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Can Matlab row reduce this matrix?

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Harel Harel Shattenstein
评论： Alexander Guillen 2018-3-2
Can we row reduce the following matrix?
  4 个评论
显示 2更早的评论
Harel Harel Shattenstein
Harel Harel Shattenstein 2018-1-9
It is not homework, I just do not know how to use x,y,z,w in matlab
Alexander Guillen
Alexander Guillen 2018-3-2
Wow, people here so f rude.English is not my first language and I can understand what this person is looking for. I am not sure if this a professional website you can ask questions. It looks like there are just over entitled, self absorbed computer geniuses. As.holes.

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回答（2 个）

John D'Errico
John D'Errico 2018-1-11
I'm not positive what you are looking for.
Make x,y,z,w symbolic variables.
syms x y z w
Now, construct your array, and do row operations on that array.
What you are not saying is what is the purpose of this task? Thus, given the array:
A = [2 -3 3 2;-2 3 -1 -1;0 3 1 2; 2 3 3 5]
A =
2 -3 3 2
-2 3 -1 -1
0 3 1 2
2 3 3 5
Then this does row reduce the combined matrix.
rref([A,[x;y;z;w]])
ans =
[ 1, 0, 0, 1, 0]
[ 0, 1, 0, 1/2, 0]
[ 0, 0, 1, 1/2, 0]
[ 0, 0, 0, 0, 1]
A clearly has rank 3. We can see that from the above computation, that there is some vector that does not lie in the row space of A. Of course, rref does not give you the linear combination of the rows that yielded that 1 in the bottom right corner. I think perhaps that is what you are looking to find. You wanted that last column to remain expressed as symbolic variables.
So, can we recover the row operations that were applied to A? A and rref(A) occupy the same row space. And while rref =does not return the actual row operations, they can be recovered.
format rat
K = rref(A)*pinv(A)
K =
-1/2 -2/3 -1/6 1/3
-1/6 -1/36 1/18 5/36
1/2 5/12 1/6 -1/12
0 0 0 0
K*A
ans =
1 * * 1
* 1 * 1/2
* * 1 1/2
0 0 0 0
The * elements are all essentially zeros, just tiny trash. As you can see, the result is essentially the same row operations that were applied to A to get rref(A).
We can now apply those same row operations to [x;y;z;w].
K*[x;y;z;w]
ans =
w/3 - x/2 - (2*y)/3 - z/6
(5*w)/36 - x/6 - y/36 + z/18
x/2 - w/12 + (5*y)/12 + z/6
0
But remember that K has rank 3, reflecting the rank of A. We can find a row that is not representable as a linear combination of the rows of A.
null(A,'r')'
ans =
-1 -1/2 -1/2 1
null(A,'r')'*[x;y;z;w]
ans =
w - x - y/2 - z/2
And I still have no idea exactly what you want.
Torsten
Torsten 2018-1-9
https://de.mathworks.com/help/symbolic/rref.html
Best wishes
Torsten.
  2 个评论
Harel Harel Shattenstein
Harel Harel Shattenstein 2018-1-9
Entered
rref([2 -3 3 2 x;-2 3 -1 -1 y;0 3 1 2 z; 2 3 3 5 w])
% code
end
and got
[ 1, 0, 0, 1, 0]
[ 0, 1, 0, 1/2, 0]
[ 0, 0, 1, 1/2, 0]
[ 0, 0, 0, 0, 1]
% code
end
Where are the variables?
Torsten
Torsten 2018-1-10
编辑：Torsten 2018-1-10
https://stackoverflow.com/questions/14449255/reduced-row-echelon-of-a-matrix-containing-unknown-constants-in-matlab
Additionally, you may want to check this link:
http://www.wolframalpha.com/input/?i=RowReduce%5B%7B%7B2,-3,3,2,x%7D,%7B-2,3,-1,-1,y%7D,%7B0,3,1,2,z%7D,%7B2,3,3,5,w%7D%7D%5D
Best wishes
Torsten.

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