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C++ Program to Perform LU Decomposition of any Matrix
The LU decomposition of a matrix produces a matrix as a product of its lower triangular matrix and upper triangular matrix. The LU in LU Decomposition of a matrix stands for Lower Upper.
An example of LU Decomposition of a matrix is given below −
Given matrix is: 1 1 0 2 1 3 3 1 1 The L matrix is: 1 0 0 2 -1 0 3 -2 -5 The U matrix is: 1 1 0 0 1 -3 0 0 1
A program that performs LU Decomposition of a matrix is given below −
Example
#include<iostream>
using namespace std;
void LUdecomposition(float a[10][10], float l[10][10], float u[10][10], int n) {
int i = 0, j = 0, k = 0;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (j < i)
l[j][i] = 0;
else {
l[j][i] = a[j][i];
for (k = 0; k < i; k++) {
l[j][i] = l[j][i] - l[j][k] * u[k][i];
}
}
}
for (j = 0; j < n; j++) {
if (j < i)
u[i][j] = 0;
else if (j == i)
u[i][j] = 1;
else {
u[i][j] = a[i][j] / l[i][i];
for (k = 0; k < i; k++) {
u[i][j] = u[i][j] - ((l[i][k] * u[k][j]) / l[i][i]);
}
}
}
}
}
int main() {
float a[10][10], l[10][10], u[10][10];
int n = 0, i = 0, j = 0;
cout << "Enter size of square matrix : "<<endl;
cin >> n;
cout<<"Enter matrix values: "<endl;
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
cin >> a[i][j];
LUdecomposition(a, l, u, n);
cout << "L Decomposition is as follows..."<<endl;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
cout<<l[i][j]<<" ";
}
cout << endl;
}
cout << "U Decomposition is as follows..."<<endl;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
cout<<u[i][j]<<" ";
}
cout << endl;
}
return 0;
}Output
The output of the above program is as follows
Enter size of square matrix : 3 Enter matrix values: 1 1 0 2 1 3 3 1 1 L Decomposition is as follows... 1 0 0 2 -1 0 3 -2 -5 U Decomposition is as follows... 1 1 0 0 1 -3 0 0 1
In the above program, the function LU decomposition finds the L and U decompositions of the given matrices. This is done by using nested for loops that calculate the L and U decompositions and store them in l[][] and u[][] matrix from the matrix a[][].
The code snippet that demonstrates this is given as follows −
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (j < i)
l[j][i] = 0;
else {
l[j][i] = a[j][i];
for (k = 0; k < i; k++) {
l[j][i] = l[j][i] - l[j][k] * u[k][i];
}
}
}
for (j = 0; j < n; j++) {
if (j < i)
u[i][j] = 0;
else if (j == i)
u[i][j] = 1;
else {
u[i][j] = a[i][j] / l[i][i];
for (k = 0; k < i; k++) {
u[i][j] = u[i][j] - ((l[i][k] * u[k][j]) / l[i][i]);
}
}
}
}In the main() function, the size of the matrix and its elements are obtained from the user. This is given as follows −
cout << "Enter size of square matrix : "<<endl; cin >> n; cout<<"Enter matrix values: "<endl; for (i = 0; i < n; i++) for (j = 0; j < n; j++) cin >> a[i][j];
Then the LU decomposition function is called and the L and U decomposition are displayed.This is given below −
LUdecomposition(a, l, u, n);
cout << "L Decomposition is as follows..."<<endl;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
cout<<l[i][j]<<" ";
}
cout << endl;
}
cout << "U Decomposition is as follows..."<<endl;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
cout<u[i][j]<<" ";
}
cout << endl;
}
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