Solving Gaussian Elimination Algorithm Problem (Matrices)

In summary, the Gaussian elimination algorithm is used to reduce a matrix into a specific form and can be used to solve systems of equations. In this case, the algorithm was used on a matrix to find the solution of x1 = -1, x2 = 1, and x3 = 0.)
  • #1
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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
[tex]\left(\begin{array}{x1x2x3k}
3 & 10 & 4 & 7\\
2 & 7 & 3 & 5\\
1 & 3 & 2 & 2
\end{array}\right)[/tex]

Gaussian Elimination or any other fancy name is simply an attempt to reduce this matrix into a [tex]\left(\begin{array}{x1x2x3k}
1 & 0 & 0 & k1\\
0 & 1 & 0 & k2\\
0 & 0 & 1 & k3
\end{array}\right)[/tex] form.

I will use notation R1, R2, and R3 to indicate Rows 1, 2, and 3. For example (R2 = R2 + R1 means you add row 1 into row 2, and after that you should have Row 1 as it was, and Row 2 will be sum of Row 1 and Row 2 from before.

Now in your case:

[tex]\left(\begin{array}{x1x2x3k}
3 & 10 & 4 & 7\\
2 & 7 & 3 & 5\\
1 & 3 & 2 & 2
\end{array}\right)[/tex] R1 = R1 - R2 -> [tex]\left(\begin{array}{x1x2x3k}
1 & 3 & 1 & 2\\
2 & 7 & 3 & 5\\
1 & 3 & 2 & 2
\end{array}\right)[/tex] R2 = R2 - 2*R3 -> [tex]\left(\begin{array}{x1x2x3k}
1 & 3 & 1 & 2\\
0 & 1 & -1 & 1\\
1 & 3 & 2 & 2
\end{array}\right)[/tex]. R3 = R3 - R1 -> [tex]\left(\begin{array}{x1x2x3k}
1 & 3 & 1 & 2\\
0 & 1 & -1 & 1\\
0 & 0 & 1 & 0
\end{array}\right)[/tex].


[tex]\left(\begin{array}{x1x2x3k}
1 & 3 & 1 & 2\\
0 & 1 & -1 & 1\\
0 & 0 & 1 & 0
\end{array}\right)[/tex] R1 = R1 - 3R2 -> [tex]\left(\begin{array}{x1x2x3k}
1 & 0 & 4 & -1\\
0 & 1 & -1 & 1\\
0 & 0 & 1 & 0
\end{array}\right)[/tex]. R1 = R1 - 4R3 -> [tex]\left(\begin{array}{x1x2x3k}
1 & 0 & 0 & -1\\
0 & 1 & -1 & 1\\
0 & 0 & 1 & 0
\end{array}\right)[/tex]. R2 = R2 + R3 -> [tex]\left(\begin{array}{x1x2x3k}
1 & 0 & 0 & -1\\
0 & 1 & 0 & 1\\
0 & 0 & 1 & 0
\end{array}\right)[/tex]

So you end up with this:
[tex]x1 = -1, x2 = 1, x3 = 0[/tex]

Your answer is: [tex](-1, 1, 0) \epsilon \mathbb{R}^3[/tex]
 
  • #3


Hi there,

Firstly, it's great that you are actively trying to solve this Gaussian elimination algorithm problem and seeking help when you encounter difficulties. It shows determination and a willingness to learn.

Regarding the first problem, it seems like you have a good understanding of the steps involved in Gaussian elimination. However, it's important to double-check your calculations, as even small errors can lead to incorrect solutions. It's also possible that you may have missed a step or made a mistake in the backwards substitution process. I would recommend checking your work and perhaps seeking assistance from your professor or classmates if you are still having trouble.

As for the precision question, using 10^-3 precision means that your final solutions should be rounded to three decimal places. This is to ensure that your answers are as accurate as possible without being overly complicated.

For the second problem, you are correct in thinking that there is no value of c that will make this system consistent. This is because the third row has an equation that is dependent on c, meaning that it can take on any value and still satisfy the system. This is known as a free variable and results in an infinite number of solutions.

I hope this helps and good luck with your future problem-solving endeavors!
 

1. What is Gaussian Elimination Algorithm?

Gaussian Elimination Algorithm is a method used to solve systems of linear equations by transforming a given matrix into a triangular form through a series of row operations, such as adding multiples of one row to another and multiplying rows by constants.

2. How is Gaussian Elimination Algorithm used to solve matrices?

To solve a matrix using Gaussian Elimination Algorithm, the matrix is first augmented with the right-hand side of the equations. Then, the matrix is transformed into a triangular form through row operations. Finally, back substitution is used to find the values of the variables and solve the system of equations.

3. What are the benefits of using Gaussian Elimination Algorithm?

Gaussian Elimination Algorithm is a powerful tool for solving systems of linear equations, as it provides an efficient and systematic approach to finding solutions. It also allows for easy computation and can be easily implemented in computer programs.

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

While Gaussian Elimination Algorithm is generally a reliable and effective method for solving matrices, it can encounter issues when dealing with matrices that are ill-conditioned or have a high degree of round-off error. In these cases, alternative methods may be more suitable for finding solutions.

5. Can Gaussian Elimination Algorithm be used for any type of matrix?

Gaussian Elimination Algorithm is most commonly used for square matrices, where the number of equations is equal to the number of variables. However, it can also be extended to rectangular matrices, as long as they are consistent and have a unique solution.

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