#include <malloc.h>
#include <stdio.h>
/// <summary>
/// 求行列式绝对值
/// </summary>
/// <param name="src">输入的矩阵</param>
/// <param name="n">矩阵的阶数</param>
/// <returns>矩阵对应行列式的值</returns>
double mat_det(double *src[], const unsigned n){
if (2 >n) return src[0][0]
int subsize = n - 1
double **subvet = (double **) malloc(sizeof(double *)*subsize)
for (int i = 0i <subsize++i)
subvet[i] = src[i+1]
double value = mat_det(subvet, subsize) * src[0][subsize] * ((subsize &1) ? -1 : 1)
for (int i = 0i <subsize++i){
subvet[i] = src[i]
value += mat_det(subvet, subsize)*src[i + 1][subsize] * ((n + i &1) ? -1 : 1)
}
free(subvet)
return value
}
/// <summary>
/// 求矩阵指定位置元素代数余子式的值
/// </summary>
/// <param name="src">输入的矩阵</param>
/// <param name="n">矩阵的阶数</param>
/// <param name="x">矩阵指定元素坐标的x值</param>
/// <param name="y">矩阵指定元素坐标的y值</param>
/// <returns>矩阵指定元素对应代数余子式的值</returns>
double mat_minor(double *src[], const unsigned n, const unsigned x, const unsigned y){
double **minmat = (double **) malloc(sizeof(double *)*(n - 1))
for (unsigned i = 0i<n - 1++i){
minmat[i] = (double *) malloc(sizeof(double)*(n - 1))
for (unsigned j = 0j<n - 1++j)
minmat[i][j] = src[i + (i >= x)][j + (j >= y)]
}
double value = mat_det(minmat, n - 1) * (x + y &1 ? -1 : 1)
for (unsigned i = 0i<n - 1++i)
free(minmat[i])
free(minmat)
return value
}
/// <summary>
/// 求矩阵的逆矩阵
/// </summary>
/// <param name="src">输入的矩阵</param>
/// <param name="des">输入的矩阵的逆矩阵</param>
/// <param name="n">矩阵的阶数</param>
void mat_inv(double *src [],double *des[],const unsigned n){
double det = mat_det(src, n)//求矩阵的行列式的值
for (unsigned i = 0i <n++i){
for (unsigned j = 0j <n++j)
des[i][j] = mat_minor(src, n, j, i) / det
}
}
int main(){
double input[][3] = { { 8, 4, 9 }, { 2, 3, 5 }, { 7, 6, 1 } }
double output[][3] = { { 0 }, { 0 }, { 0 } }
double *src[3],*des[3]
//矩阵须转换为指针数组形式
for (int i = 0i <3++i){
src[i] = input[i]
des[i] = output[i]
}
mat_inv(src,des,3)
for (int i = 0i <3++i){
for (int j = 0j <3++j)
printf("%lf\t", des[i][j])
printf("\n")
}
return 0
}
#include <stdlib.h>#include <math.h>
#include <stdio.h>
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=0k<=n-1k++)
{ d=0.0
for (i=ki<=n-1i++)
for (j=kj<=n-1j++)
{ l=i*n+jp=fabs(a[l])
if (p>d) { d=pis[k]=ijs[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=0j<=n-1j++)
{ u=k*n+jv=is[k]*n+j
p=a[u]a[u]=a[v]a[v]=p
}
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]
for (j=0j<=n-1j++)
if (j!=k)
{ u=k*n+ja[u]=a[u]*a[l]}
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]
}
for (i=0i<=n-1i++)
if (i!=k)
{ u=i*n+ka[u]=-a[u]*a[l]}
}
for (k=n-1k>=0k--)
{ 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
}
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)
}
void brmul(double a[], double b[],int m,int n,int k,double c[])
{ int i,j,l,u
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 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=0i<=3i++)
for (j=0j<=3j++)
b[i][j]=a[i][j]
i=brinv(a,4)
if (i!=0)
{ printf("MAT A IS:\n")
for (i=0i<=3i++)
{ for (j=0j<=3j++)
printf("%13.7e ",b[i][j])
printf("\n")
}
printf("\n")
printf("MAT A- IS:\n")
for (i=0i<=3i++)
{ for (j=0j<=3j++)
printf("%13.7e ",a[i][j])
printf("\n")
}
printf("\n")
printf("MAT AA- IS:\n")
brmul(b,a,4,4,4,c)
for (i=0i<=3i++)
{ for (j=0j<=3j++)
printf("%13.7e ",c[i][j])
printf("\n")
}
}
}
这个行么？不知道是不是你要的，希望对你有用
把最后的printf(" %d",a[i][j])改成printf(" %d",b[i][j]) -----------------------------------------------------------#include<stdio.h>main(){int i,j,a[3][3]={{1,2,3},{4,5,6},{7,8,9}},b[3][3]for(i=0i<=2i++){for(j=0j<=2j++)printf(" %d",a[i][j])printf("\n")}printf("\n")for(i=0i<3i++) {for(j=0j<3j++) </p><p>b[j][i]=a[i][2-j]<br>}for(i=0i<=2i++){for(j=0j<=2j++)</p><p>printf(" %d",b[i][j])</p><p>printf("\n")</p><p>}getchar()}
