# Revision history [back]

There is a good notion for dealing with such problems : the kernel of a matrix. If $K$ is the (right) kernel of $A$, if $x_0$ is a solution of the equation $Ax=b$, then the set of solutions of the equation is the set ${x_0+k \mid k in K}$.

So, you can do the following:

sage: A = random_matrix(GF(3),4,6) ; A
[2 1 2 2 0 2]
[2 2 0 1 2 0]
[0 2 2 1 2 2]
[2 0 0 1 0 2]
sage: b = vector(GF(3),(2,2,0,2)) ; b
(2, 2, 0, 2)

sage: x0 = A.solve_right(b)
sage: x0
(1, 0, 0, 0, 0, 0)

sage: K = A.right_kernel() ; K
Vector space of degree 6 and dimension 2 over Finite Field of size 3
Basis matrix:
[1 0 1 1 0 0]
[0 1 2 1 0 1]

sage: solutions = [x0 + k for k in K] ; solutions
[(1, 0, 0, 0, 0, 0),
(2, 0, 1, 1, 0, 0),
(0, 0, 2, 2, 0, 0),
(1, 1, 2, 1, 0, 1),
(2, 1, 0, 2, 0, 1),
(0, 1, 1, 0, 0, 1),
(1, 2, 1, 2, 0, 2),
(2, 2, 2, 0, 0, 2),
(0, 2, 0, 1, 0, 2)]


There is a good notion for dealing with such problems : the kernel kernel of a matrix. matrix, and i strongly suggest you to learn about it. If $K$ is the (right) kernel of $A$, if $x_0$ is a solution of the equation $Ax=b$, then the set of solutions of the equation is the set ${x_0+k \mid k in \in K}$.

So, you can do the following:

sage: A = random_matrix(GF(3),4,6) ; A
[2 1 2 2 0 2]
[2 2 0 1 2 0]
[0 2 2 1 2 2]
[2 0 0 1 0 2]
sage: b = vector(GF(3),(2,2,0,2)) ; b
(2, 2, 0, 2)

sage: x0 = A.solve_right(b)
sage: x0
(1, 0, 0, 0, 0, 0)

sage: K = A.right_kernel() ; K
Vector space of degree 6 and dimension 2 over Finite Field of size 3
Basis matrix:
[1 0 1 1 0 0]
[0 1 2 1 0 1]

sage: solutions = [x0 + k for k in K] ; solutions
[(1, 0, 0, 0, 0, 0),
(2, 0, 1, 1, 0, 0),
(0, 0, 2, 2, 0, 0),
(1, 1, 2, 1, 0, 1),
(2, 1, 0, 2, 0, 1),
(0, 1, 1, 0, 0, 1),
(1, 2, 1, 2, 0, 2),
(2, 2, 2, 0, 0, 2),
(0, 2, 0, 1, 0, 2)]


There is a good notion for dealing with such problems : the kernel of a matrix, and i strongly suggest you to learn about it. If $K$ is the (right) kernel of $A$, if $x_0$ is a solution of the equation $Ax=b$, then the set of solutions of the equation is the set ${x_0+k$\{x_0+k \mid k \in K}$.K\}$.

So, you can do the following:

sage: A = random_matrix(GF(3),4,6) ; A
[2 1 2 2 0 2]
[2 2 0 1 2 0]
[0 2 2 1 2 2]
[2 0 0 1 0 2]
sage: b = vector(GF(3),(2,2,0,2)) ; b
(2, 2, 0, 2)

sage: x0 = A.solve_right(b)
sage: x0
(1, 0, 0, 0, 0, 0)

sage: K = A.right_kernel() ; K
Vector space of degree 6 and dimension 2 over Finite Field of size 3
Basis matrix:
[1 0 1 1 0 0]
[0 1 2 1 0 1]

sage: solutions = [x0 + k for k in K] ; solutions
[(1, 0, 0, 0, 0, 0),
(2, 0, 1, 1, 0, 0),
(0, 0, 2, 2, 0, 0),
(1, 1, 2, 1, 0, 1),
(2, 1, 0, 2, 0, 1),
(0, 1, 1, 0, 0, 1),
(1, 2, 1, 2, 0, 2),
(2, 2, 2, 0, 0, 2),
(0, 2, 0, 1, 0, 2)]


