用c语言怎么编写输入一个矩阵求其逆矩阵的程序？

这是我编的一个简易矩阵计算器，C++语言，非常容易理解的，你可以参考求行列式和逆部分
#include
#include
#include
#include "windows.h"
#include
using namespace std;

void gotoxy(int x,int y) // 列x: 0~79 行y: 0~24
{ HANDLE hConsole=GetStdHandle(STD_OUTPUT_HANDLE);
COORD coordScreen={x,y};
SetConsoleCursorPosition(hConsole,coordScreen);
return;
}
void setcolor(unsigned short ForeColor,unsigned short BackColor)
// 0--黑 1--暗蓝 2--暗绿 3--暗青 4--暗红 5--暗紫 6--蟹黄 7--暗白
// 8--灰 9--亮蓝 10-亮绿 11-亮青 12-亮红 13-亮紫 14-黄 15-亮白
{ HANDLE hCon = GetStdHandle(STD_OUTPUT_HANDLE);
SetConsoleTextAttribute(hCon,(ForeColor % 16)|(BackColor % 16 * 16));
};

int main()
{
void plu();
void sub();
void amo();
void mul();
void ran();
void ord();
char sel='1';
while(sel != '0')
{ int i;
system("cls"); // 清屏
setcolor(15,0); // 下面显示黑底亮青字
gotoxy(8,1); cout<<"┌───────────────────────────┐";
for(i=2;i<20;i++)
{gotoxy(8,i);cout<<"│";gotoxy(64,i);cout<<"│";}
setcolor(15,6); // 下面显示红底白字
gotoxy(10,3); cout<<" ";
gotoxy(10,4); cout<<" 简 易 矩 阵 计 算 器 ";
gotoxy(10,5); cout<<" ";
setcolor(15,0); // 下面显示黑底亮青字
gotoxy(10,7); cout<<" 1 ---- 矩阵加法 2 ---- 矩阵减法 ";
gotoxy(10,9); cout<<" 3 ---- 矩阵数乘 4 ---- 矩阵乘法 ";
gotoxy(10,11); cout<<" 5 ---- 矩阵行列式 6 ---- 矩阵的逆 ";
gotoxy(10,13); cout<<" 0 ---- 退出 ";
gotoxy(10,15); cout<<" 请选择(0--6):";
gotoxy(8,20); cout<<"└───────────────────────────┘";
do
{ gotoxy(28,15); sel=getche( );}
while ( sel!='1' && sel!='2' && sel!='3' && sel!='4' && sel!='5' && sel!='6'&& sel!='0');

switch(sel)
{
case '1':plu(); break;
case '2':sub(); break;
case '3':amo(); break;
case '4':mul(); break;
case '5':ran(); break;
case '6':ord(); break;
case '0': break;
}
}
system("cls");
gotoxy(25,10);
cout<<"谢 谢 使 用 系 统 ！"< return 0;
}

void plu()//加法
{ char l;
system("cls"); // 清屏
setcolor(14,0); // 下面用黑底黄字
int a,b,i,j;
gotoxy(0,0);cout<<">>>>>> 矩阵加法 ";
gotoxy(0,2);cout<<"请输入矩阵的行数：";
cin>>a;
cout< cout<<"请输入矩阵的列数：";
cin>>b;
cout< double m[10][10],n[10][10];
cout<<"请输入第一个矩阵："< for(i=0;i for(j=0;j {gotoxy(6*j+20,2*i+6); cin>>m[i][j];}
cout< for(i=0;i for(j=0;j {gotoxy(6*j+20,2*a+2*i+7);cin>>n[i][j];}
cout<>>>>>>"< for(i=0;i for(j=0;j {gotoxy(6*j+20,4*a+2*i+8);cout< gotoxy(0,6*a+9);
cout<<">>>>>>>>按任意键退出:";
l=getche();

