Question 596541
<pre><font size = 4><b>

2x - 3y -  z = 0
3x + 2y + 2z = 2
 x + 5y + 3z = 2

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

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

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 3 in row 2 column 1 is,
we multiply row 1 temporarily by -3 and add it
to 2 times row 2. This is easy to do mentally
if you write -3 to the left of row 1 and 2 left
of row 2:

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

  [2  -3  -1  |  0]
  [0  13   7  |  4]
  [1   5   3  |  2]

To get a 0 where the 1 in row 3 column 1 is,
we multiply row 1 temporarily by -1 and add it
to 2 times row 3. This is easy to do mentally
if you write -1 to the left of row 1 and 2 left
of row 3:

 -1[2  -3  -1  |  0]
   [0  13   7  |  4]
  2[1   5   3  |  2]

   [2  -3  -1  |  0]
   [0  13   7  |  4]
   [0  13   7  |  4]

To get a 0 where the 13 in row 3 column 2 is,
we multiply row 2 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 2 and 1 left
of row 3:

-1[2  -3  -1 |  0]
  [0  13   7 |  4]
 2[0  13   7 |  4]


  [2  -3  -1 |  0]
  [0  13   7 |  4]
  [0   0   0 |  0]

Now that we have the three zeros, 
the above is the abbreviation of this system:

   2x -  3y - 1z = 0
   0x + 13y + 7z = 4
   0x +  0y + 0z = 0

or

   2x -  3y -  z = 0
        13y + 7z = 4
              0z = 0

The bottom equation,

              0z = 0

has as a solution "z can equal to any real number"

So we can select any letter to stand for any real number,
say the letter k.

So z = k

We substitute k for z into the middle equation 

       13y + 7z = 4
       13y + 7k = 4
            13y = 4 - 7k
              y = {{{4/13}}} - {{{7/13}}}k

We substitute k for z and {{{4/13}}}-{{{7/13}}}k for y into 
the top equation:

    2x - 3y - z = 0

2x - 3({{{4/13}}}-{{{7/13}}}k) - k = 0

2x - ({{{12/13}}} + {{{21/13}}}k) - k = 0

Clear of fractions by multiplying through by 13:

26x - 12 + 21k - 13k = 0

       26x - 12 + 8k = 0

                 26x = 12 - 8k

                   x = {{{12/26}}} - {{{8k/26}}}

                   x = {{{6/13}}} - {{{4k/13}}}

Solution:

(x, y, z) = ({{{6/13}}}-{{{4k/13}}}, {{{4/13}}}-{{{7/13}}}k, k)    


Edwin</pre>