# include<iostream>
# include<cstring>
using namespace std;

const int N=103,mod=1e9+7;

int n;
long long k;

struct jz{
	long long a[N][N];
	jz(){
		memset(a,0,sizeof(a));
	}
	void init(){
		for(int i=1;i<=n;++i)
			a[i][i]=1;
	}
	void in(){
		for(int i=1;i<=n;++i)
			for(int j=1;j<=n;++j)
				cin>>a[i][j];
	}
	void out(){
		for(int i=1;i<=n;++i){
			for(int j=1;j<=n;++j)
				cout<<a[i][j]<<" ";
			cout<<endl;
		}
	}
}A;

jz mul(jz x,jz y){
	jz z;
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
			for(int l=1;l<=n;++l)
				z.a[i][j]=(z.a[i][j]+(x.a[i][l]%mod)*(y.a[l][j]%mod)%mod)%mod;
	return z;
}

jz qpow(jz x,long long k){
	jz ans;
	ans.init();
	while(k){
		if(k&1) ans=mul(ans,x);
		x=mul(x,x);
		k>>=1;
	}
	return ans;
}

int main(){
	cin>>n>>k;
	A.in();
	qpow(A,k).out();
	return 0;
}

矩阵快速幂求斐波那契数列第 $n$ 项：

# include<iostream>
# include<cstring>
using namespace std;

const int mod=1e9+7;
long long n;

struct jz{
	long long a[3][3];
	jz(){
		memset(a,0,sizeof(a));
	}
	void init(){
		for(int i=1;i<=2;++i)
			a[i][i]=1;
	}
}A,B;

jz mul(jz x,jz y){
	jz z;
	for(int i=1;i<=2;++i)
		for(int j=1;j<=2;++j)
			for(int l=1;l<=2;++l)
				z.a[i][j]=(z.a[i][j]+(x.a[i][l]%mod)*(y.a[l][j]%mod)%mod)%mod;
	return z;
}

jz qpow(jz x,long long k){
	jz ans;
	ans.init();
	while(k){
		if(k&1) ans=mul(ans,x);
		x=mul(x,x);
		k>>=1;
	}
	return ans;
}

int main(){
	cin>>n;
	if(n==1){
		cout<<1;
		return 0;
	}
	A.a[1][1]=A.a[1][2]=A.a[2][1]=1;
	B=qpow(A,n-2);
	cout<<(B.a[1][1]+B.a[1][2])%mod;
	return 0;
}