Question 431371
<pre>
2xl+4x2-10x3=-2
3xl+9x2-2lx3=0
1xl+5x2-12x3=1
 
{{{( matrix(3,5,
 
2,4,-10,"|",-2,
3,9,-2l,"|",0,
l,5,-12,"|",1)) }}}
 
The idea is to end up with a matrix that looks like this:
 
{{{( matrix(3,5,
 
1,"#","#","|","#",
0,1,"#","|","#",
0,0,1,"|","#")) }}}
 
where there are numbers where the {{{"#"}}}'s are
 
 
 

Divide row 1 by 2 to get a 1 in the upper left corner
 
{{{( matrix(3,5,
 
1,2,-5,"|",-1,
3,9,-2l,"|",0,
l,5,-12,"|",1)) }}}
 
Get a 0 under it by multiplying row 1 by -3
and adding it to row 2 {{{ matrix(2,5,
-3,-6,15,"|",3,
3,9,-2l,"|",0) }}} getting {{{ matrix(1,5,
0,3,-6,"|",3) }}}
Then replace row 2 by that
 
{{{( matrix(3,5,
 
1,2,-5,"|",-1,
0,3,-6,"|",3,
l,5,-12,"|",1)) }}}
 
Get a 0 in the lower left by multiplying row 1 by -1
and adding it to row 3 {{{ matrix(2,5,
-1,-2,5,"|",1,
1,5,-l2,"|",1) }}} getting {{{ matrix(1,5,
0,3,-7,"|",2) }}}
Then replace row 3 by that
{{{( matrix(3,5,
1,2,-5,"|",-1,
0,3,-6,"|",3,
0,3,-7,"|",2)) }}}
 
Get a 1 where the first 3 on the second row is by
dividing the second row through by 3
 
{{{( matrix(3,5,
1,2,-5,"|",-1,
0,1,-2,"|",1,
0,3,-7,"|",2)) }}}
 
Get a 0 where the 3 is on the bottom row
 by multiplying row 2 by -3
and adding it to row 3 {{{ matrix(2,5,
0,-3,6,"|",-3,
0,3,-7,"|",2) }}} getting {{{ matrix(1,5,
0,0,-1,"|",-1) }}}
Then replace row 3 by that
{{{( matrix(3,5,
1,2,-5,"|",-1,
0,1,-2,"|",1,
0,0,-1,"|",-1)) }}}
 
Now divide the bottom row through by -1 to
get a 1, and you have the final matrix:
 
{{{( matrix(3,5,
1,2,-5,"|",-1,
0,1,-2,"|",1,
0,0,1,"|",1)) }}}
 
That means:
 
1x1 + 2x2 - 5x3 = -1
0x1 + 1x2 - 2x3 =  1
0x1 + 0x2 + 1x3 =  1
 
or simplifying,
 
x1 + 2x2 - 5x3 = -1
x2 - 2x3 = 1
x3 = 1
 
Substitute 1 for x3 in the middle equation:
 
 x2 - 2x3 = 1
x2 - 2(1) = 1
   x2 - 2 = 1
       x2 = 3
 
Substitute 1 for x3 and 3 for x1 in the top equation:
 
   x1 + 2x2 - 5x3 = -1
x1 + 2(3) - 5(1) = -1
      x1 + 6 - 5 = -1
          x1 + 1 = -1
              x1 = -2
 
Solution:  (x1,x2,x3) = (-2,3,1)
 
Edwin</pre>