用c语言怎么编写输入一个矩阵求其逆矩阵的程序？

Question

如题所述

Answer 1

下面是实现Gauss-Jordan法实矩阵求逆。
#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=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
");
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:
");
for
(i=0;
i<=3;
i++)
{
for
(j=0;
j<=3;
j++)
printf("%13.7e
",b[i][j]);
printf("
");
}
printf("
");
printf("MAT
A-
IS:
");
for
(i=0;
i<=3;
i++)
{
for
(j=0;
j<=3;
j++)
printf("%13.7e
",a[i][j]);
printf("
");
}
printf("
");
printf("MAT
AA-
IS:
");
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("
");
}
}
}参考资料：C常用算法程序集－徐士良

Answer 2

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。本回答被提问者采纳