There is a good notion for dealing with such problems : the kernel of a matrix, and i strongly suggest you to learn about it. If $K$ is the (right) kernel of $A$, if $x_0$ is a solution of the equation $Ax=b$, then the set of solutions of the equation is the set $\{x_0+k$x_0+K = \{x_0+k \mid k \in K\}$. So, you can do the following: sage: A = random_matrix(GF(3),4,6) ; A [2 1 2 2 0 2] [2 2 0 1 2 0] [0 2 2 1 2 2] [2 0 0 1 0 2] sage: b = vector(GF(3),(2,2,0,2)) ; b (2, 2, 0, 2) sage: x0 = A.solve_right(b) sage: x0 (1, 0, 0, 0, 0, 0) sage: K = A.right_kernel() ; K Vector space of degree 6 and dimension 2 over Finite Field of size 3 Basis matrix: [1 0 1 1 0 0] [0 1 2 1 0 1] sage: solutions = [x0 + k for k in K] ; solutions [(1, 0, 0, 0, 0, 0), (2, 0, 1, 1, 0, 0), (0, 0, 2, 2, 0, 0), (1, 1, 2, 1, 0, 1), (2, 1, 0, 2, 0, 1), (0, 1, 1, 0, 0, 1), (1, 2, 1, 2, 0, 2), (2, 2, 2, 0, 0, 2), (0, 2, 0, 1, 0, 2)]  There is a good an important mathematical notion for dealing with such problems : the kernel of a matrix, and i strongly suggest you to learn about it. If$K$is the (right) kernel of$A$, if$x_0$is a solution of the equation$Ax=b$, then the set of solutions of the equation is the set$x_0+K = \{x_0+k \mid k \in K\}$. So, you can do the following: sage: A = random_matrix(GF(3),4,6) ; A [2 1 2 2 0 2] [2 2 0 1 2 0] [0 2 2 1 2 2] [2 0 0 1 0 2] sage: b = vector(GF(3),(2,2,0,2)) ; b (2, 2, 0, 2) sage: x0 = A.solve_right(b) sage: x0 (1, 0, 0, 0, 0, 0) sage: K = A.right_kernel() ; K Vector space of degree 6 and dimension 2 over Finite Field of size 3 Basis matrix: [1 0 1 1 0 0] [0 1 2 1 0 1] sage: solutions = [x0 + k for k in K] ; solutions [(1, 0, 0, 0, 0, 0), (2, 0, 1, 1, 0, 0), (0, 0, 2, 2, 0, 0), (1, 1, 2, 1, 0, 1), (2, 1, 0, 2, 0, 1), (0, 1, 1, 0, 0, 1), (1, 2, 1, 2, 0, 2), (2, 2, 2, 0, 0, 2), (0, 2, 0, 1, 0, 2)]  There is an important mathematical notion for dealing with such problems : the kernel of a matrix, and i strongly suggest you to learn about it. If$K$is the (right) kernel of$A$, if$x_0$is a solution of the equation$Ax=b$, then the set of solutions of the equation is the set$x_0+K = \{x_0+k \mid k \in K\}$. So, you can do the following: Setting: sage: A = random_matrix(GF(3),4,6) ; A [2 1 2 2 0 2] [2 2 0 1 2 0] [0 2 2 1 2 2] [2 0 0 1 0 2] sage: b = vector(GF(3),(2,2,0,2)) ; b (2, 2, 0, 2)  Find parameters$x_0$and$K\$:

sage: x0 = A.solve_right(b)
sage: x0
(1, 0, 0, 0, 0, 0)

sage: K = A.right_kernel() ; K
Vector space of degree 6 and dimension 2 over Finite Field of size 3
Basis matrix:
[1 0 1 1 0 0]
[0 1 2 1 0 1]


Enumerate solutions:

sage: solutions = [x0 + k for k in K] ; solutions
[(1, 0, 0, 0, 0, 0),
(2, 0, 1, 1, 0, 0),
(0, 0, 2, 2, 0, 0),
(1, 1, 2, 1, 0, 1),
(2, 1, 0, 2, 0, 1),
(0, 1, 1, 0, 0, 1),
(1, 2, 1, 2, 0, 2),
(2, 2, 2, 0, 0, 2),
(0, 2, 0, 1, 0, 2)]


You can check:

sage: for i in solutions:
....:     print A*i
(2, 2, 0, 2)
(2, 2, 0, 2)
(2, 2, 0, 2)
(2, 2, 0, 2)
(2, 2, 0, 2)
(2, 2, 0, 2)
(2, 2, 0, 2)
(2, 2, 0, 2)
(2, 2, 0, 2)