Solving for the coefficent of linear equations with one known coefficent
- by CppLearner
clc;
clear all;
syms y a2 a3
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% [ 0.5 0.25 0.125 ] [ a2 ] [ y ]
% [ 1 1 1 ] [ a3 ] = [ 3 ]
% [ 2 4 8 ] [ 6 ] [ 2 ]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
M = [0.5 0.25 0.125; 1 1 1; 2 4 8];
t = [a2 a3 6];
r = [y 3 2];
sol = M * t'
s1 = solve(sol(1), a2) % solve for a2
s2 = solve(sol(2), a3) % solve for a3
This is what I have so far.
These are my output
sol =
conj(a2)/2 + conj(a3)/4 + 3/4
conj(a2) + conj(a3) + 6
2*conj(a2) + 4*conj(a3) + 48
s1 =
- conj(a3)/2 - 3/2 - Im(a3)*i
s2 =
- conj(a2) - 6 - 2*Im(a2)*i
sol looks like what we would have if we put them back into equation form:
0.5 * a2 + 0.25 * a3 + 0.125 * a4
a2 + a3 + a4 = 3
2*a2 + 4*a3 + 8*a4 = 2
where a4 is known == 6.
My problem is, I am stuck with how to use solve to actually solve these equations to get the values of a2 and a3.
s2 solve for a3 but it doesn't match what we have on paper (not quite).
a2 + a3 + 6 = 3 should yield a3 = -3 - a2.
because of the imaginary. Somehow I need to equate the vector solution sol to the values [y 3 2] for each row.
