Gaussian Elimination, is this method OK?

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SUMMARY

The forum discussion centers on the Gaussian elimination method applied to the matrix A, where the user performed a row operation resulting in Row2 = <0, -1, -2, 1>. The correct solution provided in the discussion is Row2 = <0, 1, 2, -1>, achieved through the operation R2 = -2R1 + R2. Both approaches yield basis vectors for the kernel of matrix A, specifically <-8, -3, 2, 1>^{T} and <8, 3, -2, -1>^{T}, which are scalar multiples of each other, indicating they span the same one-dimensional subspace of R4.

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NewtonianAlch
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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>^{T} and the solutions gave <8,3,-2,-1>^{T}, probably because of the difference in row-reducing the second row.

Am I incorrect?
 
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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>^{T} and the solutions gave <8,3,-2,-1>^{T}, 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).
 
Ah yes, that makes sense, it can be gotten by just multiplying by -1. Thanks for pointing that outt.
 

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