Question 831301
<pre>
{{{system(x[1]+x[2] = 5,
x[2]+x[3] = 5,
x[3]+x[4] = -5,
x[1]+x[4] = -5)}}}

Put in missing variables with 0 coefficients
and give single variables 1 coefficents

{{{system(1x[1]+1x[2]+0x[3]+0x[4] = 5,
0x[1]+1x[2]+1x[3]+0x[4] = 5,
0x[1]+0x[2]+1x[3]+1x[4] = -5,
1x[1]+0x[2]+0x[3]+1x[4] = -5)}}}

Put the coefficients in a 4×5 augmented matrix and do row operations
to get 0's below the diagonals:

{{{(matrix(4,6,

1,1,0,0,"|",5,
0,1,1,0,"|",5,
0,0,1,1,"|",-5,
1,0,0,1,"|",-5))}}}

-1*R1+R4->R4   (Multiply row 1 by -1, add to row 4, put it in row 4.) 

{{{(matrix(4,6,

1,1,0,0,"|",5,
0,1,1,0,"|",5,
0,0,1,1,"|",-5,
0,-1,0,1,"|",-10))}}}

R2+R4->R4    (Add row 2 and R4 and put it in row 4)  

{{{(matrix(4,6,

1,1,0,0,"|",5,
0,1,1,0,"|",5,
0,0,1,1,"|",-5,
0,0,1,1,"|",-5))}}}

-R3+R4->R4

{{{(matrix(4,6,

1,1,0,0,"|",5,
0,1,1,0,"|",5,
0,0,1,1,"|",-5,
0,0,0,0,"|",0))}}}

{{{system(1x[1]+1x[2]+0x[3]+0x[4] = 5,
0x[1]+1x[2]+1x[3]+0x[4] = 5,
0x[1]+0x[2]+1x[3]+1x[4] = -5,
0x[1]+0x[2]+0x[3]+0x[4] = 0)}}}

Eliminate the 0s and 1 coefficients, which
means we eliminate the entire bottom equation. 

{{{system(1x[1]+x[2] = 5,
x[2]+x[3] = 5,
x[3]+x[4] = -5)}}}

We let {{{x[4]=s}}}

{{{system(x[1]+x[2]+s = 5,
x[2]+x[3] = 5,
x[3]+s = -5)}}}

Solve the bottom equation of the system: {{{x[3]=-5-s}}}

Substitute in the middle equation and solve:

{{{x[2]+x[3] = 5}}}
{{{x[2]+(-5-s) = 5}}}
{{x[2]-5-s = 5}}}
{{x[2]= 10+s}}}

Substitute in the top equation

{{{x[1]+x[2] = 5}}}
{{{x[1]+(10+s) = 5}}}
{{{x[1]+10+s = 5}}}
{{{x[1]=-5-s}}}


{{{x[1]=-5-s}}}
{{{x[2]= 10+s}}}
{{{x[3]=-5-s}}}
{{{x[4]=s}}}

or in the form that you gave:

[x1] = [5] + [-1]s 
[x2] = [10] + [1]s
[x3] = [-5] + [-1]s 
[x4] = [0] + [1]s

Edwin</pre>