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5 个评论
Mushahid Rasul
2020-3-20
Walter Roberson
2020-3-20
No, the equation involves
which in MATLAB would be a(i,j)
hamza
2022-10-18
编辑:Walter Roberson
2023-5-1
Q: how can i run this code on matlab?
#include<iostream>
using namespace std;
int main() {
float A[100][100], t, x, B[100], z;
int R, C, k = 0, m = 0, l = 0, n = 0, p = 0;
cout << "Rows:";
cin >> R; //input rows in augmented matrix
cout << "Columns:";
cin >> C; //input columns in augmented matrix
for (int i = 0; i < R; i++) //augmented matrix input loop
{
for (int j = 0; j < C; j++) {
cin >> A[i][j];
}
}
cout << endl;
cout << "AUGMENTED MATRIX\n"; //Display the matrix entered
for (int i = 0; i < R; i++) {
for (int j = 0; j < C; j++) {
cout << A[i][j] << " ";
}
cout << endl;
}
cout << endl;
while (k < R) {
if (A[k][k] == 0) //Row interchange if element of primary diagonal of coefficient matrix is zero
{
for (int i = k + 1; i < R; ++i) {
if (A[i][k] != 0) {
for (int j = 0; j < C; ++j) {
t = A[i][j];
A[i][j] = A[k][j];
A[k][j] = t;
}
++n;
break;
}
}
for (int j = 0; j < C; j++) {
if (A[k][j] != 0)
++l;
}
if (l == 0) //If all the elements of a row are 0 then there are infinite solutions
{
cout << "\nThere are infinitely many solutions of the system of equations.\n";
return 0;
} else if (l == 1 && A[k][C - 1] != 0) //If all the elements, except the last one, of a row are 0 then no solution exists
{
cout << "\nNo solution exits for the system system of equations.\n";
return 0;
}
l = 0;
z = A[k][C - 1] / A[k + 1][C - 1];
for (int j = k + 1; j < C; j++)
{
if ((A[k][j] / A[k + 1][j]) != z)
++p;
}
if (n == 0 && A[k][C - 1] != 0) {
if (p != 0) {
cout << "\nNo solution exists for the system of equations.\n";
return 0;
} else {
cout << "\nThe system of equations has infinite many solutions.\n";
return 0;
}
}
n = 0;
p = 0;
}
for (int i = 0; i < R; i++) {
for (int j = 0; j < C; j++) {
cout << A[i][j] << " ";
}
cout << endl;
}
cout << endl;
x = A[k][k];
if (A[k][k] != 1) //Row transformation: Divide all the elements of the row by its first non-zero element from left side
{
for (int j = k; j < C; j++) {
A[k][j] /= x;
}
}
for (int i = k + 1; i <= R; i++) //Store values(in array B) of all the elements lying below(in same column as that element) the
{ //first non-zero element of that row(from left side).
B[i] = A[i][k];
}
for (int j = 0; j < C; j++) //Conversion of matrix to Echelon form
{
for (int i = k + 1; i <= R; i++) {
A[i][j] -= (A[k][j] * B[i]);
}
}
for (int i = 0; i < R; i++) {
for (int j = 0; j < C; j++) {
cout << A[i][j] << " ";
}
cout << endl;
}
cout << endl;
++k;
}
while (m < (R - 1)) {
for (int i = 0; i < (R - m - 1); i++) //Conversion of matrix to Row Reduced Echelon form
{
B[i] = A[i][C - m - 2];
}
for (int j = 0; j < C; j++) {
for (int i = 0; i < (R - m - 1); i++) {
A[i][j] -= (A[R - m - 1][j] * B[i]);
}
}
for (int i = 0; i < R; i++) {
for (int j = 0; j < C; j++) {
cout << A[i][j] << " ";
}
cout << endl;
}
cout << endl;
++m;
}
cout << "ROW REDUCED ECHELON FORM\n";
for (int i = 0; i < R; i++) {
for (int j = 0; j < C; j++) {
cout << A[i][j] << " ";
}
cout << endl;
}
cout << endl;
cout << "The solution to the system of equations is\n"; //Display solutions
for (int i = 0; i < R; i++) {
cout << "X" << i + 1 << " = " << A[i][C - 1];
cout << endl;
}
return 0;
}
ayman
2023-11-14
hi
x(t)=

Steven Lord
2023-11-14
This sounds like a homework assignment. If it is, show us the code you've written to try to solve the problem and ask a specific question about where you're having difficulty and we may be able to provide some guidance.
If you aren't sure where to start because you're not familiar with how to write MATLAB code, I suggest you start with the free MATLAB Onramp tutorial to quickly learn the essentials of MATLAB.
If you aren't sure where to start because you're not familiar with the mathematics you'll need to solve the problem, I recommend asking your professor and/or teaching assistant for help.
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