Coeff & constants of a symbolic equation

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Mech Princess
Mech Princess 2012 年 8 月 20 日
syms a b c
exp=[-a-2*b-3*c+2, 4*b-c, -5*a+3]
I need the coefficient matrix in the form
% Coef_matrix=[ -1, -2, -3, 2; 0, 4, -1, 0; -5, 0, 0, 3];
How do I get this?
I can get the first line from the following syntax (from a previous blog), but there is an error when I try to do the second & third lines (i.e. j=1:3) as there is a 'zero' which it wont recognize
xx = symvar(exp) %find the variables in expression 'exp'
Coef_matrix = zeros(numel(exp),numel(xx));
for j = 1
[v1,v2] = coeffs(exp(j),xx);
Coef_matrix(j,ismember(xx,v2)) = v1(1:numel(xx));
end
And when I get the coefficients, I cant get the constant term at the end.
The solution should be able to scale up for more variables, as this is a simplified version of my problem. Thanks

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Oleg Komarov
Oleg Komarov 2012 年 8 月 20 日
編集済み: Oleg Komarov 2012 年 8 月 20 日
syms a b c
exp = [-2*b-a-3*c+2; -c+4*b; -5*a+3];
xx = symvar(exp);
Coef_matrix = zeros(numel(exp),numel(xx)+1);
for j = 1:3
[v1,v2] = coeffs(exp(j),xx);
[idx,loc] = ismember(v2,xx);
idx(end) = true;
loc(loc == 0) = 4;
Coef_matrix(j,loc) = v1(idx);
end
  6 件のコメント
Mech Princess
Mech Princess 2012 年 8 月 20 日
Thanks a million. I've been working on this for weeks!
Oleg Komarov
Oleg Komarov 2012 年 8 月 20 日
You weren't far from the solution.

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