Question 592297
<pre>
  x +  y +  z = 3
 2x -  y +  z = 2
 4x - 2y + 3z = 5

 1x + 1y + 1z = 3
 2x - 1y + 1z = 2
 4x - 2y + 3z = 5

{{{(matrix(3,5,

 1, 1, 1, "|", 3,
 2, -1, 1, "|", 2,
 4, -2,  3, "|", 5))}}}

The idea is to use row operations to get it to this form {{{(matrix(3,5,

 1, 0, 0, "|", a,
 0, 1, 0, "|", b,
 0, 0, 1, "|", c))}}}

Add -2Śrow1 to 1Śrow2 to get a 0 where the 2 is on the left
of row2:

{{{matrix(3,1,-2,1,"")}}}{{{(matrix(3,5,

 1, 1, 1, "|", 3,
 2, -1, 1, "|", 2,
 4, -2,  3, "|", 5))}}} = {{{(matrix(3,5,

 1, 1, 1, "|", 3,
 0, -3, -1, "|", -4,
 4, -2,  3, "|", 5))}}}

Add -4Śrow1 to 1Śrow3 to get a 0 where the 4 is on the left
of row3:

{{{matrix(3,1,-4,"",1)}}}{{{(matrix(3,5,

 1, 1, 1, "|", 3,
 0, -3, -1, "|", -4,
 4, -2,  3, "|", 5))}}} = {{{(matrix(3,5,

 1, 1, 1, "|", 3,
 0, -3, -1, "|", -4,
 0, -6, -1, "|", -7))}}}

Add 1Śrow2 to 3Śrow1 to get a 0 where the 1 is on row1 2nd elementt

{{{matrix(3,1,3,1,"")}}}{{{(matrix(3,5,

 1, 1, 1, "|", 3,
 0, -3, -1, "|", -4,
  0, -6, -1, "|", -7))}}} = {{{(matrix(3,5,

 3, 0, 2, "|", 5,
 0, -3, -1, "|", -4,
 0, -6, -1, "|", -7))}}}

Add -2Śrow2 to 1*row3 to get a 0 where the -6 is:

{{{matrix(3,1,"",-2,1)}}}{{{(matrix(3,5,

  3, 0, 2, "|", 5,
 0, -3, -1, "|", -4,
  0, -6, -1, "|", -7))}}} = {{{(matrix(3,5,

 3, 0, 2, "|", 5,
 0, -3, -1, "|", -4,
 0, 0, 1, "|", 1))}}}

Add -2Śrow3 to 1*row1 to get a 0 where the 2 is:

{{{matrix(3,1,1,"",-2)}}}{{{(matrix(3,5,

  3, 0, 2, "|", 5,
 0, -3, -1, "|", -4,
 0, 0, 1, "|", 1))}}} = {{{(matrix(3,5,

 3, 0, 0, "|", 3,
 0, -3, -1, "|", -4,
 0, 0, 1, "|", 1))}}}

Add 1Śrow3 to 1*row2 to get a 0 where the -1 is:

{{{matrix(3,1,"",1,1)}}}{{{(matrix(3,5,

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

 3, 0, 0, "|", 3,
 0, -3, 0, "|", -3,
 0, 0, 1, "|", 1))}}}

We have all the 0's placed, so all we need do is get the 1's

Get a 1 where the first 3 on row1 is by dividing R1 by 3
Get a 1 where the first -3 on row2 is by dividing R2 by -3
{{{(matrix(3,5,

 3, 0, 0, "|", 3,
 0, -3, 0, "|", -3,
 0, 0, 1, "|", 1))}}}{{{matrix(3,1,"ś3","ś-3","")}}} = {{{(matrix(3,5,

 1, 0, 0, "|", 1,
 0, 1, 0, "|", 1,
 0, 0, 1, "|", 1))}}}

This is the row-reduced echelon form.  To get the solution, translate
it into a system of 3 equations and 3 variables:

 1x + 0y + 0z = 1
 0x + 1y + 0z = 1
 0x + 0y + 1z = 1

or

x = 1
y = 1
z = 1

So the solution is (x,y,z) = (1,1,1)

Edwin</pre>