Undergrad Are three zeros always required in the third row for a matrix in echelon form?

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In echelon form, a matrix does not require three zeros in the third row; it only needs the first non-zero term of each row to be further right than that of the row above it, with any zero rows at the bottom. The discussion clarifies that while reduced echelon form may include additional zeros, achieving echelon form is primarily about the positioning of non-zero entries. A matrix with fewer than three zeros in the third row can still satisfy the echelon form criteria. The example provided illustrates that a matrix can be in echelon form without having a specific number of zeros. Understanding these criteria helps in correctly identifying and reducing matrices to their echelon forms.
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I need to find the echelon form of:

1 1 2 8
-1 -2 3 1
3 -7 4 10

and so far I have:

1 1 2 8
0 10 -50 -90
0 0 -52 -104

I was just wondering if I was required to put another zero in my third row. Am I always required to have three zeros in the third row? I'm assuming I do, but when I looked at the solution for this problem I found it to be:

1 1 2 8
0 1 -5 -9
0 0 1 2

However this solution does not have three zeros in the third row, when all of the other problems and solutions did have three zeros.

My apologies for the misaligned matrix, I am still getting used to Physics Forum.

Thank you.
 
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no you don't need 3 zeroes in the 3rd row. as long as the first non zero term of every row is further to the right than the first non zero term in the row above it, and all identically zero rows are at the bottom, my impression is that echelon form is achieved. i sort of like reduced echelon form though, which would achieve also (in your case) zeroes in the (1,2) position, and the (1,3) and (2,3) positions, where the (n,m) entry is the one in the nth row and mth column.notice that any matrix consisting of only one row is already in echelon form, no matter how many or how few zeroes there are, if that helps.

note also the matrix whose rows are ( 1 1 1 1 1 1), (0 1 1 1 1 1) , (0 0 1 1 1 1) is in echelon form, but not reduced.
 
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mathwonk said:
no you don't need 3 zeroes in the 3rd row. as long as the first non zero term of every row is further to the right than the first non zero term in the row above it, and all identically zero rows are at the bottom, my impression is that echelon form is achieved. i sort of like reduced echelon form though, which would achieve also (in your case) zeroes in the (1,2) position, and the (1,3) and (2,3) positions, where the (n,m) entry is the one in the nth row and mth column.notice that any matrix consisting of only one row is already in echelon form, no matter how many or how few zeroes there are, if that helps.

note also the matrix whose rows are ( 1 1 1 1 1 1), (0 1 1 1 1 1) , (0 0 1 1 1 1) is in echelon form, but not reduced.
Okay thank you very much, I understand it much better now and I can now see how I can reduce the solution I was given.
 
yes, all you can guarantee about a 3xn matrix with 3 rows in echelon form is that the 3rd row has at least 2 zeroes, since the entries below both previous pivots must be zero.
 
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I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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