Solving linear systems + gaussian elimination

In summary, the conversation discusses solving a system of equations using Gaussian elimination and back substitution. It is explained that for questions 1 and 2, there are more variables than equations, resulting in an infinite number of solutions. The system is then reduced to row-echelon form and back substitution is used to find the solutions. For question 3, the same approach is used. Additionally, there is a mistake in the original solution, which is corrected in the summary.
  • #1
dcramps
43
0

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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  • #2
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.
 
  • #3
Thanks for the reply. I managed to get 2 and 3 on my own, but your advice on question 1 should be very useful. :)
 
  • #4
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?
 
  • #5
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]
 

1. What is a linear system?

A linear system is a set of equations that can be represented in the form of Ax = b, where A is a matrix of coefficients, x is a vector of variables, and b is a vector of constants. The goal of solving a linear system is to find the values of the variables that satisfy all of the equations simultaneously.

2. What is gaussian elimination?

Gaussian elimination is a method for solving linear systems by using elementary row operations to reduce the system to an equivalent system in triangular form. This makes it easier to solve the system using back substitution. The process involves systematically eliminating variables from each equation until a solution is found.

3. What is the purpose of solving linear systems using gaussian elimination?

The purpose of solving linear systems using gaussian elimination is to find the values of the variables that satisfy all of the equations simultaneously. This can be useful in many applications, such as solving systems of equations in engineering, physics, and economics.

4. What are the steps involved in gaussian elimination?

The steps involved in gaussian elimination are:

  1. Write the linear system in augmented matrix form.
  2. Use elementary row operations to eliminate variables from each equation, working from left to right.
  3. Repeat this process until the system is in triangular form.
  4. Use back substitution to solve for the values of the variables.

5. Are there any limitations to using gaussian elimination to solve linear systems?

While gaussian elimination is a powerful method for solving linear systems, it does have some limitations. It can be time-consuming for large systems, and it may not work if the matrix of coefficients is singular or has a very small pivot element. In these cases, other methods such as LU decomposition or the Gauss-Jordan method may be more effective.

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