Question 570370
<pre>
{{{system(x-2y+z=6,
2x+y-3z=-3,
x-3y+3z=10)}}}

{{{system(1x-2y+1z=6,
2x+1y-3z=-3,
1x-3y+3z=10)}}}

{{{(matrix(3,5,

1,-2,1, "|",6,
2,1,-3,"|",-3,
1,-3,3,"|",10))}}}

We want to try to end up with a matrix 
that looks like this:

{{{(matrix(3,5,

"#","#","#", "|","#",
0,"#","#","|","#",
0,0,"#","|","#"))}}}

with 0's in the three lower left hand positions:


We can get a 0 where the 2 is by multiplying Row 1 by -2
and adding it to row 2:

That instruction is written as

-2·R1+1·R2->R2

{{{matrix(3,1,-2,1,"")}}}{{{(matrix(3,5,

1,-2,1, "|",6,
2,1,-3,"|",-3,
1,-3,3,"|",10))}}}

{{{(matrix(3,5,

1,-2,1, "|",6,
0,5,-5,"|",-15,
1,-3,3,"|",10))}}}

Notice that Row 2 will be simpler if we divide it through by 5,

That instruction is written {{{1/5}}}R2->R2

{{{matrix(3,1,"",1/5,"")}}}{{{(matrix(3,5,

1,-2,1, "|",6,
0,5,-5,"|",-15,
1,-3,3,"|",10))}}}

{{{(matrix(3,5,

1,-2,1, "|",6,
0,1,-1,"|",-3,
1,-3,3,"|",10))}}}

We can get a 0 where the 1 is in the lower left corner by 
multiplying Row 1 by -1 and adding it to row 3:

That instruction is written as

-1·R1+1·R3->R3

{{{matrix(3,1,-1,"",1)}}}{{{(matrix(3,5,

1,-2,1, "|",6,
0,1,-1,"|",-3,
1,-3,3,"|",10))}}}

{{{(matrix(3,5,

1,-2,1, "|",6,
0,1,-1,"|",-3,
0,-1,2,"|",4))}}}

We can get a 0 where the -1 is in the bottom row by 
multiplying Row 2 by 1 and adding it to row 3:

That instruction is written as 1R2+1R3->R3

{{{matrix(3,1,"",1,1)}}}{{{(matrix(3,5,

1,-2,1, "|",6,
0,1,-1,"|",-3,
0,-1,2,"|",4))}}}

{{{(matrix(3,5,

1,-2,1, "|",6,
0,1,-1,"|",-3,
0,0,1,"|",1))}}}

Now that we have 0's in the lower lefthand corner,
we convert the matrix back to a system of equations
in x, y and z:

{{{system(

1x-2y+1z=6,
0x+1y-1z=-3,
0x+0y+1z=1)}}}

or just

{{{system(

x-2y+z=6,
y-z=-3,
z=1)}}}

Now we use back substitution.

From the third equation, z=1, we substitute
that into the middle equation, getting:

y-z = -3
y-1 = -3
  y = -2

Then substitute y=-2 and z=1 in the 1st equation:

x-2(-2)+(1) = 6
      x+4+1 = 6
        x+5 = 6
          x = 1

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

Edwin</pre>