Linear System problem help?
have a linear system problem, it's very similiar to this one http://answers.yahoo.com/question/index;… that I've asked about before. I tried the method explained in those problems but I'm struggling to get the correct anser so if anyone could please help me it would be very much appreciated 
Note: x_1 means x subscript 1 etc
Solve the system:
5(x_1)-6(x_2)+ 4(x_3)+2(x_4)=1
-x_1+x_2+3(x_3)+3(x_4)=3
4(x_1)-5(x_2)+7(x_3)+5(x_4)=4
2(x_1)-2(x_2)-6(x_3)-6(x_4)=-6
and the form of the answer needs to be:
x(subscript1)= ___ + ___[s] + ___[t]
x(subscript2)= ___ + ___[s] + ___[t]
x(subscript3)= ___ + ___[s] + ___[t]
x(subscript4)=___ + ___[s] + ___[t]
So I basically know I need to solve one term in terms of two other terms and substitute back into oneof the eqns etc.. I think thats what I need to do at least lol and this is what my professor said about the solutions form "the parametric form is point on the plain plus all multiples of 2 linearly independent vectors in the plane" like i said any help would be appreciated Thanks!
Fantastic! I appreciate you're help, I'm so glad you took the time to explain the steps too because I really need to understand it, not just get an answer. I wish I could give you more than 10 pts for best answer!!
Comments
I notice you reposted this question because there were no answers, I suppose I'm the only one interested in it, I gave you a bad answer last time because I wasn't quite clear on what you were trying to do. I've done this kind of math before but a long time ago, it's come back slowly. I think I've got it now. Check out the links to better explain the procedure.
Ignore all the ``` marks, they're for spacing.
Setup a matrix and use Gauss-Jordan elimination to reduce it...
[ +5 -6 +4 +2 | +1 ]
[ -1 +1 +3 +3 | +3 ]
[ +4 -5 +7 +5 | +4 ]
[ +2 -2 `-6 `-6 | `-6 ]
Reorder the lines...
[ -1 +1 +3 +3 | +3 ]
[ +2 -2 `-6 `-6 | `-6 ]
[ +4 -5 +7 +5 | +4 ]
[ +5 -6 +4 +2 | +1 ]
Divide the first line by -1...
[ +1 -1 `-3 `-3 | `-3 ]
[ +2 -2 `-6 `-6 | `-6 ]
[ +4 -5 +7 +5 | +4 ]
[ +5 -6 +4 +2 | +1 ]
Clear the first column by multiplying the first line by the appropriate number
and adding to the affected line (-2, then -4, then -5)...
[ +1 `-1 `-3 `-3 | `-3 ]
[ ``0 ``0 ``0 ``0 | ``0 ]
[ ``0 `-1 19 17 | `16 ]
[ ``0 `-1 19 17 | `16 ]
Divide the third line by -1...
[ +1 `-1 `-3 `-3 | `-3 ]
[ ``0 ``0 ``0 ``0 | ``0 ]
[ ``0 +1 -19 -17 | -16 ]
[ ``0 `-1 `19 `17 | `16 ]
Add the third line to the fourth...
[ +1 `-1 `-3 `-3 | `-3 ]
[ ``0 ``0 ``0 ``0 | ``0 ]
[ ``0 +1 -19 -17 | -16 ]
[ ``0 ``0 ``0 ``0 | ``0 ]
Clear the second column by adding the third line to the first...
[ +1 ``0 -22 -20 | -19 ]
[ ``0 ``0 ``0 ``0 | ``0 ]
[ ``0 +1 -19 -17 | -16 ]
[ ``0 ``0 ``0 ``0 | ``0 ]
Notice the third and the fourth columns are not able to be cleared
since two of these equations are linearly dependent, this means there is
not an exact solution, but it can be written in parametric form.
This gives...
x1 - 22x3 - 20x4 = -19
x2 - 19x3 - 17x4 = -16
x1 = -19 +22x3 + 20x4
x2 = -16 +19x3 + 17x4
We now have two variable in terms of the other two, let x3 = s and x4 = t
x1 = -19 + 22s + 20t
x2 = -16 + 19s + 17t
x3 = s
x4 = t
Thank you for posting such a good question :-)