}

void sub()//减法
{ char l;
system("cls"); // 清屏
setcolor(14,0); // 下面用黑底黄字
int a,b,i,j;
gotoxy(0,0);cout<<">>>>>矩阵减法";
gotoxy(0,2);cout<<"请输入矩阵的行数：";
cin>>a;
cout< cout<<"请输入矩阵的列数：";
cin>>b;
cout< double m[10][10],n[10][10];
cout<<"请输入第一个矩阵："< for(i=0;i for(j=0;j {gotoxy(6*j+20,2*i+6); cin>>m[i][j];}
cout< for(i=0;i for(j=0;j {gotoxy(6*j+20,2*a+2*i+7);cin>>n[i][j];}
cout<>>>>>>"< for(i=0;i for(j=0;j {gotoxy(6*j+20,4*a+2*i+8);cout< gotoxy(0,6*a+9);
cout<<">>>>>>>>按任意键退出:";
l=getche();

}

void amo()//数乘
{ char h;
system("cls"); // 清屏
setcolor(14,0); // 下面用黑底黄字
int a,b,i,j;
gotoxy(0,0);cout<<">>>>>>矩阵数乘";
gotoxy(0,2);cout<<"请输入矩阵的行数：";
cin>>a;
cout< cout<<"请输入矩阵的列数：";
cin>>b;
cout< double m[10][10],c;
cout<<"请输入矩阵："< for(i=0;i for(j=0;j {gotoxy(6*j+20,2*i+6);cin>>m[i][j];}
cout< cin>>c;
cout< for(i=0;i for(j=0;j {gotoxy(8*j+20,2*a+2*i+9);cout< gotoxy(0,4*a+12);
cout<<">>>>>>>按任意键退出：";h=getche();
}

void mul()//乘法
{
char k;
system("cls"); // 清屏
setcolor(14,0); // 下面用黑底黄字
int a,b,c,i,j,q;
gotoxy(0,0);cout<<">>>>>>矩阵乘法";
gotoxy(0,2);cout<<"请输入第一个矩阵的行数：";
cin>>a;
cout< cin>>b;
cout< cout< cin>>c;
cout< double m[10][10],n[10][10],p[10][10]={0};
cout<<"请输入第一个矩阵："< for(i=0;i for(j=0;j {gotoxy(6*j+18,2*i+10); cin>>m[i][j];}
cout< for(i=0;i for(j=0;j {gotoxy(6*j+18,2*a+2*i+11);cin>>n[i][j];}
cout<>>>>>>"< for(i=0;i for(j=0;j for(q=0;q for(i=0;i for(j=0;j {gotoxy(10*j+18,2*a+2*b+2*i+12);cout< gotoxy(16,2*a+2*b+2*i+15);
cout<<">>>>>>>按任意键退出：";k=getche();

}
//===================================================行列式

float Fun(int n1,float a1[10][10]);
void ran()
{
system("cls"); // 清屏
setcolor(15,0); // 下面用黑底黄字
char k;
int n,i,j;
cout<<">>>>>矩阵行列式"< cin>>n;
cout< float a[10][10];
for(i=0;i for(j=0;j {gotoxy(6*j+12,2*i+4);cin>>a[i][j];}
cout< cout<<">>>>>>按任意键退出：";
k=getche();

}
float Fun(int n1,float a1[10][10])//求行列式的递归函数
{
int i_1,j_1,c;//c为数组b的行
float b[10][10];
int p=0,q=0;
float sum=0;
if(n1==1) return a1[0][0];
for(i_1=0;i_1 {
for(c=0;c {if(c else p=1;
for(j_1=0;j_1 {b[c][j_1]=a1[c+p][j_1+1];}
}
if(i_1%2==0)
q=1;
else q=(-1);
sum=sum+a1[i_1][0]*q*Fun(n1-1,b);
}return sum;
}

//================================================================

void ord()
{
char g;
system("cls"); // 清屏
setcolor(15,0); // 下面用黑底黄字
int i,j,n;
gotoxy(0,0);cout<<">>>>>矩阵的逆";
gotoxy(0,2);cout<<"请输入矩阵的阶数：";
cin>>n;
cout< cout<<"请输入矩阵：";
float l[10][10],m[10][10],p;
for(i=0;i for(j=0;j {gotoxy(4*j+12,2*i+4); cin>>l[i][j];}
if(Fun(n,l)==0) cout< else
{p=Fun(n,l);
cout< for(i=0;i for(j=0;j {{float f[10][10];
int r,w,e,d;//e为数组f的行数
for(int j_1=0,e=0;j_1 for(int i_1=0,d=0;i_1 {if(e if(d f[i_1][j_1]=l[i_1+w][j_1+r];};
if((i+j)%2==0) m[i][j]=Fun(n-1,f)/p;
else m[i][j]=-Fun(n-1,f)/p;
};
gotoxy(9*j+12,2*n+2*i+4);cout< cout<>>>>>按任意键退出：";g=getche();
}

