In input1, we can assume the matrix 
3 1 5
7 2 3
5 4 3
In the first step:
    The smallest integer in row 1 is 1
    The smallest integer in row 2 is 2
    The smallest integer in row 3 is 3

After subtracting, the matrix becomes
2 0 4
5 0 1
2 1 0

And the second step is performed on this matrix. In the second step:
    The smallest integer in column 1 is 2
    The smallest integer in column 2 is 0
    The smallest integer in column 3 is 0

After subtracting, the resulting matrix is
0 0 4
3 0 1
0 1 0

And this is the matrix given on the input, so the answer is "YES".

In Input2: 
We cannot assume any such matrix from 
which we can obtain the given matrix.

Note: Before going to solution please try it by yourself.

Short description of solution approach

In this problem, whatever is talked about the second matrix from which you have to obtain the given matrix is just to puzzle you.

If you run behind the second matrix so there is a maximum probability that you are not going to get the correct answer for all input.

What you have to check is:

1. Check if there is at least one 0 present in each row of the matrix
2. Check if there is at least one 0 present in each column of the matrix

If these two conditions satisfy the matrix then your answer is YES otherwise NO.

Because the minimum value we can assume is 0. Now, if we subtract 0 from each row and each column of the matrix then we will be always getting the given matrix.

C++ implementation:

#include <iostream>
#define ll long long int
#define FASTIO ios_base::sync_with_stdio(false); cin.tie(NULL)
using namespace std;

int main()
{
	FASTIO;  //For fast I/O
	//I have defined long long type as ll above
	ll n;
	cin>>n;
	
	//decalring matrix of size N*N
	ll a[n][n],i,j; 
	for(i=0;i<n;i++)
	{
		for(j=0;j<n;j++)
			//taking input from user of the matrix
			cin>>a[i][j];
	}
	//these are flag variable
	ll f1=0,f2=0;
	//to count the number of rows and column which have 0
	ll cn1=0,cn2=0;
	for(i=0;i<n;i++)
	{
		f1=0;
		for(j=0;j<n;j++)
		{
			//Traversing the matrix row by 
			//row and checking of there is any 0
			if(a[i][j]==0)
			{
				//if 0 found stop at that row and give 0 tp f1
				f1=1;
				break;
			}
		}
		if(f1==1) //count that row              
			cn1++;
	}
	for(i=0;i<n;i++)
	{
		f2=0;
		for(j=0;j<n;j++)
		{
			//Traversing the matrix column by column 
			//and checking of there is any 0
			if(a[j][i]==0)
			{
				f2=1;
				break;
			}
		}
		if(f2==1) //count that column
			cn2++;
	}
	
	//check if those count are equal 
	//to the total number of row and column
	if(cn1==n && cn2==n)    
		cout<<"YES\n";
	else
		cout<<"NO\n";

	return 0;
}
Output