Recovering row reduced matrix given known column information
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I'm not the best at MATLAB but to try to learn further, I tried to see if I could recover the matrix A given the RREF form U and information regarding columns a1 and a2. I successfully formed matrix A including columns a1 and a2 while leaving columns a3 and a4 as unknown variables, but using and equivalent expression to equate U and rref(A) yields an incorrect answer. What am I doing incorrectly here?
Interestingly, when isolating rref(A) I noticed that entry a(3,3) is 1. I find this intriguing since I'm not sure how MATLAB can assemble column a3 at all. An explanation as to why MATLAB (seemingly) assumes column a3 would be super helpful.
Code is pasted below...
%%currently doesn't work...
syms A
A = sym('A%d%d', [3 4])
A(:,1) = [2,-1,1]; A(:,2) = [-2,-3,-1]
U = sym([1, 0, 4, 1; 0, 1, 1, 2; 0, 0, 0, 0])
rref(A) == U
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MULI
el 28 de Mzo. de 2024
Editada: MULI
el 28 de Mzo. de 2024
Hi Zad,
I understand that you are trying to reconstruct the original matrix from its reduced row echelon form.
When you are directly equating “rref(A)” to U it does not automatically solve the unknowns in A. The “rref” function simplifies A to facilitate the solving of linear systems. However it does not work the other way around i.e., it cannot be used to reconstruct unknown elements in A from U.
The symbolic equations from rref(A) == U set conditions for the unknowns but solving these may be complex and does not always lead to a unique solution.
The reason behind the entry “a(3,3)” being 1 in rref(A) is that the row operations in “rref” rules naturally led to this result but not because of MATLAB assumption.
Please refer to the following documentation for more information on “rref” function and examples related to it.
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