Question 313934
<pre><b>
{{{system(x+y+z+u=1,
2x+3y-4z+5u=2,
3x-4y-3z+6u=0)}}} 

The augmented matrix is:

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

-2*R1+R2->R2
-3*R1+R3->R3

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

-1*R2+R1->R1
 7*R2+R3->R3

{{{(matrix(3,6, 
          1,  0,   7, -2, "|", 1,
          0,  1,  -6,  3, "|", 0,
          0,  0, -48, 24, "|",-3))}}}

{{{-1/48}}}*R3->R3

{{{(matrix(3,6, 
          1,  0,   7, -2, "|", 1,
          0,  1,  -6,  3, "|", 0,
          0,  0,   1, -1/2, "|",1/16))}}}

-7*R3+R1->R1
 6*R3+R2->R2

{{{(matrix(3,6, 
          1,  0,  0, 3/2, "|", 9/16,
          0,  1,  0,  0, "|", 3/8,
          0,  0,  1, -1/2, "|",1/16))}}}

Rewrite as this system of equations:

{{{system(1x+0y+0z+(3/2)u=9/16,
          0x+1y+0z+    0u=3/8,
          0x+0y+1z-(1/2)u=1/16)}}}

Simplify:

{{{system(x+(3/2)u=9/16,
          y=3/8,
          z-(1/2)u=1/16)}}}

Solve for x, y and z

{{{system(x=9/16-(3/2)u,
          y=3/8,
          z=1/16+(1/2)u)}}}


Let {{{u=k}}}, where k can represent any arbitrary
value, and we have this solution:

{{{(matrix(1,7, x, ",", y, ",", z, ",", u))=  


(matrix(1,7,

9/16-(3/2)k, ",",
          3/8, ",",
          1/16+(1/2)k,",",k))}}}

Edwin</pre>