Gaussian Elimination, is this method OK?

In summary, the conversation discusses the calculation of the basis of the kernel for a given matrix and the difference in row-reducing methods used. The solution provided uses R2 = -2R1 + R2, while the person asking the question used Row2 = 2*Row1 - Row2. Both methods result in the same basis vector, with one being a multiple of the other.
  • #1
NewtonianAlch
453
0

Homework Statement



A=
<1, -3, -1, 1>
<2, -5, 0, 1>
<-3, 5, -6, 3>


What I did was Row2 = 2*Row1 - Row2 which renders Row2 as: <0, -1, -2, 1>

However in the solutions, Row2 was given as: <0, 1, 2, -1>, which appears to be R2 = -2R1 + R2

I'm guessing it makes no real difference, we were asked to calculate the basis of the kernel(A), to which I got <-8,-3,2,1>[itex]^{T}[/itex] and the solutions gave <8,3,-2,-1>[itex]^{T}[/itex], probably because of the difference in row-reducing the second row.

Am I incorrect?
 
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  • #2
NewtonianAlch said:

Homework Statement



A=
<1, -3, -1, 1>
<2, -5, 0, 1>
<-3, 5, -6, 3>


What I did was Row2 = 2*Row1 - Row2 which renders Row2 as: <0, -1, -2, 1>

However in the solutions, Row2 was given as: <0, 1, 2, -1>, which appears to be R2 = -2R1 + R2

I'm guessing it makes no real difference, we were asked to calculate the basis of the kernel(A), to which I got <-8,-3,2,1>[itex]^{T}[/itex] and the solutions gave <8,3,-2,-1>[itex]^{T}[/itex], probably because of the difference in row-reducing the second row.

Am I incorrect?

No, your work is fine.
I prefer the method used in in the solutions for this problem, though, with R2 being replaced by -2R1 + R2.

The basis vector you got is a multiple of the one shown in the solutions. Both vectors span exactly the same space (a one-dimension subspace of R4).
 
  • #3
Ah yes, that makes sense, it can be gotten by just multiplying by -1. Thanks for pointing that outt.
 

What is Gaussian Elimination?

Gaussian Elimination is a method used to solve systems of linear equations by systematically eliminating unknown variables. It involves transforming the system into an equivalent system with a triangular matrix, making it easier to solve for the unknown variables.

How does Gaussian Elimination work?

Gaussian Elimination involves three main steps: elimination, back substitution, and solution. First, the system is augmented with a column vector and transformed into an upper triangular matrix through elimination. Then, back substitution is used to solve for the unknown variables. Finally, the solution is obtained by substituting the values back into the original equations.

When should I use Gaussian Elimination?

Gaussian Elimination is a useful method for solving systems of linear equations where the number of equations is equal to or greater than the number of variables. It is also commonly used when solving equations with large matrices, as it can be easily implemented using computer programs.

What are the advantages of Gaussian Elimination?

Gaussian Elimination is a reliable and efficient method for solving systems of linear equations. It allows for easy manipulation of the equations, making it useful for solving complex systems with many variables. Additionally, it can easily be implemented using computer programs, making it a convenient choice for solving large equations.

Are there any limitations to Gaussian Elimination?

Although Gaussian Elimination is a powerful method, it does have some limitations. It can only be used for systems of linear equations, and it may become computationally intensive for very large matrices. Additionally, it does not work for systems with no solutions or infinite solutions.

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