Loads of solutions...like [14,7,-1,0] and [99,58,-18,0]...
where did you get this?
Take the first 2; add them:
(x1) – 2(x2) – (x3)+(x4)=1
2(x1) – 3(x2) + (x3) – (x4)=6
===================
3x1 - 5x2 = 7
Did you notice that?
Find all solutions to the following system of linear equations:
(x1) – 2(x2) – (x3)+(x4)=1
2(x1) – 3(x2) + (x3) – (x4)=6
3(x1) – 3(x2) + 6(x3))=15
(x1) + 5(x3)+(x4)=9
Using a system of linear equations, I found:
1 -2 -1 0 1
0 1 3 -3 4
0 0 0 6 0
0 0 0 0 0
so three solutions are:
(x1)=9, (x2)=4, (x3)=0, (x4)=0
(x1)=4, (x2)=1, (x3)=1, (x4)=0
(x1)=-1, (x2)=-2, (x3)=2, (x4)=0
How do I write my final solution (ie:what form)?
Loads of solutions...like [14,7,-1,0] and [99,58,-18,0]...
where did you get this?
Take the first 2; add them:
(x1) – 2(x2) – (x3)+(x4)=1
2(x1) – 3(x2) + (x3) – (x4)=6
===================
3x1 - 5x2 = 7
Did you notice that?
A system of n equations in n unknowns either has one solution or an infinite number of solutions. If an infinite number of solutions, then you need to write a general formula for the equations, not just list some.
What you have at this point is correct:
1 -2 -1 0 1
0 1 3 -3 4
0 0 0 6 0
0 0 0 0 0
The last line, of course, is true no matter what the "x"s are so we really only have three equations for the four unknowns. The third line is equivalent to the equation 6x4= 0 so x4= 0. The second line is equivalent to x2+ 3x3- 3x4= 4. Since x4= 0, that says x2+ 3x3= 4 so x2= 4- 3x3 for any x3. The first line is equivalent to x1- 2x2- x3= 1. Since x2= 4- 3x3, 2x2= 8- 6x3, that becomes x1- (8- 6x3)- x3= x1- 8+ 5x3= 1 so that x1= 9- 5x3.