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#include <iostream>
#include <cstdlib>
using namespace std;
void printmatrix(float A[][4]);
void RowReduce (float A[][4]);
int main()
{
float A[3][4] = {{5, -6, -7, 7}, {3, -2, 5, -17}, {2, 4, -3, 29}}; //Using a
//matrix with
//a known answer of
//{2, 4, -3}
printmatrix(A); //print out the matrix
RowReduce(A); //perform the row reduced echelon form on matrix--
// it prints out again in the function itself
}
void printmatrix(float A[][4]){ // Outputs the matrix
int p=3;
int q=4;
for (int i=0; i<p; i++)
{
for (int j=0; j<q; j++)
{
cout << A[i][j] << " ";
}
cout << endl;
}
}
void RowReduce (float A[][4]){
int p=3; //rows
int q=4; //columns
int lead = 0; //"lead" determines the column we are at that
// holds the diagonal "pivot". This helps us to zero out all numbers
//above and below the pivot.
cout << endl;
while (lead<q-1)
{
for (int i=0; i<p; i++) //for each row . . .
{
if (i!=lead) //When the pivot column and row are equal,
//we do not want to zero out the diagonal,
//so we skip this line and only non-"lead" rows.
{
cout << A[lead][lead] << " " << A[i][lead]; //ignore this as this
//was part of my debugging attempts
for (int j=0; j<q; j++)
{
A[i][j]=A[lead][lead]*A[i][j]; // Here is where the math is and
//probably where the error is. I am not looking for
//a full identity matrix, but a fully reduced with a
//non-zero diagonal. Making them 1 is easy.
A[i][lead]=A[i][lead]*A[lead][j];
A[i][j]=A[i][j]-A[i][lead];
}
cout << endl;
}
}
lead++; // Now go to the next column pivot
cout << endl;
}
printmatrix(A);
}
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