division of two vectors

67 views (last 30 days)
mimalo salina
mimalo salina on 9 Oct 2017
Answered: mimalo salina on 9 Oct 2017
What is the result of A/B ???? (if A and B are same 3x1 row vectors) Example:
A=[-1;1;7]
B=[5;7;-2]
A/B
in Matlab gives: ans =
0 -0.1429 0
0 0.1429 0
0 1.0000 0
  5 Comments
mimalo salina
mimalo salina on 9 Oct 2017
Edited: mimalo salina on 9 Oct 2017
yes, but A ,B are same 3x1 row vectors, it seems false, because , we can't divide two vectors? i want to undersand the meaning of this result !
Guillaume
Guillaume on 9 Oct 2017
Edited: Guillaume on 9 Oct 2017
"we can't divide two vectors". Well, obviously you can apply the / operators to two vectors since matlab give you a result.
As Adam said in his comment and as Jan showed in its answer, what it does is fully documented. It solves a system of linear equation. If the right-hand side is not a square matrix as is the case here, it is solved using least-square method. As documented:
If A is a rectangular m-by-n matrix with m ~= n, and B is a matrix with n columns, then x = B/A returns a least-squares solution of the system of equations x*A = B
I'm not sure what other answer you are looking for.

Sign in to comment.

Accepted Answer

Jan
Jan on 9 Oct 2017
C = A / B determines the matrix C such that:
C * B = A
Try it:
A = [-1;1;7];
B = [5;7;-2];
C = A / B;
C * B
>> [-1; 1; 7]

More Answers (2)

John D'Errico
John D'Errico on 9 Oct 2017
A=[-1;1;7];
B=[5;7;-2];
First of all, A and B are NOT row vectors. They are column vectors. Thus, A is a vector in column orientation.
A
A =
-1
1
7
Next, what is C=A/B?
C = A/B
C =
0 -0.142857142857143 0
0 0.142857142857143 0
0 1 0
It is a matrix, such that if possible, we will have A=C*B.
C*B
ans =
-1
1
7
In some cases of vectors or matrices of varying sizes, that will not be possible.

mimalo salina
mimalo salina on 9 Oct 2017
Action closed.
Thanks a lot for all your answers!
Kind regards.

Categories

Find more on Creating and Concatenating Matrices in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!