for(q=0;q

printf(\if((q+1)%i==0) printf(\} }

;break; case 5:

float a[M][2*M]; float b[N][2*N]; float t,x; int k,T;

printf(\输入方阵的维数：\\n\请输入方阵，即行和列相等的矩阵。 scanf(\

printf(\请输入矩阵:\\n\for(i=0;i

for (j=0;j

printf(\ printf(\ }

for(i=0;i

a[i][j]=0.0; }

for(i=0;i

for(k=0;k

t=a[k][i]/a[i][i]; for(j=0;j<(2*T);j++) {

x=a[i][j]*t;

a[k][j]=a[k][j]-x; } } } }

for(i=0;i

t=a[i][i];

for(j=0;j<(2*T);j++) a[i][j]=a[i][j]/t; }

for(i=0;i

printf(\对不起，您输入的矩阵没有逆矩阵，请重新输入。\\n\else {

for(i=0;i

printf(\逆矩阵为：\\n\ for (i=0;i

for (j=0;j

printf(\ printf(\ }

};break; case 6:;break; default:

printf(\您选择错误，请重试.********\\n\break; }

printf(\再次感谢您使用本系统,合作愉快！############\printf(\

printf(\}