循环输入矩阵元素，
你想想求行列式的算法，改一改就是求逆矩阵
通过(A E) ~(E A^-1)这个初等变换来求逆矩阵。

下面是实现Gauss-Jordan法实矩阵求逆。

#include
#include
#include
int brinv(double a[], int n)
{ int *is,*js,i,j,k,l,u,v;
double d,p;
is=malloc(n*sizeof(int));
js=malloc(n*sizeof(int));
for (k=0; k<=n-1; k++)
{ d=0.0;
for (i=k; i<=n-1; i++)
for (j=k; j<=n-1; j++)
{ l=i*n+j; p=fabs(a[l]);
if (p>d) { d=p; is[k]=i; js[k]=j;}
}
if (d+1.0==1.0)
{ free(is); free(js); printf("err**not inv\n");
return(0);
}
if (is[k]!=k)
for (j=0; j<=n-1; j++)
{ u=k*n+j; v=is[k]*n+j;
p=a[u]; a[u]=a[v]; a[v]=p;
}
if (js[k]!=k)
for (i=0; i<=n-1; i++)
{ u=i*n+k; v=i*n+js[k];
p=a[u]; a[u]=a[v]; a[v]=p;
}
l=k*n+k;
a[l]=1.0/a[l];
for (j=0; j<=n-1; j++)
if (j!=k)
{ u=k*n+j; a[u]=a[u]*a[l];}
for (i=0; i<=n-1; i++)
if (i!=k)
for (j=0; j<=n-1; j++)
if (j!=k)
{ u=i*n+j;
a[u]=a[u]-a[i*n+k]*a[k*n+j];
}
for (i=0; i<=n-1; i++)
if (i!=k)
{ u=i*n+k; a[u]=-a[u]*a[l];}
}
for (k=n-1; k>=0; k--)
{ if (js[k]!=k)
for (j=0; j<=n-1; j++)
{ u=k*n+j; v=js[k]*n+j;
p=a[u]; a[u]=a[v]; a[v]=p;
}
if (is[k]!=k)
for (i=0; i<=n-1; i++)
{ u=i*n+k; v=i*n+is[k];
p=a[u]; a[u]=a[v]; a[v]=p;
}
}
free(is); free(js);
return(1);
}

void brmul(double a[], double b[],int m,int n,int k,double c[])
{ int i,j,l,u;
for (i=0; i<=m-1; i++)
for (j=0; j<=k-1; j++)
{ u=i*k+j; c[u]=0.0;
for (l=0; l<=n-1; l++)
c[u]=c[u]+a[i*n+l]*b[l*k+j];
}
return;
}

int main()
{ int i,j;
static double a[4][4]={ {0.2368,0.2471,0.2568,1.2671},
{1.1161,0.1254,0.1397,0.1490},
{0.1582,1.1675,0.1768,0.1871},
{0.1968,0.2071,1.2168,0.2271}};
static double b[4][4],c[4][4];
for (i=0; i<=3; i++)
for (j=0; j<=3; j++)
b[i][j]=a[i][j];
i=brinv(a,4);
if (i!=0)
{ printf("MAT A IS:\n");
for (i=0; i<=3; i++)
{ for (j=0; j<=3; j++)
printf("%13.7e ",b[i][j]);
printf("\n");
}
printf("\n");
printf("MAT A- IS:\n");
for (i=0; i<=3; i++)
{ for (j=0; j<=3; j++)
printf("%13.7e ",a[i][j]);
printf("\n");
}
printf("\n");
printf("MAT AA- IS:\n");
brmul(b,a,4,4,4,c);
for (i=0; i<=3; i++)
{ for (j=0; j<=3; j++)
printf("%13.7e ",c[i][j]);
printf("\n");
}
}
}参考资料：C常用算法程序集－徐士良