Linear Algebra 10c: A Row Reduced Echelon Form (RREF) Challenge

[skrisna]

This problem is amazing. I solved it only because I was able to understand the concepts you enunciated in the previous lectures. I thought I knew LA before but now I think I am beginning to really "get" it. Thank you!

For hints, let me say that it is possible to do everything by hand to find the RREF of the two column Null set shown in the video. Also, it involves computing inverse of a 2x2 matrix.

[gijsschroder]

This course is amazing. I found a solution to the challenge, while yesterday I did not even know what RREF is.

Thanks a lot for these lectures.

[dannuttle]

This challenge was, well, challenging. But satisfying once I finished it. I found that it helped to use Sympy to do the grunt work, like finding rref. Too tedious and too prone to human error. (I work in Jupyter so I get nice formatted output.) Also using Sympy let me just look at things and ponder, what exactly am I looking for here--what's the concept behind the problem. Sympy is fairly easy to learn, though if you're brand-new to it, you might want to set aside a little time just to learn how to do various things like get a transpose, or rref, or nullspace. Some may say this is cheating, because I'm skipping a whole bunch of manual steps. But I am honest--if I have to go through a large number of calculations by hand, especially when they involve fractions, and changing signs, I will make a small mistake somewhere, and then my final answer is a million miles off.

One more thing, there are two additional problems paired with this video at lem.ma. Do them first! The first one there is quite a bit more straightforward and simple than the one in the video, and it is hard enough.

[benmurray]

Ok; I've spent a while on this problem but I'm basically stuck solid. I understand that the row count of the null-space elements determines the number of columns in the (RREF) matrix. What I don't get is how you can know the number of rows based on the null vectors. Specifically, for the first example, I can't see how you get four rows without prior knowledge that there were four rows, as it seems that one could arbitrarily add all zero rows to the bottom of the matrix without that changing the null space vectors. I'd be grateful for any clues.

[robensonlarokulu]

Solution: row space is unchanged by gaussian elimination. So apply GE to the transpose of the two columns that form the null space. This is a matrix of 2 x 6. Get the reduced row echelon form of this 2 x 6 matrix. Since row space is unchanged we now have the exact same null space but in terms of zeros and ones, which is easy to interpret and get first original reduced row echelon form that we after for.

[ekaptsv]

Thank you for the great challenge, Pavel! It looks like solution doesn't exist, does it?
I've tried to solve a problem by multiplying N by all possible column-permutations of 4x6 matrix with 4 basis cols and 2 arbitrary cols. If a solution does exist then there should be a nontrivial relation between alpha, beta and arbitrary elements of one or more such permuted matrices. But there is not.

[ramleo]

where can I get solution and explanation for this challenge?