C ++程序实现高斯约旦消除
这是一个实现高斯约旦消除的C++程序。用于分析联立方程的线性系统。它主要集中在通过行运算将方程组简化为对角矩阵形式，以便直接获得解。
算法
Begin
n = size of the input matrix
To find the elements of the diagonal matrix:
Make nested for loops j = 0 to n and i = 0 to n
The element in the first row and the first column is made 1
and then the remaining elements in the first column are made 0.
Similarly, the elements in the second row and the second
column is made 1, and then the other elements in the second
column are reduced to 0 and so on.
Print all calculated solution values.
End示例
#include<iostream>
using namespace std;
int main() {
int i,j,k,n; // declare variables and matrixes as
input
float a[10][10],b,x[10];
printf("\nEnter the size of matrix: ");
scanf("%d",&n);
printf("\nEnter the elements of augmented matrix (rowwise):\ n");
for(i=1; i<=n; i++) {
for(j=1; j<=(n+1); j++) {
cout << "A[" << i << ", " << j << " ]=";
cin >> a[i][j];
}
}
//找到对角矩阵的元素
for(j=1; j<=n; j++) {
for(i=1; i<=n; i++) {
if(i!=j) {
b=a[i][j]/a[j][j];
for(k=1; k<=n+1; k++) {
a[i][k]=a[i][k]-b*a[j][k];
}
}
}
}
cout<<"\nThe solution is:\n";
for(i=1; i<=n; i++) {
x[i]=a[i][n+1]/a[i][i];
cout<<"x"<<i << "="<<x[i]<<" ";
}
return(0);
}输出结果
Enter the size of matrix: 3 Enter the elements of augmented matrix row-wise: A[1, 1 ]=1 A[1, 2 ]=2 A[1, 3 ]=-4 A[1, 4 ]=2 A[2, 1 ]=7 A[2, 2 ]=6 A[2, 3 ]=-2 A[2, 4 ]=-5 A[3, 1 ]=0 A[3, 2 ]=-3 A[3, 3 ]=-5 A[3, 4 ]=-8 The solution is: x1=-2.89831 x2=2.5678 x3=0.059322
热门推荐
10 圣诞祝福语简短小学
11 祖国七十华诞简短祝福语
12 老师送的祝福语简短
13 生日祝福语大全女生简短
14 祝女性生日祝福语简短
15 牛年女神节祝福语简短
16 情人表白祝福语简短大气
17 老公开业祝福语简短
18 官宣新年祝福语简短