Matrix inverse with LU decomposition in C++

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Discussion Overview

The discussion revolves around finding the inverse of a 3x3 matrix using LU decomposition in C++. Participants share code snippets and results, focusing on discrepancies in the output values of the matrix elements, particularly A11 and A21, which are expected to be zero.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a C++ program to compute the inverse of a matrix using LU decomposition and notes that the output for A11 and A21 is not zero as expected, but rather very small floating-point numbers.
  • Another participant questions whether this is a repost of a previous problem, emphasizing the importance of not reposting without responses.
  • One participant suggests changing the matrix variable types from float to double for increased precision, which results in values closer to zero but still not exactly zero.
  • There is a discussion about the inherent imprecision of floating-point arithmetic, with one participant noting the presence of negative values in the outputs after switching to double.
  • A suggestion is made to use the precision function in C++ to control the display of floating-point numbers, which leads to a participant achieving a display of A11 and A21 as -0.000 and 0.000, respectively.
  • Another participant comments that the setprecision function is merely truncating or rounding the numbers, rather than resolving the underlying precision issues.

Areas of Agreement / Disagreement

Participants express varying opinions on the implications of using different data types for matrix calculations and the effects of floating-point precision. There is no consensus on how to achieve exact zero values for A11 and A21, and the discussion remains unresolved regarding the best approach to handle these precision issues.

Contextual Notes

Participants note that floating-point arithmetic can lead to roundoff and truncation errors, which may affect the results. The discussion highlights the limitations of using float and double types in C++ for precise calculations.

KWKWII
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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'

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

KWKWII said:

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'

Code: 
#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.
 
Mark44 said:
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
 
KWKWII said:
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.
KWKWII said:
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.
 
Mark44 said:
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 said:
using the precision function

Code: 
#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
 
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.
 

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