Question 1061938
Let A= [1 2 3 5;2 4 8 12;3 6 7 13] I have already reduced
the matrix to [1 2 3 5;0 0 1 1;0 0 0 0] I need help 
solving the next step which is to find the general 
solution in the form x=xh + p of Ax= (0, 6, -6)T
<pre><b> 
Matrix A is an abbreviation for this system of equations:

{{{system(

matrix(3,9,

1x, ""+"", 2y, ""+"", 3z, "",""="","", 5,
2x, ""+"", 4y, ""+"", 8z, "",""="","", 12,
3x, ""+"", 6y, ""+"", 7z, "",""+"","", 13))}}}
{{{matrix(1,3,
A,
""="", 
(matrix(3,5,1, 2, 3,"|",5,2, 4, 8,"|", 12,3, 6, 7,"|", 13))

)}}}
You have reduced the matrix down to this, call it B:

{{{matrix(1,3,
B,
""="", 
(matrix(3,5,1, 2, 3,"|",5,0, 0, 1,"|", 1,0, 0, 0,"|", 0))

)}}}

That reduced matrix is an abbreviation for this system of equations:

{{{system(

matrix(3,9,

1x, ""+"", 2y, ""+"", 3z, "",""="","", 5,
0x, ""+"", 0y, ""+"", 1z, "",""="","", 1,
0x, ""+"", 0y, ""+"", 0z, "",""="","", 0))}}}

Eliminating all the 0 terms and the 1 coefficients:

{{{system(

matrix(2,9,

x, ""+"", 2y, ""+"", 3z, "",""="","", 5,
"", "", "", "", z, "",""="","", 1))}}}

Substitute 1 for z in the first equation:

{{{matrix(1,9,

x, ""+"", 2y, ""+"", 3(1), "",""="","", 5)}}}

{{{matrix(1,7,

x, ""+"", 2y,  "",""="","", 2)}}}

{{{matrix(1,5,

x, "",""="","", 2-2y)}}}


So the general solution is

x = 2-2y
y = y
z = 1

or

(x, y, z) = (2-2y, y, 1)

That's the way most books like the general solution in.

I've never seen 
</pre>
       "the form x=xh + p of Ax= (0, 6, -6)T"
<pre>
though I've taught college mathematics for many years.  I'm sure
it's equivalent to that, though. Some books will put a k for the y,
like this:

(x, y, z) = (2-2k, k, 1)

Edwin</pre></b>