Question 87119
<pre>
2x -  y + 2z = -8
-x + 3y - 4z = 15
 x + 2y - 3z =  9

[ 2  -1   2 | -8]
[-1   3  -4 | 15]
[ 1   2  -3 |  9] 

Get a zero where the -1 is on the 2nd row is by
multiplying the first row by 1 and adding it
to 2 times the second row:

1[ 2  -1   2 | -8]
2[-1   3  -4 | 15]
 [ 1   2  -3 |  9]

[ 2  -1   2 | -8]
[ 0   5  -6 | 22]
[ 1   2  -3 |  9]

Get a zero where the 1 is on the 3rd row is by
multiplying the first row by 1 and adding it
to -2 times the third row: 

 1[ 2  -1   2 | -8]
  [ 0   5  -6 | 22]
-2[ 1   2  -3 |  9]

[ 2  -1   2 | -8]
[ 0   5  -6 | 22]
[ 0  -5   8 |-26]

Get a 0 where the -5 is on the third row by
multiplying the second row by 1 and adding it
to 1 times row 3:

 [ 2  -1   2 | -8]
1[ 0   5  -6 | 22]
1[ 0  -5   8 |-26]

[ 2  -1   2 | -8]
[ 0   5  -6 | 22]
[ 0   0   2 | -4]

-1,2,-2

Get a 1 where the 2 is in the first row by
dividing the first row through by 2

Get a 1 where the 5 is in the second row by
dividing the second row through by 5

Get a 1 where the 2 is in the third row by
dividing the third row through by 2

[ 1   {{{-1/2}}}    1 |  -4]
[ 0     1  {{{-6/5}}} | {{{22/5}}}]
[ 0     0    1 |  -2]

Convert back to three equations:

{{{x-(1/2)y + z}}} = {{{-4}}}
   {{{y-(6/5)z}}} = {{{22/5}}}
          {{{z = -2}}}

Substitute z = -2 into the second equation

   {{{y-(6/5)(-2)}}} = {{{22/5}}}

   {{{y+12/5}}} = {{{22/5}}}

       {{{y}}} = {{{22/5-12/5}}}
       {{{y}}} = {{{10/5}}}
       {{{y}}} = {{{2}}}

Finally substitute {{{y=2}}} and {{{z=-2}}}
into the first equation:

{{{x-(1/2)y + z}}} = {{{-4}}}

{{{x-(1/2)(2) + (-2)}}} = {{{-4}}}

{{{x - 1 - 2 = -4}}}

{{{x = -4+1+2}}}

{{{x = -1}}}

So the solution is

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

Edwin</pre>



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