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tính tổng - tích - hiệu của 2 ma trận:

Code:

#include <conio.h>
#include <iostream.h>
#include <math.h>
class MT
{
private:
int m;
int n;
float a[10][10];
public:
MT()
{
m=1;
n=1;
for (int i=1;i<=m;i++)
  for (int j=1;j<=n;j++)
a[i][j]=0;
}

friend ostream& operator<<(ostream& os,const MT &x)
{
 for (int i=1;i<=x.m;i++)
  {
  for (int j=1;j<=x.n;j++)
os<<x.a[i][j]<<" ";
os<<"\n";
  }
 return os;
 }
friend istream& operator>>(istream& is,MT &x)
{
cout<<"Nhap so hang :";is>>x.m;
cout<<"Nhap so cot :";is>>x.n;
 for (int i=1;i<=x.m;i++)
  for (int j=1;j<=x.n;j++)
  {
cout<<"Nhap phan tu hang "<<i<<", cot "<<j<<" : ";
is>>x.a[i][j];
}
 return is;
}
MT operator+(MT x2)
{
 if (m!=x2.m||n!=x2.n)
  {
  cout<<"Khong thuc hien duoc phep cong ma tran \n";
  return x2;
  }
 else
 {
 MT x3;
 x3.m = x2.m;
 x3.n = x2.n;
 for (int i=1;i<=m;i++)

  for (int j=1;j<=n;j++)
  {
x3.a[i][j]=a[i][j]+x2.a[i][j];
}

 return x3;
 }
}
MT operator-(MT x2)
{
 if (m!=x2.m||n!=x2.n)
  {
  cout<<"Khong thuc hien duoc phep tru ma tran \n";
  return x2;
  }
 else
 {
 MT x3;
 x3.m = x2.m;
 x3.n = x2.n;
 for (int i=1;i<=m;i++)

  for (int j=1;j<=n;j++)
  {
x3.a[i][j]=a[i][j]-x2.a[i][j];
}

 return x3;
 }
}
MT operator*(MT x2)
{
  if (this->n!=x2.m)
  {
cout<<"Khong nhan duoc 2 ma tran \n";
  return x2;
  }
  else
  {
  MT x3;
  x3.m=m;
  x3.n=x2.n;
  for (int i=1;i<=m;i++)
  for (int j=1;j<=n;j++)
{
 x3.a[i][j]=0;
 for (int k=1;k<=x2.m;k++)
x3.a[i][j]+=a[i][k]*x2.a[k][j];
}
 return x3;
 }
 }
};
void main()
{
clrscr();
MT d,b,c,e,f;
cin>>c>>b;
d=c+b;
e=c-b;
f=c*b;
cout<<"Tong 2 ma tran :\n"<<d;
cout<<"Hieu 2 ma tran :\n"<<e;
cout<<"Tich 2 ma tran :\n"<<f;
getch();
}

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