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大家好,小跳来为大家解答以上的问题 。求逆矩阵c语言代码，矩阵求逆c语言这个很多人还不知道,现在让我们一起来看看吧！
1、下面是实现Gauss-Jordan法实矩阵求逆 。
2、#include#include
#includeint 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("");}}} 。
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