c语言的方阵求逆的一个完整的程序如下，在tc2.0下运行良好：
#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
#include <math.h>
#define MAX 255
void MatrixMul(a,b,m,n,k,c) /*实矩阵相乘*/
int m,n,k/*m:矩阵A的行数, n:矩阵B的行数, k:矩阵B的列数*/
double a[],b[],c[]/*a为A矩阵, b为B矩阵, c为结果，即c = AB */
{
int i,j,l,u
/*逐行逐列计算乘积*/
for (i=0i<=m-1i++)
for (j=0j<=k-1j++)
{
u=i*k+jc[u]=0.0
for (l=0l<=n-1l++)
c[u]=c[u]+a[i*n+l]*b[l*k+j]
}
return
}
int brinv(a,n) /*求矩阵的逆矩阵*/
int n/*矩阵的阶数*/
double a[]/*矩阵A*/
{
int *is,*js,i,j,k,l,u,v
double d,p
is=malloc(n*sizeof(int))
js=malloc(n*sizeof(int))
for (k=0k<=n-1k++)
{
d=0.0
for (i=ki<=n-1i++)
/*全选主元，即选取绝对值最大的元素*/
for (j=kj<=n-1j++)
{
l=i*n+jp=fabs(a[l])
if (p>d) { d=pis[k]=ijs[k]=j}
}
/*全部为0，此时为奇异矩阵*/
if (d+1.0==1.0)
{
free(is)free(js)printf(" >>This is a singular matrix, can't be inversed!\n")
return(0)
}
/*行交换*/
if (is[k]!=k)
for (j=0j<=n-1j++)
{
u=k*n+jv=is[k]*n+j
p=a[u]a[u]=a[v]a[v]=p
}
/*列交换*/
if (js[k]!=k)
for (i=0i<=n-1i++)
{
u=i*n+kv=i*n+js[k]
p=a[u]a[u]=a[v]a[v]=p
}
l=k*n+k
a[l]=1.0/a[l]/*求主元的倒数*/
/* a[kj]a[kk] ->a[kj] */
for (j=0j<=n-1j++)
if (j!=k)
{
u=k*n+ja[u]=a[u]*a[l]
}
/* a[ij] - a[ik]a[kj] ->a[ij] */
for (i=0i<=n-1i++)
if (i!=k)
for (j=0j<=n-1j++)
if (j!=k)
{
u=i*n+j
a[u]=a[u]-a[i*n+k]*a[k*n+j]
}
/* -a[ik]a[kk] ->a[ik] */
for (i=0i<=n-1i++)
if (i!=k)
{
u=i*n+ka[u]=-a[u]*a[l]
}
}
for (k=n-1k>=0k--)
{
/*恢复列*/
if (js[k]!=k)
for (j=0j<=n-1j++)
{
u=k*n+jv=js[k]*n+j
p=a[u]a[u]=a[v]a[v]=p
}
/*恢复行*/
if (is[k]!=k)
for (i=0i<=n-1i++)
{
u=i*n+kv=i*n+is[k]
p=a[u]a[u]=a[v]a[v]=p
}
}
free(is)free(js)
return(1)
}
print_matrix(a,n)/*打印的方阵a的元素*/
int n/*矩阵的阶数*/
double a[]/*矩阵a*/
{
int i,j
for (i=0i<ni++)
{
for (j=0j<nj++)
printf("%13.7f\t",a[i*n+j])
printf("\n")
}
}
main()
{
int i,j,n=0
double A[MAX],B[MAX],C[MAX]
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]
clrscr()
puts("**********************************************************")
puts("* This program is to inverse a square matrix A(nxn). *")
puts("**********************************************************")
while(n<=0)
{
printf(" >>Please input the order n of the matrix (n>0): ")
scanf("%d",&n)
}
printf(" >>Please input the elements of the matrix one by one:\n >>")
for(i=0i<n*ni++)
{
scanf("%lf",&A[i])
B[i]=A[i]
}
for(i=0i<4i++)
for(j=0j<4j++)
b[i][j]=a[i][j]
i=brinv(A,n)
if (i!=0)
{
printf(" Matrix A:\n")
print_matrix(B,n)
printf("\n")
printf(" A's Inverse Matrix A-:\n")
print_matrix(A,n)
printf("\n")
printf(" Product of A and A- :\n")
MatrixMul(B,A,n,n,n,C)
print_matrix(C,n)
}
printf("\n Press any key to quit...")
getch()
}
/*** @author 熙熙
* @2011-4-20
*/
public class JuZhenZhuanZhi {
/**
* @param args
*/
public static void main(String[] args) {
int a[][]={
{11,12,13}
,{21,22,23}
,{31,32,33}
}
System.out.println("转置前的矩阵：")
print(a)
zhuanZhi(a)
System.out.println("转置后的矩阵：")
print(a)
}
public static void zhuanZhi(int a[][]){
for(int i=0i<a.lengthi++){
for(int j=i+1j<a[i].lengthj++){
int tem = a[i][j]
a[i][j] = a[j][i]
a[j][i] = tem
}
}
}
public static void print(int a[][]){
for(int i=0i<a.lengthi++){
int j=0
for(j<a[i].length-1j++){
System.out.print(a[i][j]+",")
}
System.out.println(a[i][j])
}
}
}
#include "stdio.h"
int main()
{
int a[3][3]={1,2,3,4,5,6,7,8,9}
int i,j,sum1=1,sum2=1
printf("Matrix is:\n")
for(i=0i<3i++)
{
for(j=0j<3j++)
printf("%d ",a[i][j])
printf("\n")
}
for(i=0i<3i++)
for(j=0j<3j++)
{
if(i==j)
sum1*=a[i][j]
if(i+j==2)
sum2*=a[i][j]
}
printf("主对角线元素乘积:%d\n副对角线元素乘积:%d\n",sum1,sum2)
}
主对角线：1*5*9==45正确
副对角线：3*5*7==105
