solving implicit equations and saving positive roots in "double" array

Hi - I need to find the solutions of several implicit equations, then save the positive roots in a matrix (possibly a "double" array).

When I try to do so I get an error, "Assignment has more non-singleton rhs dimensions than non-singleton subscripts".

I'd greatly appreciate some help. Here's a simplified version of my code.

Notes:

• I understand that using "solve" like I'm doing here below will return 2 solutions for each equation. I tried using vpasolve instead, specifying an interval, but I still get the same error message;
• I need to do this for different equations and some functional forms make it impossible to find the solutions explicitly, hence my need for a general procedure

clear all; close all; clc;

GAMMA =1.1;
ALOW=0.6;

data=[0 0 7.34 9.39 0 9.12 7.95 8.06 8.47 9.16;
8.26 0 0 0 7.5 0 7.82 7.24 9.54 9.13;
0 8.31 8.99 9.46 0 0 0 8.16 7.85 7.23;
7.96 0 7.53 8.47 0 8.62 0 0 7.31 7.13;
8.83 8.72 8.65 9.42 0 8.87 7.86 0 8.34 8.89;
0 7.79 0 8.27 9.45 7.15 0 8.19 0 0;
0 0 9.34 7.08 8.72 0 0 0 0 0;
7.59 0 7.53 7.48 7.78 0 8.86 0 7.74 8.95;
7.37 0 8.07 0 7.64 0 8.9 9.27 9.15 7.89;
0 7.09 8.34 8.55 9.33 8.89 0 0 6.96 8.95]; 

output=zeros(10,10);
for j = 1 : 10
    for t=1:10
        if data(t,j) >0
            syms x
            output(t,j)=solve(ALOW*(1/(x^2+1)^(3/2))*GAMMA*data(t, j)-1, x);
        else
            output(t,j)=0;
        end
    end
end

Answer 1

[Walter Roberson]

Use

syms x positive

The difficulty you are having is that solve() is finding two solutions, and two solutions do not fit into one slot. You describe wanting to find positive solutions but you are not assigning to a temporary variable and selecting the positive solutions out of the list returned. There is only at most one positive solution for positive data, so adding the constraint when declaring x makes it select that one solution if it exists.

Note: there are no positive solutions when data(i,j) <= 50/33 . That never happens with that particular data matrix you have, but it is a possibility.

[Luigi Pisano]

Thanks. This works perfectly in the "simplified" code I posted; for the "real" one I had to do some extra work but I was able to find a solution, building on what you suggested