Originally Posted by

**Deveno** assign parameters to x3,x5 and x6 (s,t and u are the traditional choices). then express x1, x2, and x4 in terms of s,t,and u....it will look like this:

(5-2s-t-u,3s+2t-3u-1,s,1-t-u,t,u) although it looks cleaner if you write it like so:

s(-2,3,1,0,0,0) + t(-1,2,0,-1,1,0) + u(-1,-3,0,-1,0,1) + (5,-1,0,1,0,0,0) (these are often written as column vectors)

your solution set is a 3-dimensional subspace "translated" by a constant vector =

general solution of homogeneous system + one particular solution of non-homogeneous system.

note that (5,-1,0,1,0,0,0) is indeed "a" solution: 5(5) +(-1)(1) + 0(7) + 1(-1) + 0(2) + 0(6) = 23

5(2) + (-1)(9) + 0(-23) + 1(-2) + 0(-18) + 0(25) = -1,

5(1) + (-1)(3) + 0(-7) + 1(2) + 0(-3) + 0(14) = 4

5(6) + (-1)(2) + 0(6) + 1(1) + 0(3) + 0(14) = 29