Solving Unknowns with Symbolic Toolbox

I have a relatively simple problem to solve with the symbolic toolbox which works fine with X = lsqminnorm(a,b), but not this toolbox. The knowns are as follows (A=1,B=2,C=3,D=4), which I am trying to solve with the relationships below:

syms A B C D
eqn_1 = 3 == A + B;
eqn_2 = 7 == C + D;
eqn_3 = 4 == A + C;
eqn_4 = 6 == B + D;

eqns = [eqn_1
        eqn_2
        eqn_3
        eqn_4];

[a,b] = equationsToMatrix(eqns, [A B C D]);   
X = linsolve(a,b); 

I get the following error:

Warning: Solution is not unique because the system is rank-deficient. 
> In symengine
In sym/privBinaryOp (line 1214)
In sym/linsolve (line 63) 
 
X =
 
-3
 6
 7
 0

Works fine though if I use X = lsqminnorm(a,b)

 >> X = lsqminnorm(double(a),double(b))

X =

    1.0000
    2.0000
    3.0000
    4.0000

The issue is that I can't use lsqminnorm with double() as the other similar equations organized in this format contain other symbolic variables/unknowns. Any help would be greatly appreciated. I apologize in advance if this is a silly question.

Answer 1

[KSSV]

A = [1 1 0 0 ;
     0 0 1 1 ;
     1 0 1 0 ;
     0 1 0 1] ;
 b = [3;7;4;6] ;
 x = pinv(A)*b 

x = 4×1

1.0000 2.0000 3.0000 4.0000

Note that rank of A is 3, so with A\b you will not get solution.

With symbolic tool box:

syms A B C D
eqn_1 = 3 == A + B;
eqn_2 = 7 == C + D;
eqn_3 = 4 == A + C;
eqn_4 = 6 == B + D;

eqns = [eqn_1
        eqn_2
        eqn_3
        eqn_4];

[a,b] = equationsToMatrix(eqns, [A B C D]);   
X = pinv(a)*b

X = 

[J.D. Johnston]

Thank you so much. I never would have considered pinv.

Answer 2

[Walter Roberson]

syms A B C D real
eqn_1 = 3 == A + B;
eqn_2 = 7 == C + D;
eqn_3 = 4 == A + C;
eqn_4 = 6 == B + D;

eqns = [eqn_1
        eqn_2
        eqn_3
        eqn_4];

sol = solve(eqns, [A, B, C, D], 'returnconditions', true)

sol = struct with fields:
             A: x - 3
             B: 6 - x
             C: 7 - x
             D: x
    parameters: x
    conditions: in(x, 'real')

sol.A

ans = 

sol.B

ans = 

sol.C

ans = 

sol.D

ans = 

x

sol.parameters

ans = 

x

Your system is not full rank. It can be rewritten in terms of a parameter, here written as x which is equivalent to D. So every different D value has a different solution.

lsqminnorm() "minimizes the value of norm(A*X-B). If several solutions exist to this problem, then lsqminnorm returns the solution that minimizes norm(X)" and those are things you can potentially compute yourself

[a, b] = equationsToMatrix(eqns)

a = 

b = 

syms X [4 1]
n = norm(a*X - b)

n = 

We can minimize the norm by minimizing the individual parts... but that is the same system of equations as we solved earlier, so we get

Xbest = [sol.A; sol.B; sol.C; sol.D]

Xbest = 

and we can proceed to minimize it in accordance with the second clause about which solution is chosen. For ease of use we will restrict to real values

nXbest = norm(Xbest)

nXbest = 

syms FREE real
dnXbest = simplify(subs(diff(nXbest, sol.parameters), sol.parameters, FREE))

dnXbest = 

best_free_parameter_value = solve(dnXbest, FREE)

best_free_parameter_value = 

4

subs(Xbest, sol.parameters, best_free_parameter_value)

ans =