Well, how about that! It looks as though LongJonSilver 
started to do it, found it was too messy and quit. He
suggested Cramer's rule, but that's the determinant
method, not the matrix method.  OK, I'm game.  
I'll do it by matrices

Solution by Edwin McCravy:

Erase all the letters, plus signs,
and equal signs and put a line 
where the equal signs were: 

[12 -4 -7 |8]
[-8 -6  9 |7]
[34  6 -2 |5]

The plan is seven-fold, in this order:

1. Use the 1st row to get a 0 where the -8 is in the 2nd row
2. Use the 1st row to get a 0 where the 34 is in the 3rd row
3. Use the new 2nd row to get a 0 where the 6 is in the 3rd row

Your matrix will then look like this:

    [ #  #  # | #]
    [ 0  #  # | #]
    [ 0  0  # | #]

There will be numbers where the #'s are. This form is called
the reduced echelon form.

4. Rewrite the matrix as three equations.
5. Solve the 3rd equation for z
6. Substitute that value for z into the 2nd equation and solve for y
7. Substitute both the value of z and the value for y in the 1st
   equation and solve for x. 

Here we go:

[12  -4  -7 | 8]
[-8  -6   9 | 7]
[34   6  -2 | 5]

To get a zero where the -8 is, to the side, 
multiply the 1st row by 2 and the 
2nd row by 3, and add them vertically 

        1st row x2     24  -8 -14 | 16
        2nd row x3    -24 -18  27 | 21
                      ----------------
        new 2nd row     0 -26  13  | 37

Replace only the 2nd row by that bottom line
in the matrix

[12  -4  -7 |  8]
[ 0 -26  13 | 37]
[34   6  -2 |  5]

To get a zero where the 34 is, to the side, 
multiply the 1st row by -17 and the 
3rd row by 6, and add them vertically 

        1st row x -17    -204  68 119 | -136
        2nd row x 6       204  36 -12 |   30 
                      ----------------------
        new 3rd row         0 104 107 | -106

Replace only the 3rd row by that bottom line
in the matrix

[12  -4   -7 |    8]
[ 0 -26   13 |   37]
[ 0 104  107 | -106]

To get a zero where the 104 is, to the side, 
multiply the 2nd row by 4 and the 
3rd row by 1, (that is, leave it as it is),
and add them vertically 

        2nt row x 4       0 -104  52 |  148
        2nd row x 1       0  104 107 | -106 
                      ----------------------
        new 3rd row       0    0 159 |   42

Replace only the 3rd row by that bottom line
in the matrix

[12  -4   -7 |   8]
[ 0 -26   13 |  37]
[ 0   0  159 |  42]

This is reduced echelon form.  If your teacher
wants you to get it all the way to row-reduced
echelon form, you'll have to get 0's where the
-4, the -7 and the 4 are, too.  But it can be 
solved from the reduced echelon form as
follows.  

Erase the brackets and the line, and put the 
letters, plus signs, and equal signs back in:

 12x -  4y -   7z =  8
  0x - 26y +  13z = 37
  0x +  0y + 159z = 42

Erase the 0 terms

 12x -  4y -   7z =  8
      -26y +  13z = 37
             159z = 42
  
Solve the 3rd equation for z

             159z = 42
                z = 42/159
                z = 14/53

Sunstitute 14/53 for z in the 2nd equation

      -26y +  13z = 37
 -26y + 13(14/53) = 37
    -26y + 182/53 = 37

Clear of fractions by multiplying through by 53

     -1378y + 182 = 1961
           -1378y = 1779
                y = -1779/1378

Now substitute y = -1779/1378 and z = 14/53 in
the 1st equation:

               12x -  4y -   7z =  8  
 12x - 4(-1779/1378) - 7(14/53) = 8
        12x + 7116/1378 - 98/53 = 8

Reduce the fraction 7116/1378 to 3558/689

         12x + 3558/689 - 98/53 = 8

Clear of fractions by multiplying through by 689

            8268x + 3558 - 1274 = 5512
                   8268x + 2284 = 5512
                          8268x = 3228
                              x = 3228/8268
                              x = 269/689

So the solution is

(x, y, z) = (269/689, -1779/1378, 14/53)

and I'll bet anything you copied it wrong, for
no respectable teacher would expect you to do a 
problem that would come out with such a god-awful 
answer!  No wonder LongJonSilver gave it up.
But what I got is correct for the problem you posted.

Edwin