Can Matlab row reduce this matrix?

Can we row reduce the following matrix?

[Birdman]

What do you mean?

[John D'Errico]

Sure. What did you try? What makes this not a doit4me, apparently on your homework?

[Harel Harel Shattenstein]

It is not homework, I just do not know how to use x,y,z,w in matlab

[Alexander Guillen]

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.

Answer 1

[John D'Errico]

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.

Answer 2

[Torsten]

https://de.mathworks.com/help/symbolic/rref.html

Best wishes

Torsten.

[Harel Harel Shattenstein]

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]

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.