Coeff & constants of a symbolic equation

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

 Accepted Answer

Oleg Komarov
Oleg Komarov on 20 Aug 2012
Edited: Oleg Komarov on 20 Aug 2012
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 Comments

Thanks, it works for the symbols.
-1 -2 -3
0 4 -1
-5 0 0
How do I get the constant term at the end?
-1 -2 -3 2
0 4 -1 0
-5 0 0 3
Thanks in advance
why don't you change your expression
syms a b c d
exp = [-2*b-a-3*c+2*d; -c+4*b; -5*a+3*d]
then just ignore d
Thanks. This seems great if it was a straight fwd Qn. I dont have control of the immediate expression. My actual problem is way more complicated.
"exp" is the derivative of another expression "LD_new".
LD_new is arrived at by substituting expression "st" in "LD".
LD is arrived at by summing all the expressions of "LD0".
LD0 is my input vector multiplied by the variables.
Is there an easier way to do this? Thanks again
@Mech: I edited the code to generate the coeff matrix with the constant term as well.
Thanks a million. I've been working on this for weeks!
You weren't far from the solution.

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