This is easy to do in several ways. I'll start with a rather arbitrary array of data as an example.
A = rand(3,4)
b = rand(3,1)
A =
0.14189 0.79221 0.035712 0.67874
0.42176 0.95949 0.84913 0.75774
0.91574 0.65574 0.93399 0.74313
b =
0.39223
0.65548
0.17119
Now, we can see that backslash solves this linear system, picking one of the unknowns to set to zero.
A\b
ans =
-0.68744
0.59853
0.43708
0
Or, I could use pinv to solve it for a different style of solution, but arguably equally as good.
pinv(A)*b
ans =
-0.69486
0.58673
0.44058
0.015139
A simple way to solve it where any given element will be selected to set at a given value is to use LSE, a code I have placed on the file exchange.
So I'll choose variable 4 to set to zero.
zind = zeros(1,size(A,2));
zind(4) = 1;
lse(A,b,zind,0)
ans =
-0.68744
0.59853
0.43708
0
The nice thing is, I could have picked a different variable, setting it to any given number. So, if I wanted to set the first unknown to pi, that takes nothing more than a simple change:
zind = zeros(1,size(A,2));
zind(1) = 1;
lse(A,b,zind,pi)
ans =
3.1416
6.6954
-1.3746
-7.8213
In fact, this is entirely doable without needing LSE for these simple cases, by dropping out the appropriate column of A and then solving. (I can add that solution if you wish to see it.) Note that in any case, if A is rank deficient (singular) once you remove a column, then there will be a problem.
Finally, you could also have used lsqlin from the optimization toolbox.
