Question 378061
<pre>
  3x -  6y + 1z = 0
  1x +  1y + 0z = 0
-12x + 24y - 4z = 0

We make the augmented matrix by eliminating the variables:

{{{(matrix(3,5,

  3,-6, 1,"|",0,
  1, 1, 0,"|",0,
-12,24,-4,"|",0))}}}

Swap R1 and R2  (row 1 and row 2)
That instruction is written
R1<-->R2

{{{(matrix(3,5,
  1, 1, 0,"|",0,
  3,-6, 1,"|",0,
-12,24,-4,"|",0))}}}

Get a 0 where the 3 is by multiplying R1 by -3
temporarily and add it to row 2, then restore R1

That instruction is written

-3*R1+R2 -> R2

{{{(matrix(3,5,
  1, 1, 0,"|",0,
  0,-9, 1,"|",0,
-12,24,-4,"|",0))}}}

Get a 0 where the -12 is by multiplying R1 by 12
temporarily and add it to row 2, then restore R1

That instruction is written

12*R1+R3 --> R3

{{{(matrix(3,5,
  1, 1, 0,"|",0,
  0,-9, 1,"|",0,
  0,36,-4,"|",0))}}}

Get a 1 where the -9 is by multiplying row 2 by {{{-1/9}}}

That instruction is written:

{{{-1/9}}}R2 --> R2

{{{(matrix(3,5,
  1, 1, 0,"|",0,
  0,1, -1/9,"|",0,
  0,36,-4,"|",0))}}}

Get a 0 where the 36 is by multiplying R2 by -36
temporarily and add it to row 3, then restore R2

That instruction is written

-36R2+R3-->R3

{{{(matrix(3,5,
  1, 1, 0,"|",0,
  0,1, -1/9,"|",0,
  0,0,0,"|",0))}}}

Write that as a system of equations:

{{{system(  1x +  1y + 0z = 0,
  0x +  1y - expr(1/9)z = 0,
  0x +  0y + 0z = 0)}}}

or 

{{{system(  x +  y = 0,
  y - expr(1/9)z = 0)}}}

We let z be arbitrary.  Solve the second equation for y

{{{y=expr(1/9)z}}}

Solve the first equation for x:

{{{x = -y}}}

Substitute {{{expr(1/9)z}}} for y

{{{x = -expr(1/9)z}}}

So the set of solutions is

(x,y,z) = ({{{-expr(1/9)z}}},{{{expr(1/9)z}}},{{{z}}})

where z is any arbitrary number.

Edwin</pre>