Solve the system using an augmented matrix. Show you work 

5x + 4y - z = 1
2x - 2y + z = 1
-x -  y + z = 2

Put in the coefficients of all
the unknowns that don't show them:

 5x + 4y - z = 1
 2x - 2y + z = 1
 -x -  y + z = 2

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

[ 5  4 -1 | 1 ]
[ 2 -2  1 | 1 ]
[-1 -1  1 | 2 ]

It's easiest but not really necessary
to rearrange the rows so that the 
simplest row is at the top.  So
swap the 1st and 3rd rows

[-1 -1  1 | 2 ]
[ 2 -2  1 | 1 ]
[ 5  4 -1 | 1 ]


The plan is seven-fold, in this order:

1. Use the 1st row to get a 0 where the 2 is in the 2nd row
2. Use the 1st row to get a 0 where the 5 is in the 3rd row
3. Use the new 2nd row to get a 0 where the 4 is in the 3rd row

Your matrix will then look like this:

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

There will be numbers where the #'s are.

4. Divide each row through by the first non-zero number
in each row, and the matrix will then look like this:

    [ 1  #  # | #]
    [ 0  1  # | #]
    [ 0  0  1 | #]

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:

[-1 -1  1 | 2 ]
[ 2 -2  1 | 1 ]
[ 5  4 -1 | 1 ]


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

        1st row x 2     -2  -2   2 | 4
        2nd row x 1      2  -2   1 | 1
                       ----------------
        new 2nd row      0  -4   3 | 5

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

[-1 -1  1 | 2 ]
[ 0 -4  3 | 5 ]
[ 5  4 -1 | 1 ]


To get a zero where the 5 is, to the side, 
multiply the 1st row by 5 and the 
3rd row by 1, and add them vertically 

        1st row x 5    -5 -5  5 | 10
        3rd row x 1     5  4 -1 |  1
                      ---------------
        new 3rd row     0 -1  4 | 11

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

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

[-1 -1  1 |  2 ]
[ 0 -4  3 |  5 ]
[ 0 -1  4 | 11 ]

To get a zero where the -1 is, to the side, 
multiply the 2nd row by 1 and the 
3rd row by -4, and add them vertically 

        2nd row x 1     [ 0 -4   3 |   5 ]
        3rd row x -4    [ 0  4 -16 | -44 ] 
                      ----------------------
        new 3rd row       0  0 -13 | -39 

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

[-1 -1    1 |   2 ]
[ 0 -4    3 |   5 ]
[ 0  0  -13 | -39 ]

Now we want to get the first non-zero number
in each row to be 1. so we divide the 1st row
through by -1, the 2nd row through by -4 and
the third row through by -13:

[ 1  1   -1  |  -22]
[ 0  1 -3/4  | -5/4]
[ 0  0    1  |    3]

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
1 and the -1 are in the 1st row and the -3/4
in the 2nd row.  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:

  1x +  1y -   1z =   -2
  0x +  1y - 3/4z = -5/4
  0x +  0y +   1z =    3

Erase the 0 terms and the 1 coefficients:

   x +   y -    z =   -2
         y - 3/4z = -5/4
                z =    3

The third equation tells us that z = 3.

Subtitute 3 for z in the 2nd equation

        y - {3/4}(3) = -5/4
             y - 9/4 = -5/4
                   y = -5/4 + 9/4
                   y = 4/4
                   y = 1

Now substitute y = 1 and z = 3 in
the 1st equation:

          x + y - z = -2
          x + 1 - 3 = -2
              x - 2 = -2
                  x = 0

So the solution is

(x, y, z) = (0, 1, 3)

Edwin