[ 2   6   8 |  5]
 [-2   9 -12 | -1]
 [ 4   6  -4 |  3]

We must get 0's where the -2, 
the 4, and the 6 are now. We 
make three 0's in the lower 
left hand corner

The top row will never change. 
The other two will. To get a 0 
where the -2 is, We multiply 1 
times the first row and add it 
to 1 times the 2nd row. It's a 
good idea to put what you're
going to multiply a row by out
to the left of the row, even 
when it is just 1, like this:

1[ 2   6   8 |  5]
1[-2   9 -12 | -1]
 [ 4   6  -4 |  3]

The next matrix is then

 [ 2   6   8 |  5]
 [ 0  15  -4 |  4]
 [ 4   6  -4 |  3]

To get a 0 where the 4 is, 
We multiply -2 times the first 
row and add it to 1 times the 
3rd row. 

-2[ 2   6   8 |  5]
  [ 0  15  -4 |  4]
 1[ 4   6  -4 |  3]

  [ 2   6   8 |  5]
  [ 0  15  -4 |  4]
  [ 0  -6 -20 | -7]

To get a 0 where the -6 is, 
we notice that the least common
multiple of 15 and 6 is 30.  So,
We multiply 2 times the second 
row and add it to 5 times the 
3rd row.

  [ 2   6   8 |  5]
 2[ 0  15  -4 |  4]
 5[ 0  -6 -20 | -7]

We get this:

  [ 2   6    8|  5]
  [ 0  15   -4|  4]
  [ 0   0 -108|-27]

Now that we have three 0's in the
lower left corner, we go back to
a system of equations:



Start at the bottom, and go back up.
This is called back-substitution.
Start with the bottom equation, solve for
z:





Substitute  for  in the
second equation:








Finally sibstitute both  for 
and  for 








So the solution is

(x,y,z)=(,,)

Edwin