hi, i hv to do the inverse of a matrix by row elimination

plz

help

Jump to PostThis has a step by step method:

http://www.ehow.com/how_5079317_calculate-inverse-matrix.html

Jump to PostThis has a step by step method:

http://www.ehow.com/how_5079317_calculate-inverse-matrix.htmlThat was a nice link enlightening the procedure.

I would like to add more on it by saying that u could use matrix template library

http://www.osl.iu.edu/research/mtl/intro.php3It might be helpful to you.

jonsca
1,059
Quantitative Phrenologist
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rahul8590
71
Posting Whiz

This has a step by step method:

http://www.ehow.com/how_5079317_calculate-inverse-matrix.html

That was a nice link enlightening the procedure.

I would like to add more on it by saying that u could use matrix template library

http://www.osl.iu.edu/research/mtl/intro.php3

It might be helpful to you.

jonsca
1,059
Quantitative Phrenologist
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StuXYZ
731
Practically a Master Poster

If you do write your own Gauss-Jorden elimination procedure, be ** very** careful to deal with the huge numerical rounding errors. Re- stabalization of matrix after most steps is essential. Also please not

the obvious test matrix [1,2,3][4,5,6][7,8,9] has zero determinate so has no inverse.

Simple approaches in books like numerical recipeshttp://www.nr.com/, e.g. pivoting are a start but much better methods exist (e.g. partial row summation).

Note the NR book is freely viewable.

KASHIF JAMAL
0
Newbie Poster

```
//PennyBoki @ </dream.in.code>
#include <stdio.h>
int main()
{
float A[3][3];// the matrix that is entered by user
float B[3][3];//the transpose of a matrix A
float C[3][3];//the adjunct matrix of transpose of a matrix A not adjunct of A
double X[3][3];//the inverse
int i,j;
float x,n=0;//n is the determinant of A
printf("========== Enter matrix A =============================================\n");
for(i=0;i<3;i++)
{ printf("\n");
for(j=0;j<3;j++)
{
printf(" A[%d][%d]= ",i,j);
scanf("%f", &A[i][j]);
B[i][j]=0;
C[i][j]=0;
}
}
for(i=0,j=0;j<3;j++)
{
if(j==2)
n+=A[i][j]*A[i+1][0]*A[i+2][1];
else if(j==1)
n+=A[i][j]*A[i+1][j+1]*A[i+2][0];
else
n+=A[i][j]*A[i+1][j+1]*A[i+2][j+2];
}
for(i=2,j=0;j<3;j++)
{
if(j==2)
n-=A[i][j]*A[i-1][0]*A[i-2][1];
else if(j==1)
n-=A[i][j]*A[i-1][j+1]*A[i-2][0];
else
n-=A[i][j]*A[i-1][j+1]*A[i-2][j+2];
}
printf("\n========== The matrix A is ==========================================\n");
for(i=0;i<3;i++)
{
printf("\n");
for(j=0;j<3;j++)
{
printf(" A[%d][%d]= %.2f ",i,j,A[i][j]);
}
}
printf("\n \n");
printf("=====================================================================\n\n");
printf("The determinant of matrix A is %.2f ",n);
printf("\n\n=====================================================================\n");
if(n!=0) x=1.0/n;
else
{
printf("Division by 0, not good!\n");
printf("=====================================================================\n\n");
return 0;
}
printf("\n========== The transpose of a matrix A ==============================\n");
for(i=0;i<3;i++)
{
printf("\n");
for(j=0;j<3;j++)
{
B[i][j]=A[j][i];
printf(" B[%d][%d]= %.2f ",i,j,B[i][j]);
}
}
printf("\n\n");
C[0][0]=B[1][1]*B[2][2]-(B[2][1]*B[1][2]);
C[0][1]=(-1)*(B[1][0]*B[2][2]-(B[2][0]*B[1][2]));
C[0][2]=B[1][0]*B[2][1]-(B[2][0]*B[1][1]);
C[1][0]=(-1)*(B[0][1]*B[2][2]-B[2][1]*B[0][2]);
C[1][1]=B[0][0]*B[2][2]-B[2][0]*B[0][2];
C[1][2]=(-1)*(B[0][0]*B[2][1]-B[2][0]*B[0][1]);
C[2][0]=B[0][1]*B[1][2]-B[1][1]*B[0][2];
C[2][1]=(-1)*(B[0][0]*B[1][2]-B[1][0]*B[0][2]);
C[2][2]=B[0][0]*B[1][1]-B[1][0]*B[0][1];
printf("\n========== The adjunct matrix of transpose of the matrix A ==========\n");
for(i=0;i<3;i++)
{
printf("\n");
for(j=0;j<3;j++)
{
printf("C[%d][%d]= %.2f",i,j,C[i][j]);
}
}
printf("\n\n");
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
{
X[i][j]=C[i][j]*x;
}
}
printf("\n========== The inverse matrix of the matrix you entered!!! ==========\n");
for(i=0;i<3;i++)
{ printf("\n");
for(j=0;j<3;j++)
{
printf(" X[%d][%d]= %.2f",i,j,X[i][j]);
}
}
printf("\n\n");
return 0;
}
```

Edited
by mike_2000_17 because:
*
Fixed formatting *

rahul8590
commented:
please donot post source code if the OP asks a problem .
+0

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