mldivide returns matrix with multiple rows = 0 when they should not be

I'm trying to solve the equation Ax=B where A is a 1-by-m matrix and B is a 1-by-n matrix. By the rules of linear algebra I know that this means that x should be a m-by-n matrix. When I use the \ operator, I get a correctly-sized matrix, except that only one row contains non-zero numbers. If I instead write a for loop to \ each element of A individually into the entire B matrix, I get the correct answer. For example:

EDU>> A=1:4;
EDU>> B=1:6;
EDU>> x=A\B

x =

           0         0         0         0         0         0
           0         0         0         0         0         0
           0         0         0         0         0         0
      0.2500    0.5000    0.7500    1.0000    1.2500    1.5000

EDU>> for m = length(A):-1:1
x(m,:) = A(m)\B;
end
EDU>> x

x =

      1.0000    2.0000    3.0000    4.0000    5.0000    6.0000
      0.5000    1.0000    1.5000    2.0000    2.5000    3.0000
      0.3333    0.6667    1.0000    1.3333    1.6667    2.0000
      0.2500    0.5000    0.7500    1.0000    1.2500    1.5000

I have tried this several times, with different sized A and B matrices. In some instances the first row of x is correct, and the remaining rows are zero, while in others (as above) the last row of x is correct, and the remaining rows are zero. Can anyone explain why this is happening?

Answer 1

[Walter Roberson]

bxsfun(@rdivide, B.', A)

or

bxsfun(@ldivide, A.', B)

Please have another look at the documentation for what you used, mldivide

x = A\B solves the system of linear equations A*x = B.

That is quite different than the output that you want for output.

[Lauren]

any suggestions for how to achieve the output I'm trying to get (without the for loop)?

Answer 2

[Roger Stafford]

Lauren, your equations are highly underdetermined. Given the sizes of A and B, you necessarily have six equations and twenty-four unknowns. In each equation there are four unknowns which don't appear anywhere else. Consequently in each equation matlab is free to arbitrarily set three of them to zero and solve for the fourth one. It could have set three of them to any other arbitrary values and solved for the fourth. Why do you find this surprising? What else would you expect for a solution? If you want unique solutions you need in general to have an equal number of equations and unknowns.