大家好，我是小环，我来为大家解答以上问题。矩阵求逆的方法，矩阵求逆很多人还不知道，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 #include 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(" "); } } }

本文到此讲解完毕了，希望对大家有帮助。