c语言编写一个子函数求矩阵的逆矩阵

2022-04-20 教育 78阅读
#include
#include
#include
intbrinv(doublea[],intn)
{int*is,*js,i,j,k,l,u,v;
doubled,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**notinv\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);
}

voidbrmul(doublea[],doubleb[],intm,intn,intk,doublec[])
{inti,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;
}

intmain()
{inti,j;
staticdoublea[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}};
staticdoubleb[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("MATAIS:\n");
for(i=0;i<=3;i++)
{for(j=0;j<=3;j++)
printf("%13.7e",b[i][j]);
printf("\n");
}
printf("\n");
printf("MATA-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("MATAA-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");
}
}
}
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