之前竟然把这个落了。

高斯-约旦消元

$$A^{-1}\times [AI]=[IA^{-1}] $$

把 $[AI]$ 左半边消成单位矩阵即可。

const int N = 410, mod = 1e9 + 7;

inline ll Read() {
	ll x = 0, f = 1;
	char c = getchar();
	while (c != '-' && (c < '0' || c > '9')) c = getchar();
	if (c == '-') f = -f, c = getchar();
	while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
	return x * f;
}

namespace Main {
	int n;
	ll a[N][N << 1];
	ll qpow (ll a, ll b) {
		ll ans = 1ll;
		for (; b; b >>= 1, a = a * a % mod)
			if (b & 1) ans = ans * a % mod;
		return ans;
	}
	int main () {
		n = Read();
		for (int i = 1; i <= n; a[i][i + n] = 1, i++) 
			for (int j = 1; j <= n; j++) a[i][j] = Read();

		for (int i = 1; i <= n; i++) {
			int mxi = i;
			for (int j = i + 1; j <= n; j++) 
				if (a[j][i] > a[mxi][i]) mxi = j;
			if (mxi != i) swap(a[i], a[mxi]);
			if (!a[i][i]) { puts("No Solution"); return 0; }

			int inv = qpow(a[i][i], mod - 2);
			for (int k = 1; k <= n; k++) {
				if (k == i) continue;
				int p = a[k][i] * inv % mod;
				for (int j = i; j <= (n << 1); j++) 
					a[k][j] = ((a[k][j] - p * a[i][j]) % mod + mod) % mod;
			}

			for (int j = 1; j <= (n << 1); j++) a[i][j] = a[i][j] * inv % mod;
		}

		for (int i = 1; i <= n; i++, puts(""))
			for (int j = n + 1; j <= (n << 1); j++) printf ("%lld ", a[i][j]);
		return 0;
	}
}

int main () {
//	freopen(".in", "r", stdin);
//	freopen(".out", "w", stdout);
	Main::main();
	return 0;
}