What is A/B when A and B are 1x3 row vectors

 This is so simple...i think i'm missing some thing

What is the result of A/B ???? (if A and B are same length row vectors)

ex: A=[1 2 3]; B=[4 5 6];

A/B in MAtlab gives me 0.4156

Answer 1

[Matt J]

I want to know how this gives me 0.4156 ??? A/B is A*inv(B) ....

A/B is always defined as the least squares solution to the equation,

     A=X*B

or

     min. f(X) = norm(A-X*B)

In this case, the solution X has to be a scalar, since that is the matrix shape that can be left multiplied with B to produce another 1x3 matrix A. You can readily confirm that X=0.4156 is the minimizing value by plotting f(X),

    >> fplot(@(X)norm(A-X*B),[0.41,.42])

Answer 2

[Star Strider]

You’re likely not missing anything other than an slightly more inclusive interpretation of that operation. You’re doing a linear least-squares fit through the origin on ‘A’ as a function of ‘B’.

To illustrate symbolically:

A=[1 2 3]; B=[4 5 6];

P = A/B;                        % ‘P’ = Parameter ‘Vector’ (1x1 Here)
R = P*B;                        % Regression Line

figure(1)
plot(B, A, 'p')                 % Plot Data
hold on
plot(B, R, 'LineWidth',1.5)     % Plot Regression
plot([0 4], P*[0 4], '--r')     % Extend To Origin
hold off
axis([0  8    0  4])
grid

and to illustrate literally:

[Ashish Sheikh]

I want to know how this gives me 0.4156 ??? A/B is A*inv(B) ....

But in this case Inverse (B) doesn't exist as it is not a Square matrix

[Star Strider]

But the pesudo-inverse exists:

P = A*pinv(B)

giving:

P =

     415.5844e-003

[Ashish Sheikh]

Thanks @Star strider ....Now i got it ,...

[Star Strider]

My pleasure!

You can now Accept two answers, if you want to Accept mine as well.