Let’s consider the cell (1,2), that is, cell in Row 1 and Column 2.
Since Row 1 has 4 and Column 2 has 1 already, we can’t take these two values for this cell.

Since the top left major square consisting of cells (1, 1), (1, 2), (2, 1) and (2, 2) doesn't contain any number other than '4' and '1', no other number is precluded for Cell (1,2). So, from the possible numbers 1,2,3 and 4, excluding '4' and '1' for reasons above, the only possible numbers that can get into Cell (1,2) are ‘2' and '3

Suppose Cell (2,1) had the number ‘3’ in it (which is not the case here), Cell (1, 2) can't take the value '3' too. And the Cell would have been forced to take only the number ‘2'.

Likewise, cell (1,3) can’t take the values ‘2’ and ‘4’, since Row 1 contains 4 and Column 3 contains 2 already. Also, there is no fixed number in the top right major square. So, the values permitted in the cell (1,3) are 1 and 3.

Similarly, consider the cell (2,3).Now, we see that Row 2 has 1 and Column 3 has 2 already. Hence, these two values ‘1’ and ‘2’ can be excluded from the possibilities and as a result, we get ‘3’ and ‘4’ as possible values for the cell (2,3).

Also, consider the cell (3,4). Row 3 has 2 in it and Column 4 has 3 already, they can be removed from the possibilities 1 to 4, for the cell (3,4). Since the Bottom Right Major Square consisting of the cells (3,3), (3,4), (4,3) and (4,4), doesn’t contain any other number than these two values ‘2’ and ‘3’, the cell (3,4) take ‘1’ and ‘4’ as its possible values.

Note that the cell (3,1) can take only the value ‘3’ as ‘2’ is there in Row 3 and ‘1’ and ‘4’ in Column 1 already. So, all the three possibilities 1,2 and 4 except 3 are ruled out. Thus, 3 becomes the unique value of the cell (3,1).
Similarly, you can see that 4 becomes the unique value of the cell (4,3).

In the same way, let's fill up the entire table.

We have solved the puzzle partly. Now, let’s continue solving the puzzle, by the process of Reduction.

That is reducing the Matrix, removing exclusions newly created, if any. We see that the cell (3,1) takes the value '3'.

Since cell (3, 1) takes the value '3', we can't have any more '3' in row 3, column 1, and the Bottom Left Major Square. So, delete '3'
from the possibilities in the other cells of Row 3, column 1 and
Bottom Left Major Square.

Now, we get the values for the cells (2,1) and (3,2) which are
'2' and '4' respectively.

Similarly, 4 can be removed from the possible values of other cells in Row 4 and Column 3 and Bottom Right Major Square as 4 is the fixed value of the cell (4,3).

As a result, We get the value ‘2’ for the cell (4,2), ‘3’ for the cell (2,3) and ‘1’ for the cell (3,4).

The reduced puzzle is given below:

Now, consider the columns 2 and 3. All the cells have been filled up except one in both case.

The numbers 1,4 and 2 are there in Column 2 and the left out number is 3 and thus 3 is assigned to the cell (1,2). In column 3, we see that 1 has to be taken for the cell (1,3). We can also get the values for these two cells by considering the corresponding Major Square and Row as most of the cells have been filled up already.

Only, two unresolved cells are there now, which are (1,4) and (2,4). As Row 2 contains the value ‘2’ already, 2 gets eliminated from the cell (2,4), thus leaving the value ‘4’ to this cell.

Also, you can see that the Column 4 has 1 already which gives way to remove the value ‘1’ from the cell (1,4) and enter ‘2’ as its fixed value.

Thus, the puzzle is solved completely and the final solution is :