How Can I Solve System Of Linear Equations In Sympy?

Sorry, I am pretty new to sympy and python in general. I want to solve the following underdetermined linear system of equations: x + y + z = 1 x + y + 2z = 3

Solution 1:

SymPy recently got a new Linear system solver: linsolve in sympy.solvers.solveset, you can use that as follows:

In [38]: from sympy import *

In [39]: from sympy.solvers.solvesetimport linsolve

In [40]: x, y, z = symbols('x, y, z')

List of Equations Form:

In [41]: linsolve([x + y + z - 1, x + y + 2*z - 3 ], (x, y, z))
Out[41]: {(-y - 1, y, 2)}

Augmented Matrix Form:

In [59]: linsolve(Matrix(([1, 1, 1, 1], [1, 1, 2, 3])), (x, y, z))
Out[59]: {(-y - 1, y, 2)}

A*x = b Form

In [59]: M = Matrix(((1, 1, 1, 1), (1, 1, 2, 3)))

In [60]: system = A, b = M[:, :-1], M[:, -1]

In [61]: linsolve(system, x, y, z)
Out[61]: {(-y - 1, y, 2)}

Note: Order of solution corresponds the order of given symbols.

Solution 2:

In addition to the great answers given by @AMiT Kumar and @Scott, SymPy 1.0 has added even further functionalities. For the underdetermined linear system of equations, I tried below and get it to work without going deeper into sympy.solvers.solveset. That being said, do go there if curiosity leads you.

from sympy import *
x, y, z = symbols('x, y, z')
eq1 = x + y + zeq2= x + y + 2*z
solve([eq1-1, eq2-3], (x, y,z))

That gives me {z: 2, x: -y - 1}. Again, great package, SymPy developers!

Solution 3:

import sympy as sp
x, y, z = sp.symbols('x, y, z')
eq1 = sp.Eq(x + y + z, 1)             # x + y + z  = 1
eq2 = sp.Eq(x + y + 2 * z, 3)         # x + y + 2z = 3
ans = sp.solve((eq1, eq2), (x, y, z))

this is similar to @PaulDong answer with some minor changes

1. its a good practice getting used to not using import * (numpy has many similar functions)
2. defining equations with sp.Eq() results in cleaner code later on

Solution 4:

Another example on matrix linear system equations, lets assume we are solving for this system:

In SymPy we could do something like:

>>>import sympy as sy...sy.init_printing()>>>a, b, c, d = sy.symbols('a b c d')...A = sy.Matrix([[a-b, b+c],[3*d + c, 2*a - 4*d]])...A

⎡ a - b     b + c  ⎤
⎢                  ⎥
⎣c + 3⋅d  2⋅a - 4⋅d⎦


>>>B = sy.Matrix([[8, 1],[7, 6]])...B

⎡8  1⎤
⎢    ⎥
⎣7  6⎦


>>>A - B

⎡ a - b - 8     b + c - 1  ⎤
⎢                          ⎥
⎣c + 3⋅d - 7  2⋅a - 4⋅d - 6⎦


>>>sy.solve(A - B, (a, b, c, d))
{a: 5, b: -3, c: 4, d: 1}

Solution 5:

You can solve in matrix form Ax=b (in this case an underdetermined system but we can use solve_linear_system):

from sympy importMatrix, solve_linear_system

x, y, z = symbols('x, y, z')
A = Matrix(( (1, 1, 1, 1), (1, 1, 2, 3) ))
solve_linear_system(A, x, y, z)

{x: -y - 1, z: 2}

Or rewrite as (my editing, not sympy):

[x]=  [-1]   [-1]
[y]= y[1]  + [0]
[z]=  [0]    [2]

In the case of a square A we could define b and use A.LUsolve(b).