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Let's say I have a linear system of equations $$ Ax = \biggl( \begin{matrix} 1 & 2\\ 0 &1\\ \end{matrix} \biggr) \biggl(\begin{matrix} x_1 \\ x_2 \end{matrix} \biggr) = \biggl( \begin{matrix} 11 \\ 4 \end{matrix} \biggr).$$ The answer is $$ x = \biggl( \begin{matrix} 3 \\ 4 \end{matrix} \biggr). $$ But what if I couldn't solve for x directly because A is large and badly conditioned, but I am 95% sure that $$ 1 \leq x_1 \leq 10 \\ 1 \leq x_2 \leq 7, $$ and both $x_1, x_2$ are integers.

How can I use Mathematica to solve or guess for x, given A, b, and $$\\$$ a) an arbitrary set of constraints for an range for each $x_i,$ and/or $$\\$$b) an arbitrary integer range for each $x_i,$ given that all $x_i$ are integers? $$\\$$ Thank you!

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The template for solving A.x = b with constraints is:

Solve[A.x == b && constraints..., variables, domain]

For your example, Solve finds solutions over the domain of integers, with constraints for x1 and x2, where && is logical AND.

A = {{1, 2}, {0, 1}};
x = {{x1}, {x2}};
b = {{11}, {4}};

Solve[A.x == b && 1 <= x1 <= 10 && 1 <= x2 <= 7, {x1, x2}, Integers]

{{x1 -> 3, x2 -> 4}}
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  • $\begingroup$ x and b can be defined more simply as x = {x1, x2}; b = {11, 4}; You will get the same result. $\endgroup$
    – Bob Hanlon
    Commented Oct 29, 2017 at 4:50
  • $\begingroup$ It's a linear problem though, so the constraints are superfluous. $\endgroup$
    – Daniel Lichtblau
    Commented Oct 29, 2017 at 15:01
  • $\begingroup$ @DanielLichtblau Yes, but the issue is how to apply constraints and integer domain to a large and badly conditioned A. The simple example is not the reason for the question. Have I missed something? $\endgroup$
    – creidhne
    Commented Oct 29, 2017 at 21:16
  • $\begingroup$ This solves my problem! Thanks so much! $\endgroup$
    – BJParks
    Commented Oct 30, 2017 at 11:13
  • $\begingroup$ (1) If the system is linear then most (maybe all) methods will be unable to use constraints to the solutions. Possible exceptions are indirect methods e.g. Krylov where constraints might be used for initial guesses. $\endgroup$
    – Daniel Lichtblau
    Commented Oct 30, 2017 at 15:24

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