Using a numeric solver or the symbolic toolbox is like using a sledge-hammer to crack open a tiny walnut!
It is very simple to rearrange those equations and get the actual algebraic solution, which works for any data set (as opposed to using a numeric solver or Ameer Hamza's method, which you will need to run for every set of data).
yf = M11*y0 + M12*a0
af = M21*y0 + M22*a0
We note that the first equation can be rearranged to give a0 entirely in terms of known values:
which we can then simply substitute into the second equation to get that last unknown af.
Lets try it with some numbers:
>> M = [1,2;3,4];
>> yf = (5e-3)/2;
>> y0 = (34e-3)/2;
>> a0 = (yf - M(1,1)*y0) / M(1,2)
a0 = -0.0072500
>> af = M(2,1)*y0 + M(2,2)*a0
af = 0.022000
So we get exactly the same output as Ameer Hamza's answer shows, just much more efficiently. And because this is the actual algebraic solution this works for any values that you have, there is no need to run inefficient solvers again and again and again and again...
"Computers are useless. They can only give you answers." Pablo Picasso