Matrix inverse with LU decomposition in C++

[KWKWII]

Homework Statement

the problem is to find the inverse of a 3x3 matrix using LU Decomposition with C++ command, with the numbers designated. in my case, my numbers for the matrix are
'306
410
780'

#include <stdio.h>
#include <iostream>
#include <stdlib.h>
#include <math.h>
using namespace std;

int main(void)
{
    int e, i, j, k, y;
    float A[3][4] = {{3,0,6,1},{4,1,0,2},{7,8,0,3}};
    float x[3][3], c, sum;
    float L[3][4] = {{1,0,0,0},{0,1,0,0},{0,0,1,0}};
        for(j=0; j<=2; j++)
        {
            for(i=0; i<=2; i++)
            {
                if(i>j)
                {
                    c=A[i][j]/A[j][j];
                    L[i][j] = c;
                    for(k=0; k<=2; k++)
                    {
                        A[i][k]=A[i][k]-c*A[j][k];
                    }
                }
            }
        }
             
        for(y=0; y<=2; y++)
        {
            if(y==0)
            {
                    L[0][3] = 1;
                L[1][3] = 0;
                L[2][3] = 0;
            }
            else if (y==1)
            {
                    L[0][3] = 0;
                L[1][3] = 1;
                L[2][3] = 0;
            }
            else if (y==2)
            {
                    L[0][3] = 0;
                L[1][3] = 0;
                L[2][3] = 1;
            }
            A[0][3]=L[0][3]/L[0][0];
            for(i=1; i<=2; i++)
            {
                sum=0;
                for(j=i-1; j>=0; j--)
                {
                    sum=sum+L[i][j]*A[j][3];
                }
                A[i][3]=(L[i][3]-sum)/L[i][i];
            }
                x[2][y]=A[2][3]/A[2][2];
                for(i=1; i>=0; i--)
                {
                    sum=0;
                    for(j=i+1; j<=2; j++)
                    {
                        sum=sum+A[i][j]*x[j][y];
                    }
                    x[i][y]=(A[i][3] - sum)/A[i][i];
                }
        }
    cout<<"Matriks X"<<endl;
    for(int i=0;i<=2;i++)
    {
        for(int j=0;j<=2;j++)
        {
            cout<< x[i][j]<<"  ";
        }
    cout<<endl;
    }
    return(0);
}

Homework Equations

The Attempt at a Solution

the result from the program will show most numbers correct, except the numbers on X₁₁ and X₂₁

it's supposed to be
A₁₁ = A₂₁ = 0
but it became
A₁₁ = -3.97364e-08
A₂₁ = 1.19209e-07

with loop, i can find out the numbers from each matrix, and all of them are correct

only the calculations from the X matrix seems wrong
need help in making A₁₁ and A₂₁ become 0

 

[Mark44]

Is this the same as the one you posted yesterday? If so, do not repost a problem just because no one has yet responded.

Homework Statement

the problem is to find the inverse of a 3x3 matrix using LU Decomposition with C++ command, with the numbers designated. in my case, my numbers for the matrix are
'306
410
780'

#include <stdio.h>
#include <iostream>
#include <stdlib.h>
#include <math.h>
using namespace std;

int main(void)
{
    int e, i, j, k, y;
    float A[3][4] = {{3,0,6,1},{4,1,0,2},{7,8,0,3}};
    float x[3][3], c, sum;
    float L[3][4] = {{1,0,0,0},{0,1,0,0},{0,0,1,0}};
        for(j=0; j<=2; j++)
        {
            for(i=0; i<=2; i++)
            {
                if(i>j)
                {
                    c=A[i][j]/A[j][j];
                    L[i][j] = c;
                    for(k=0; k<=2; k++)
                    {
                        A[i][k]=A[i][k]-c*A[j][k];
                    }
                }
            }
        }
            
