Question

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.

Answer 1

You need to provide solve function with an equation. Just as simple as that:

 sol = solve(M * t' == r');

As a result you have

sol.a2 = 17
sol.a3 = -20
sol.y = 17/4

This works for MATLAB R2012b. In general this can be solved just in one line:

solve('a2 / 2 + a3 / 4 - y + 3 / 4 = 0', 'a2 + a3 + 3 = 0', '2 * a2 + 4 * a3 + 46 = 0')

P.S. I have checked, this works for MATLAB R2011b.