I have the following matrix and RREF:
[
[ 2 1 0 -1 0| 4] [ 1 0 0 -1 -1/2| 0]
[ 0 1 0 1 1| 4] [ 0 1 0 1 1| 4]
[ 1 0 -1 2 0| 0], [ 0 0 1 -3 -1/2| 0]
]
Is it possible to convert the set of solutions:
x == 1/2*u + w
y == -u - w + 4
z == 1/2*u + 3*w
into vector form:
[ x ] [ 0 ] [ 1 ] [ 1/2 ] |
[ y ] [ 4 ] [ -1 ] [ -1 ] |
[ z ] = [ 0 ] + [ 3 ] w + [ 1/2 ]u | w, u ∈ ℝ
[ w ] [ 0 ] [ 1 ] [ 0 ] |
[ u ] [ 0 ] [ 0 ] [ 1 ] |
Is this possible with Sagemath?
The solutions are created using:
def my_solve(augmented_matrix, vars=None):
A = augmented_matrix[:, :-1]
Y = augmented_matrix[:, -1]
m, n = A.dimensions()
p, q = Y.dimensions()
if m!=p:
raise RuntimeError("The matrices have different numbers of rows")
if vars and len(vars) != n:
raise RuntimeError(f"Provided variables '{vars}' != number of columns '{n}'")
if vars:
# don't include the free variables in solve
X = vector([var(vars[i]) for i in [0..n-1]])
X_pivots = vector([var(X[i]) for i in [0..n-1] if i in A.pivots()])
X_free = vector([var(X[i]) for i in [0..n-1] if i not in A.pivots()])
else:
X = vector([var(f"x_{i}") for i in [0..n-1]])
X_pivots = vector([var(f"x_{i}") for i in [0..n-1] if i in A.pivots()])
X_free = vector([var(f"x_{i}") for i in [0..n-1] if i not in A.pivots()])
sols = []
for j in range(q):
system = [A[i]*X==Y[i,j] for i in range(m)]
sols += solve(system, *X_pivots)
return sols, X, X_pivots, X_free
M = matrix(QQ, 3, [2,1,0,-1,0, 0,1,0,1,1, 1,0,-1,2,0])
v = vector(QQ, [4,4,0])
Maug = M.augment(v, subdivide=True)
my_solve(Maug, var('x y z w u'))[0]
