Question 923799
<pre><font size = 4 color = "indigo"><b>
{{{system(10x + y + z= 12,
x + 10y + z= 12,
x + y + z= 12)}}}

or rather

{{{system(10x + 1y + 1z= 12,
1x + 10y + 1z= 12,
1x + 1y + 1z= 12)}}}

Write that as a matrix by dropping the letters
and putting a vertical line instead of equal signs:

{{{(matrix(3, 5,
     10, 1, 1,"|",12,
     1,10, 1,"|",12,
    1,1, 1,"|",12  ))}}}

Things are easier if it is possible to get a 1 in 
the upper left corner by swapping rows. So let's 
begin by swapping rows 1 and 3.  That operation is
written as 

R1<->R3

{{{(matrix(3, 5,
     1, 1, 1,"|",12,
     red(1),10, 1,"|",12,
    red(10),red(1), 1,"|",12  ))}}}

The idea is to get three zeros in the three positions
in the lower left corner of the matrix, where the elements
I've colored red are:

To get a 0 where the red 1 on the left of the middle row is,
multiply R1 by -1 and add it to 1 times R2, and put it in place 
of the present R2.  That's written as

-1R1+1R2->R2

To make it easy, write the multipliers to the left of the two
rows you're working with; that is, put a -1 by R1 and a 1 by R2

{{{matrix(3,1,-1,1,"")}}}{{{(matrix(3, 5,
     1, 1, 1,"|",12,
     red(1),10, 1,"|",12,
    red(10),red(1), 1,"|",12  ))}}}


We are going to change only R2.  Although R1 gets multiplied
by -1 we are going to just do that mentally and add it to R2, but
not really change R1.

{{{(matrix(3, 5,
     1, 1,1,"|",12,
     red(0),9, 0,"|",0,
    red(10),red(1), 1,"|",12  ))}}}

-----

To get a 0 where the lower left red 10 is, multiply R1
by -10 and add it to 1 times R3.  That's written as

-10R1+1R3->R3

Write the multipliers to the left of the two rows you're 
working with; that is, put a -10 by R1 and a 1 by R3

{{{matrix(3,1,-10,"",1)}}}{{{(matrix(3, 5,
     1, 1,1,"|",12,
     red(0),9, 0,"|",0,
    red(10),red(1), 1,"|",12  ))}}}


We are going to change only R3. 

{{{(matrix(3, 5,
     1, 1,1,"|",12,
     red(0),9, 0,"|",0,
    red(0),red(-9), -9,"|",-108  ))}}}


---------------

To get a 0 where the red -9 is, multiply R2
by 1 (leave it as it is) and add it to 1 times R3.
We're just adding row 2 as is to row 3 as is. 
That's written as

1R2+1R3->R3

Write the multipliers to the left of the two
rows you're working with; that is, put a 1 by R2 
and a 1 by R3:

{{{matrix(3,1,"","1",1)}}}{{{(matrix(3, 5,
     1, 1,1,"|",12,
     red(0),9, 0,"|",0,
    red(0),red(-9), -9,"|",-108  ))}}}

We are going to change only R3. 

{{{(matrix(3, 5,
     1, 1,1,"|",12,
     red(0),9, 0,"|",0,
    red(0),red(0), -9,"|",-108  ))}}}

Now that we have 0's in the three positions in the
lower left corner of the matrix, we change the matrix
back to equations:

{{{system(x+y+z=12,0x+9y+0z=0,0x+0y-9z=-108)}}}

or just

{{{system(x+y+z=12,9y=0,-9z=-108)}}}

Solve the third equation for z:

{{{-9z=-108}}}
{{{z=(-108)/(-9)}}}
{{{z=12}}}

Solving the middle system:

{{{9y=0}}}
{{{y=0}}}

Substitute 12 for z and 0 for y in the top equation:

{{{x+y+z=12}}}
{{{x+(0)+(12)=12}}}
{{{x+0+12=12}}}
{{{x+12=12}}}
{{{x=0}}}

So the solution is {{{"(x,y,z)"="(0,0,12)"}}}

Edwin</pre>