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.
part i. I need to get rid of the imaginary
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.
I need to equate the vector solution sol to the values [y 3 2] for each row so that imaginary can go away.
part ii.
I may figure out part ii to solve a2 and a3 if part i is solvable.... can anyone provide hints?
Thanks!
Thanks.
