| 1  2  3  4|
| 5  6  7  8|
| 9 10 11 12|
|13 14 15 16|

Get a 0 where the 2 is.  To do that, 
multiplying the first column through 
by -2:

 -2 
-10 
-18
-26

then add that to the second column:

 -2 +  2 =   0 
-10 +  6 =  -4
-18 + 10 =  -8 
-26 + 14 = -12

Replace the 2nd column by that:

| 1   0  3  4|
| 5  -4  7  8|
| 9  -8 11 12|
|13 -12 15 16|

===============================

Get a 0 where the 3 is.  To do that, 
multiplying the first column through 
by -3:

 -3 
-15 
-27
-39

then add that to the third column:

 -3 +  3 =   0 
-15 +  7 =  -8
-27 + 11 = -16 
-39 + 15 = -24

Replace the 3rd column by that:

| 1   0   0  4|
| 5  -4  -8  8|
| 9  -8 -16 12|
|13 -12 -24 16|

===================================

Get a 0 where the 4 is.  To do that, 
multiplying the first column through 
by -4:

 -4 
-20 
-36
-52

then add that to the fourth column:

 -4 +  4 =   0 
-10 +  8 =  -2
-18 + 12 =  -6 
-26 + 16 = -10

Replace the 4th column by that:

| 1   0   0   0|
| 5  -4  -8  -2|
| 9  -8 -16  -6|
|13 -12 -24 -10|

Now expand about the top row.
Since there are 3 0's in the
top row, we only need multiply
the 1 by its signed minor.

The 1 is in row 1 and column 1,
adding the row and column numbers
gives 2, which is even, so the
sign of the minor is +. The 
minor of 1 is the 3x3 determinant
formed by taking all the elements
which are NOT in the same row or
column as the 1, so the above becomes

    | -4  -8  -2|
1 × | -8 -16  -6|
    |-12 -24 -10|

Now you can evaluate that 3×3 determinant,
and you'll find it is 0.  So the value of the
original 4×4 matrix is 0.

Edwin