: 8
Rep Power: 3
Quote:
 Originally Posted by wenxu hi，Pedro24。but A=3 3((-3 10 -4)(2 3 10)(2 6 1)) multiply with x1=(-0.452599 1.12232 -0.149847) (the result got by the test code) do not equal to b1=(2 1 3). why? i have misunderstand this? please help me. Thank you!!!
Hi wenxu, have you gotten the answer for this problem?

September 20, 2017, 06:06
#5
New Member

Pierre HORGUE
Join Date: May 2009
Posts: 29
Rep Power: 10
Quote:
 Originally Posted by bigorange Hi wenxu, have you gotten the answer for this problem?
Hi all,

I didn't see this message few years ago but you cannot separate one solution vector from the others (the system has 9 dependent unknowns, and not 3 x 3 unknwowns). When you run the code, you get :
Code:
`3((-0.452599 1.12232 -0.149847) (0.125382 0.452599 0.345566) (0.152905 0.0397554 0.2263))`
as solution. I you multiply :
Code:
```A =

-3   10   -4
2    3   10
2    6    1```
with the given solution :
Code:
```x =

-0.452599   1.122320  -0.149847
0.125382   0.452599   0.345566
0.152905   0.039755   0.226300```
you will get the source term (which is also a matrix, i.e. a vector of vectors) :
Code:
```A * x =

2.0000e+00   1.0000e+00   3.0000e+00
1.0000e+00   4.0000e+00   3.0000e+00
-1.0000e-06   5.0000e+00   2.0000e+00```
Sincerely,

Pierre

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