Solving linear systems + gaussian elimination

AI Thread Summary
The discussion focuses on solving linear systems using Gaussian elimination and back substitution. Participants address specific homework questions, noting that questions 1 and 2 have more variables than equations, indicating an infinite number of solutions. The process of reducing the matrix to reduced row-echelon form (RREF) is emphasized, allowing for the expression of variables in terms of others. Back substitution is explained as a method to derive solutions from the RREF. Corrections are made regarding matrix entries, highlighting the collaborative nature of problem-solving in linear algebra.
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Homework Statement


http://img.rflz.net/problems.jpg

Homework Equations


The Attempt at a Solution


Question 1
x4 = 1 + 4x5
but from there I am not sure where to go or if there is even a solution?

Question 2
I eventually got the matrix down to this:

Code:
1  2 -3  1   4     1
0  1  1  0   1     1
0  0  1  0  -1/3  -1/3
0  0  0  1  -1/3   4/3

but again I am not sure if that is enough? I don't understand how back substitution is possible.

Question 3
Similar to 3. I sort of got it to row-echelon, but am having trouble. Any help on the previous question should help here.

Thanks
 
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For questions 1 and 2 you have more variables than equations, so you're going to have an infinite number of solutions.

For 1, you can get the system to reduced row-echelon form, which is a step beyond Gaussian elimination. In RREF, each row that has a nonzero first entry has 1 in that position, and all rows above and below have 0 in the same position. Doing this, I was able to get the system in this form:
Code:
[1 0 -5 0 36 | 3]
[0 1 1 0 -15 |-3]
[0 0 0 1 -4 | 1]
From this you can write x1 in terms of x3 and x5, x2 in terms of x3 and x5, and x4 in terms of x5.

If you do this by back substitution, you'll start with the 3rd equation, and write x4 in terms of x5. Substitute for x4 in the 2nd equation, so you'll eventually get to x2 in terms of x3 and x5. Substitute what you have here for x2 in the 1st equation to get x1 in terms of x3 and x5. IOW, the same as what I did before using RREF, which is essentially Gauss elimination with back substitution.

For question 2, do the same thing.
 
Thanks for the reply. I managed to get 2 and 3 on my own, but your advice on question 1 should be very useful. :)
 
I tried putting the system in the form you have, and I think you have made a mistake. The -3 in the last column should be a -1, no?
 
Yes, you're right. The last entry in the 2nd row should be -1.
Code:
[1 0 -5 0 36 | 3]
[0 1 1 0 -15 |-1]
[0 0 0 1 -4 | 1]
 
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