Question 155400
<pre><font size = 4 color = "indigo"><b>
{{{system(2x+2y+4z=40,0x-2y+2z=12,2x+3y+3z=34)}}} 

We form the augmented coefficient matrix

 [ 2   2   4 | 40]
 [ 0  -2   2 | 12]
 [ 2   3   3 | 34]

We must get three 0's in the lower 
left hand corner, where the numbers
are that I have colored red below:

 [ 2   2   4 | 40]
 [ <font color="red">0</font>  -2   2 | 12]
 [ <font color="red">2   3</font>   3 | 34]


Notice that one of them is already
0, so we get a zero where the red 2
is.

The top row will never change. 
The other two will. To get a 0 
where the red 2 is, We multiply -1 
times the first row and add it 
to 1 times the 3rd row. It's a 
good idea to put what you're
going to multiply a row by out
to the left of the row, even 
when it is just 1, like this:

-1[ 2   2   4 | 40]
  [ <font color="red">0</font>  -2   2 | 12]
 1[ <font color="red">2   3</font>   3 | 34]

The next matrix is then

 [ 2   2   4 | 40]
 [ <font color="red">0</font>  -2   2 | 12]
 [ <font color="red">0   1</font>  -1 | -6]

To get a 0 where the 1 is, 
we multiply 1 times the second 
row and add it to 2 times the 
3rd row.

 [ 2   2   4 | 40]
1[ <font color="red">0</font>  -2   2 | 12]
2[ <font color="red">0   1</font>  -1 | -6]

We get this:

 [ 2   2   4 | 40]
 [ <font color="red">0</font>  -2   2 | 12]
 [ <font color="red">0   0</font>   0 |  0]

Now that we have three 0's in the
lower left corner, the red places,
we go back to a system of equations:

{{{system(2x+2y+4z=40,0x-2y+2z=12,0x+0y+0z=0 )}}}

or just:

{{{system(2x+2y+4z=40,-2y+2z=12,0z=0 )}}}

Start at the bottom, and go back up.
This is called back-substitution.
Start with the bottom equation, solve for
z:

{{{0z=0}}}

That's a strange sort of equation.  We see
that any arbitrarily chosen value of z will 
satisfy that equation.  This means that z 
is any arbitrarily chosen number.  Some books
use the letter "c" and others simply keep {{{z}}}
and write {{{z=z}}}.

I will write as its solution

{{{z=a}}} because "a" stands for "arbitrary"

Substitute {{{a}}} for {{{z}}} in the
second equation:

{{{-2y+2z=12}}}
{{{-2y+2(a)=12}}}
{{{-2y+2a=12}}}
{{{-2y=12-2a}}}
Divide through by -2
{{{y=-6+a}}}

Finally substitute both {{{a}}} for {{{z}}}
and {{{-6+a}}} for {{{y}}} in the original
first equation:

{{{2x+2y+4z=40}}}
{{{2x+2(-6+a)+4(a)=40}}}
{{{2x+2(-6+a)+4a=40}}}
{{{2x-12+2a+4a=40}}}
{{{2x-12+6a=40}}}
{{{2x=52-6a}}}
Divide through by 2
{{{x=26-3a}}}

So the solution is

(x,y,z)=(26-3a, -6+a, a)

Edwin</pre>