Solving Gaussian Elimination Algorithm Problems (Matrices)

The system will be inconsistent and have no solution.In summary, The conversation discusses solving a Gaussian elimination algorithm problem involving matrices. The procedure involves scaling the first row, making entries below the first pivot equal to 0, repeating the algorithm for the submatrix, and using backwards substitution. There is also a discussion about solving problems with a specified precision and determining the value of a variable in order to have an inconsistent system.
  • #1
adc85
35
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I tried to solve this Gaussian elimination algorithm problem (matrices) but for some reason when I plug in the x variables it doesn't work. The problem is:

[ 3 10 4 : 7 ]
[ 2 7 3 : 5 ]
[ 1 3 2 : 2 ]

Alright so the first thing I did was divide the 1st row by 1/3 (scaling). Then I made the entries below the first pivot equal to 0 using:

Row2 = Row2 - Second Row, First Column * Row 1
Row3 = Row3 = Third Row, First Column * Row 1

Then I repeat this algorithm for the submatrix created afterwards (ignoring the first row and first column). Afterwards, I used backwards substitution (even tried using reduced echelon form). But I am not getting quite the right answers (very close for row 3 and the other two rows are fine though). Any input appreciated.

Also, I am having trouble understanding what my professor is saying when he says to solve a certain problem like this using 10^-3 precision for example. Do you just use the same method except placing decimal places at the end of each number or whatever?

Then he has this other weird problem that goes like:

[1 1 1 : 0]
[3 4 8 : 1]
[4 5 c^2 : c - 2]

What value of c would make this inconsistent (in other words, no solution)? I'm thinking that x3 could equal anything by making the last row full of zeros. Not sure though.

Thanks for any help.
 
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  • #2
As to your first question, can't tell you where you're going wrong if you don't show your work. Just a thought though: rather than starting off dividing (and immediately dealing with fractions) why not just swap the 1st & 3rd rows?

As to the second, I don't understand what you're asking.

As to the third, if you end up with all zeroes in the last row (i.e. by making c=3), you end up with a free variable, NOT an inconsistent system. So keep thinking about what value you can give to c, so you end up with zeroes in the first 3 columns but a non-zero value in the last column. THAT will be a system with no solution.
 
  • #3
Thank you gnome.

So if any row has this set-up:

[ 0 0 0 : Non-Zero Number ]

Then it means that it has no solution? Even if it's the second row?
 
  • #4
That's right.
 

1. What is Gaussian Elimination Algorithm?

Gaussian Elimination Algorithm is a method used to solve systems of linear equations by using elementary row operations to transform a matrix into reduced row-echelon form. This method is commonly used in linear algebra and is an efficient way to solve large systems of equations.

2. How do I solve a system of equations using Gaussian Elimination Algorithm?

To solve a system of equations using Gaussian Elimination Algorithm, you first need to write the system of equations in matrix form. Then, use elementary row operations to transform the matrix into reduced row-echelon form. The final matrix will represent the solution to the system of equations.

3. What is the purpose of reducing a matrix to row-echelon form?

Reducing a matrix to row-echelon form makes it easier to solve systems of equations. It eliminates any redundant or unnecessary information and allows us to see the relationships between the variables more clearly.

4. Can Gaussian Elimination Algorithm be used to solve any system of equations?

Yes, Gaussian Elimination Algorithm can be used to solve any system of linear equations. However, it may not always be the most efficient method, and other methods such as Cramer's rule or using a matrix calculator may be more suitable for certain types of systems.

5. Are there any limitations to using Gaussian Elimination Algorithm?

One limitation of Gaussian Elimination Algorithm is that it cannot be used for systems of equations with infinite solutions or no solutions. In these cases, the algorithm will either fail or produce inconsistent results. Additionally, the algorithm may become computationally expensive for very large matrices.

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