The other tutor doesn't know about
solving dependent systems of equations
with Gaussian elimination.  Here's how
to solve that using matrices and 
Gaussian elimination:

2x - y + z = 8
3x + y - 6z = -28
x - y + 2z = 12

Line up the terms vertically:

2x -  y +  z =   8
3x +  y - 6z = -28
 x -  y + 2z =  12

Fill in all invisible 1's for
coefficients:

2x - 1y + 1z =   8
3x + 1y - 6z = -28
1x - 1y + 2z =  12

Erase all the letters and
equal signs:

 2  - 1  + 1    8
 3  + 1  - 6  -28
 1  - 1  + 2   12 

Erase the plus signs
and move the minus
signs close to the
numbers as negative
signs:

 2  -1   1    8
 3   1  -6  -28
 1  -1   2   12 

Draw a vertical line
where the equal signs
were and put brackets
around the whole thing.

[2  -1   1  |   8]
[3   1  -6  | -28]
[1  -1   2  |  12]

This is called the
augmented matrix

The idea is to get 0's
in the three lower left
corner positions, 
underneath the diagonal:

[2  -1   1  |   8]
[3   1  -6  | -28]
[1  -1   2  |  12]

It's easier when there is a 1
in the upper left corner, the
1st number on the diagonal, so
we will swap row 1 and row 3,
so we'll have a 1 on the first
diagonal element:

[1  -1   2  |  12]
[3   1  -6  | -28]
[2  -1   1  |   8]

To get a 0 where the 3 is,
mentally multiply each of the
numbers in the top row by -3 
and add them to 1 times the 
middle row. It makes it easy 
if you write the numbers to 
multiply the two rows by left 
of the matrix beside the rows:

-3[1  -1   2  |  12]
 1[3   1  -6  | -28]
  [2  -1   1  |   8]

We get:
 
  [1  -1   2  |  12]
  [0   4 -12  | -64]
  [2  -1   1  |   8]

To get a 0 where the 2 is in the
bottom left corner, mentally multiply each of the
numbers in the top row by -2 
and add them to 1 times the 
bottom row:

-2[1  -1   2  |  12]
  [0   4 -12  | -64]
 1[2  -1   1  |   8]
 
  [1  -1   2  |  12]
  [0   4 -12  | -64]
  [0   1  -3  | -16]

It's easier when there is a 1
on the diagonal, so
we will swap row 2 and row 3,
so we'll have a 1 on the middle
diagonal element: 

  [1  -1   2  |  12]
  [0   1  -3  | -16]
  [0   4 -12  | -64]

To get a 0 where the 4 is,
multiply the middle row by -4
and add to 1 times the 
bottom row:

  [1  -1   2  |  12]
-4[0   1  -3  | -16]
 1[0   4 -12  | -64]

We get

  [1  -1   2  |  12]
-4[0   1  -3  | -16]
 1[0   0   0  |   0]


Now that there are 0's
in those three positions,
we rewrite the augmented
matrix as a system of
equations, by putting the
variables and equal signs
back in:

  [ 1x  -1y   2z =  12]
  [ 0x   1y  -3z = -16]
  [ 0x   0y   0z =   0]

Erase the brackets, the
terms with 0 coefficients,
the 1's, and move the negative signs
left as minus signs:

     x -  y + 2z =  12 
          y - 3z = -16 
              0z =   0 

The bottom equation 0z = 0 is true
for any number we choose for z, so we
will represent it by the constant k.
[There are infinitely many solutions].
So we will set z equal to k

               z = k

Substitute z = k into the middle
equation:

          y - 3z = -16
          y - 3k = -16
               y = 3k - 16 

Substitute y = 3k - 16 and z = k into
the top equation:

        x - y + 2z = 12
x - (3k - 16) + 2k = 12
  x - 3k + 16 + 2k = 12
        x - k + 16 = 12
                 x = k - 4

So the solutions are given by:

(x, y, z) = (k-4, 3k-16, k)

For example the solution when k=0
is (x, y, z) = (-4, -16, 0)

The solution for k=1 is
(x, y, z) = (1-4, 3(1)-16, 1) or
(x, y, z) = (-3, -13, 1)

etc., etc., etc.

Edwin