SOLUTION: I am having problems figuring out matrices and row operations. If anyone can help with this it would be a huge help. Use matricies to find the general solution of the following

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