I am new to Mathematica and I apologize if this is a very basic question.
I am attempting to use the solve function to derive multiple solutions for a variable, however it seems as though I can only have the same number of equations as I do unknowns, which then leads to only one solution per unknown.
E.g.:
Solve[{m == ab, n == ac, o == b*c}, {a, b, c}]
What I am trying to solve for is this:
Solve[{m == ab, n == ac, o == ad, p == bc, q == bd, r == cd}, {a, b, c, d}]
This indicates that each unknown (a,b,c,d) will have more than one unique solution. However, this statement leads to an empty solution.
I can try to break up the over-identified (determined?) statement and eventually produce all unique solutions, however this could get complicated and tiresome with a larger model. I was wondering, is there a way to get Mathematica to produce all unique solutions in a situation like this?
(2) The second system is overdetermined. Might instead use Reduce for it.
Reduce[{m == a*b, n == a*c, o == a*d, p == b*c, q == b*d, r == c*d}, {a, b, c, d}]
(* (o == 0 && n == 0 && m == 0 && a == 0 &&
r != 0 && (b == -((Sqrt[p] Sqrt[q])/Sqrt[r]) ||
b == (Sqrt[p] Sqrt[q])/Sqrt[r]) && b != 0 && c == p/b &&
d == q/b) || (q == 0 && p == 0 && m == 0 &&
r != 0 && (a == -((Sqrt[n] Sqrt[o])/Sqrt[r]) ||
a == (Sqrt[n] Sqrt[o])/Sqrt[r]) && b == 0 && a != 0 && c == n/a &&
d == o/a) || (r == 0 && p == 0 && n == 0 &&
q != 0 && (a == -((Sqrt[m] Sqrt[o])/Sqrt[q]) ||
a == (Sqrt[m] Sqrt[o])/Sqrt[q]) && a != 0 && b == m/a && c == 0 &&
d == o/a) || (r == 0 && q == 0 && o == 0 &&
p != 0 && (a == -((Sqrt[m] Sqrt[n])/Sqrt[p]) ||
a == (Sqrt[m] Sqrt[n])/Sqrt[p]) && a != 0 && b == m/a &&
c == n/a && d == 0) || (r == 0 && q == 0 && p == 0 && o == 0 &&
n == 0 && m == 0 && a == 0 && b == 0 && c == 0) || (r == 0 &&
q == 0 && p == 0 && o == 0 && n == 0 && a != 0 && b == m/a &&
c == 0 && d == 0) || (q == 0 && p == 0 && o == 0 && n == 0 &&
m == 0 && a == 0 && b == 0 && c != 0 && d == r/c &&
r != 0) || (r == 0 && p == 0 && o == 0 && n == 0 && m == 0 &&
a == 0 && c == 0 && b != 0 && d == q/b && q != 0) || (r == 0 &&
q == 0 && o == 0 && n == 0 && m == 0 && a == 0 && b != 0 &&
c == p/b && d == 0 && p != 0) || (r == 0 && q == 0 && p == 0 &&
n == 0 && m == 0 && b == 0 && c == 0 && a != 0 && d == o/a &&
o != 0) || (r == 0 && q == 0 && p == 0 && o == 0 && m == 0 &&
b == 0 && a != 0 && c == n/a && d == 0 && n != 0) || (r == 0 &&
q == 0 && p == 0 && o == 0 && n == 0 && m == 0 && a == 0 &&
b == 0 && c != 0 && d == 0) || (r == 0 && q == 0 && p == 0 &&
o == 0 && n == 0 && m == 0 && a == 0 && b != 0 && c == 0 &&
d == 0) || (q != 0 && n == (o p)/q && r != 0 && m == (o p)/r &&
p != 0 && (a == -((Sqrt[m] Sqrt[n])/Sqrt[p]) ||
a == (Sqrt[m] Sqrt[n])/Sqrt[p]) && a != 0 && b == m/a &&
c == n/a && d == o/a) *)
Thank you, Daniel.
I think this will work for my purposes! Looks like I can just replace the first unknown in the second phrase to get the multiple solutions for it.