        for(y=0; y<=2; y++)
        {
            if(y==0)
            {
                    L[0][3] = 1;
                L[1][3] = 0;
                L[2][3] = 0;
            }
            else if (y==1)
            {
                    L[0][3] = 0;
                L[1][3] = 1;
                L[2][3] = 0;
            }
            else if (y==2)
            {
                    L[0][3] = 0;
                L[1][3] = 0;
                L[2][3] = 1;
            }
            A[0][3]=L[0][3]/L[0][0];
            for(i=1; i<=2; i++)
            {
                sum=0;
                for(j=i-1; j>=0; j--)
                {
                    sum=sum+L[i][j]*A[j][3];
                }
                A[i][3]=(L[i][3]-sum)/L[i][i];
            }
                x[2][y]=A[2][3]/A[2][2];
                for(i=1; i>=0; i--)
                {
                    sum=0;
                    for(j=i+1; j<=2; j++)
                    {
                        sum=sum+A[i][j]*x[j][y];
                    }
                    x[i][y]=(A[i][3] - sum)/A[i][i];
                }
        }
    cout<<"Matriks X"<<endl;
    for(int i=0;i<=2;i++)
    {
        for(int j=0;j<=2;j++)
        {
            cout<< x[i][j]<<"  ";
        }
    cout<<endl;
    }
    return(0);
}

Homework Equations

The Attempt at a Solution

the result from the program will show most numbers correct, except the numbers on X₁₁ and X₂₁

it's supposed to be
A₁₁ = A₂₁ = 0
but it became
A₁₁ = -3.97364e-08
A₂₁ = 1.19209e-07

with loop, i can find out the numbers from each matrix, and all of them are correct

only the calculations from the X matrix seems wrong
need help in making A₁₁ and A₂₁ become 0

All of your matrix variables are type float, which can hold only about 6 or 7 decimal places. You will have a lot more precision if you make them double.

 

[KWKWII]

do not repost a problem just because no one has yet responded

it is a repost, but i think i need to because when i re-read the guidelines, i must use proper English
which, i think, is not proper on my last post
so i tried to simplify my words on the new post (here)

and by using double for all matrix variables, it indeed became more precise, but also became worse
after changing float -> double
A₁₁ = 7.40149e-17
A₂₁ = -2.22045e-16

 

[Mark44]

it is a repost, but i think i need to because when i re-read the guidelines, i must use proper English
which, i think, is not proper on my last post
so i tried to simplify my words on the new post (here)

Your English was fine in the other post. If you need to edit a post, and you're no longer able to do it, report the post using the Report button, and a mentor will take care of it.

and by using double for all matrix variables, it indeed became more precise, but also became worse
after changing float -> double
A₁₁ = 7.40149e-17
A₂₁ = -2.22045e-16

Those aren't worse. They're much closer to 0 than the results when you used float variables.

Floating point arithmetic on computers is inherently imprecise due to roundoff and truncation errors.

 

[KWKWII]

If you need to edit a post, and you're no longer able to do it, report the post using the Report button, and a mentor will take care of it

oh, i thought it wasn't fine...
i see, i'll do so next time

and also
although it is closer to 0, it is not 0
and the actual answer (done manually) is supposed to be 0
tried changing into integer, but it's even worse than using float or double
because the solution become numbers without decimals
is it also because of the rounding and truncation errors?

and i also notice that there's a minus on A₂₁ and no minus on A₁₁
a weird change considering the previous one is reversed

 

[Mark44]

You can set the precision of the numbers you display using the precision function in the iosbase class. Look at the example code on this page: http://www.cplusplus.com/reference/ios/ios_base/precision/.

 

[KWKWII]

using the precision function

#include <iomanip>

cout<< setprecision(3) << fixed << x[i][j]<<"  ";

(kind of different, but it works the same
i set it to only spell out 3 digits behind . )

using that, the result now becomes

A₁₁ = -0.000
A₂₁ = 0.000

well, although it's still weird for 0 to have (-) signs
but at least better for it shows the (seemingly) same amount as the actual answer

thanks

 

[Mark44]

The setprecision function is just truncating (or possibly rounding - I don't know) the fractional parts of numbers like 0.000000587 and -0.000342, and displaying only the first three decimal places.