代码来源见洛谷题解一楼，据作者说此解法非常快，但他也不知道原理。
不太喜欢大佬的代码风格自己敲了一遍。
#include <bits/stdc++.h>
#define int64 long long
using namespace std;
const int64 mod = 1e9 + 7;
int64 a[410][410];
int n, is[410], js[410];
void exgcd(int a, int b, int& x, int& y) {
if (!b)return x = 1, y = 0, void();
exgcd(b, a % b, y, x); y -= x * (a / b);
}
int inv(int p) {
int x, y; exgcd(p, mod, x, y);
return (x + mod) % mod;
}
void inv() {
for (int k = 1; k <= n; k++) {
for (int i = k; i <= n; i++) {
for (int j = k; j <= n; j++) {
if (a[i][j]) {
is[k] = i, js[k] = j; break;
}
}
}
for (int i = 1; i <= n; i++)
swap(a[k][i], a[is[k]][i]);
for (int i = 1; i <= n; i++)
swap(a[i][k], a[i][js[k]]);
if (!a[k][k]) {
puts("No Solution");
exit(0);
}
a[k][k] = inv(a[k][k]);
for (int j = 1; j <= n; j++) {
if (j != k) {
(a[k][j] *= a[k][k]) %= mod;
}
}
for (int i = 1; i <= n; i++) {
if (i != k) {
for (int j = 1; j <= n; j++) {
if (j != k) {
(a[i][j] += mod - a[i][k] * a[k][j] % mod) %= mod;
}
}
}
}
for (int i = 1; i <= n; i++) {
if (i != k) {
a[i][k] = (mod - a[i][k] * a[k][k] % mod) % mod;
}
}
}
for (int k = n; k; k--) {
for (int i = 1; i <= n; i++) swap(a[js[k]][i], a[k][i]);
for (int i = 1; i <= n; i++) swap(a[i][is[k]], a[i][k]);
}
}
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
scanf("%lld", a[i] + j);
inv();
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
printf("%lld%c", a[i][j], j == n ? '\n' : ' ');
return 0;
}