x +  y + 6z = 3
x +  y + 3z = 3
x + 2y + 4z = 7

I will assume you have had matrices. If not,
post again and I will show you how to solve
it without matrices:

The augmented matrix is

[1   1   6  |  3]
[1   1   3  |  3]
[1   2   4  |  7]

We need to get 0's in the three lower left
positions, that is, in the positions below
the upper left to lower-right diagonal:

To get a 0 where the 1 is in row 2 column 1 is,
we multiply row 1 temporarily by -1 and add it
to 1 times row 2. This is easy to do mentally
if you write -1 to the left of row 1 and 1 left
of row 2:

-1[1   1   6  |  3]
 1[1   1   3  |  3]
  [1   2   4  |  7]

  [1   1   6  |  3]
  [0   0  -3  |  0]
  [1   2   4  |  7]

To get a 0 where the 1 is in row 3 column 1 is,
we multiply row 1 temporarily by -1 and add it
to 1 times row 3. This is easy to do mentally
if you write -1 to the left of row 1 and 1 left
of row 3:

 -1[1   1   6  |  3]
   [0   0  -3  |  0]
  1[1   2   4  |  7]

   [1   1   6  |  3]
   [0   0  -3  |  0]
   [0   1  -2  |  4]

To get a 0 where the 1 is in row 3 column 2 is,
we merely need to swap rows 2 and 3

   [1   1   6  |  3]
   [0   1  -2  |  4]
   [0   0  -3  |  0]
 
Now we need to get 1's on the upper left to
lower right diagonal.  Two of the diagonal 
elements are already 1's.  To get a 1 where 
the -3 is, we multiply row 3 by -1/3

    [1   1   6  |  3]
    [0   1  -2  |  4]
-1/3[0   0  -3  |  0]

   [1   1   6  |  3]
   [0   1  -2  |  4]
   [0   0   1  |  0]

This is the system:

   1x + 1y + 6z = 3
   0x + 1y - 2z = 4
   0x + 0y + 1z = 0

or

    x +  y + 6z = 3
         y - 2z = 4
              z = 0

The bottom equation tells us that
z = 0

Substitute z = 0 into the 2nd equation

      x + y + 6z = 3
          y - 2z = 4
         y - 2·0 = 4
               y = 4

Substitute z = 0 and y = 4 into the 1st
equation

     x + 4 + 6·0 =  3
       x + 4 + 0 =  3
               x = -1

So the solution is (x, y, z) = (-1, 4, 0)

Edwin