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



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





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

That instruction is written R2->R2





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





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





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:



or just



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