矩阵求逆

前言

以前拖了好久，知道现在才知道怎么求矩阵求逆。果然是因为我是个大菜鸡。

正言

其实方法很简单，直接在原矩阵右边加上一个单位矩阵，然后高斯-约旦消元之后右边的那个矩阵就是逆矩阵了。

正确性怎么说呢？应该很显然吧。。。

\(\text {Code}\)

#include 
using namespace std;

#define Int register int
#define mod 1000000007
#define MAXN 405

template  inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template  inline void read (T &t,Args&... args){read (t);read (args...);}
template  inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}

int n,a[MAXN][MAXN << 1]; 

int quick_pow (int a,int b){
	int res = 1;for (;b;b >>= 1,a = 1ll * a * a % mod) if (b & 1) res = 1ll * res * a % mod;
	return res;
}

signed main(){
	read (n);
	for (Int i = 1;i <= n;++ i){
		for (Int j = 1;j <= n;++ j) read (a[i][j]);
		a[i][n + i] = 1;
	}
	for (Int i = 1;i <= n;++ i){
		int fir = 0;
		for (Int j = i;j <= n;++ j) if (a[j][i]){fir = j;break;}
		if (!fir) return puts ("No Solution"),0;
		if (fir ^ i) swap (a[fir],a[i]);
		int kk = quick_pow (a[i][i],mod - 2);
		for (Int j = 1;j <= n;++ j){
			if (j == i) continue;
			int Inv = 1ll * a[j][i] * kk % mod;
			for (Int k = i;k <= n << 1;++ k) a[j][k] = (a[j][k] + mod - 1ll * a[i][k] * Inv % mod) % mod;
		}
		for (Int j = 1;j <= n << 1;++ j) a[i][j] = 1ll * a[i][j] * kk % mod;
	}
	for (Int i = 1;i <= n;++ i){
		for (Int j = 1;j <= n;++ j) write (a[i][j + n]),putchar (' ');
		putchar ('\n');
	}
	return 0;
}