Ask Your Question
1

How to use `solve()` for a system of equations using matrices?

asked 2019-05-23 11:11:47 +0100

torito_verdejo gravatar image

I know I can solve a system of equations by inputing independently each equation in a same solve() expression using the syntax solve([[exp1],[exp2], ... [expn]], x1,x2, ... xn), but what should I do if, having defined a matrix A and two columns vectors x and y, I want to express the system of equations as A*x == y?

Each instance of A[i] * x returns a list, that is (of course) not treated as a valid argument for an equation in solve(). How could I "tranform" that list in order to use it with solve()?

edit retag flag offensive close merge delete

Comments

thank you for asking that. not knowing how to answer this question,(I did not dare to ask the question) I did like that but I'm sure it must be easier way ! see this post and the cocalc link in it

ortollj gravatar imageortollj ( 2019-05-23 11:56:25 +0100 )edit

2 Answers

Sort by ยป oldest newest most voted
3

answered 2019-05-23 16:51:26 +0100

Emmanuel Charpentier gravatar image

updated 2019-05-23 17:00:12 +0100

Try this :

sage: A=matrix([[var("a_{}_{}".format(u,v), latex_name="a_{{{},{}}}".format(u,v)) for v in (1..2)] for u in (1..2)]);A
[a_1_1 a_1_2]
[a_2_1 a_2_2]
sage: X=vector([var("x_{}".format(u)) for u in (1..2)]);X
(x_1, x_2)
sage: Y=vector([var("y_{}".format(u)) for u in (1..2)]);Y
(y_1, y_2)
sage: Sol=solve(map(lambda u,v:u==v, A*X, Y),[u for u in X]);Sol
[[x_1 == -(a_2_2*y_1 - a_1_2*y_2)/(a_1_2*a_2_1 - a_1_1*a_2_2), x_2 == (a_2_1*y_1 - a_1_1*y_2)/(a_1_2*a_2_1 - a_1_1*a_2_2)]]

i. e. $$\left[\left[x_{1} = -\frac{{a_{2,2}} y_{1} - {a_{1,2}} y_{2}}{{a_{1,2}} {a_{2,1}} - {a_{1,1}} {a_{2,2}}}, x_{2} = \frac{{a_{2,1}} y_{1} - {a_{1,1}} y_{2}}{{a_{1,2}} {a_{2,1}} - {a_{1,1}} {a_{2,2}}}\right]\right]$$.

which looks entirely reasonable...

HTH,

edit flag offensive delete link more

Comments

Actually I don't understand how this actually works. A*x returns a list, as map() does, but A*x won't work while you're construction does. Why?

torito_verdejo gravatar imagetorito_verdejo ( 2019-05-23 19:00:55 +0100 )edit
1

A*x returns a list

Nope :

sage: (A*X).parent()
Vector space of dimension 2 over Symbolic Ring

But A*x is an iterable, as well as Y, and map iterates through those, producing at each step one equation, and returns the list of those equations. Similarly, I do not use X as the second argument of solve, but the list of its components.

?map should enlighten you :-).

Emmanuel Charpentier gravatar imageEmmanuel Charpentier ( 2019-05-23 21:23:44 +0100 )edit
1

POST-SCRIPTUM : It turns out that vectors have a list` method, that you can use to lighten the notations :

sage: solve((A*X-Y).list(),X.list())
[[x_1 == -(a_2_2*y_1 - a_1_2*y_2)/(a_1_2*a_2_1 - a_1_1*a_2_2), x_2 == (a_2_1*y_1 - a_1_1*y_2)/(a_1_2*a_2_1 - a_1_1*a_2_2)]]

Lighter and cleaner, indeed...

Emmanuel Charpentier gravatar imageEmmanuel Charpentier ( 2019-05-23 21:41:13 +0100 )edit
1

answered 2019-05-25 03:38:38 +0100

Juanjo gravatar image

updated 2019-06-20 21:03:49 +0100

I suppose you are aware that, in fact, you don't need to use solve: once defined $A$ and $Y$, the solution of the linear system $AX=Y$ is given by A\Y. For example,

sage: A = matrix([[1,2,3], [4,5,6], [-7,8,9]])
sage: Y = vector([3,9,1])
sage: print "Solution:", A\Y
Solution: (1, 1, 0)

Anyway, if you really need to use solve and write the system in the form $AX=Y$, you can continue the above example as follows:

sage: X = vector([var("x_{}".format(i)) for i in [1..3]])
sage: system = [A[i]*X==Y[i] for i in range(len(Y))]
sage: solve(system, *X)
[[x_1 == 1, x_2 == 1, x_3 == 0]]

Post-scriptum. The following function covers the cases considered in the comments below:

def my_solve(A, Y):
    m, n = A.dimensions()
    p, q = Y.dimensions()
    if m!=p:
        raise RuntimeError("The matrices have different numbers of rows")
    X = vector([var("x_{}".format(i)) for i in [1..n]])
    sols = []
    for j in range(q):
        system = [A[i]*X==Y[i,j] for i in range(m)]
        sols += solve(system, *X)
    return sols

Of course, A and Y should be two matrices with the same number of rows. Let us check the examples:

sage: A = matrix(QQ, [[1, 2, 2, 2], [2, 4, 6, 8], [3, 6, 8, 10], [0, 0, 0, 0]])
sage: Y = matrix(QQ, 4, 1, [0, 0, 0, 0])
sage: my_solve(A, Y)
[[x_1 == 2*r1 - 2*r2, x_2 == r2, x_3 == -2*r1, x_4 == r1]]

Please, note the way Y is given.

sage: A = matrix(QQ, [[2, 4, -3], [-3, -2, 4], [2, -5, -2]])
sage: Y = matrix(QQ, [[-5, -2], [3, 1], [0, 3]])
sage: my_solve(A, Y)
[[x_1 == 51, x_2 == 4, x_3 == 41], [x_1 == -5, x_2 == -1, x_3 == -4]]

Update. The code had an error in the way the number of unknowns was determined. I have modified the code to correct this bug. Let us check it with this new example:

sage: A = matrix([[1, 2, 3], [5, 7, 11]])
sage: Y = matrix([[13 ,17], [19, 23]])
sage: my_solve(A, Y)
[[x_1 == -1/3*r11 - 53/3, x_2 == -4/3*r11 + 46/3, x_3 == r11],
 [x_1 == -1/3*r12 - 73/3, x_2 == -4/3*r12 + 62/3, x_3 == r12]]

You can see these examples in this Sagemath Cell.

edit flag offensive delete link more

Comments

Juanjo, the code i put herein cocalc works for this 2 types of Matrix: not your solution ?

N=matrix(QQ,[[ 1 , 2 ,  2 , 2 ],[ 2 , 4 , 6 , 8 ],[ 3, 6 , 8 , 10 ],[ 0, 0 , 0 , 0 ]])
M=matrix(QQ,[0,0,0,0]).transpose()

or

N=matrix(QQ,[[ 2  2 -2],[ 0  2 -4],[-5  4 -2]]

M=matrix(QQ,[[-4 -5],[-5 -4],[ 1  0]]

Y=\left(\begin{array}{rr}
-4 & 1 \\
0 & 2 \\
2 & 2
\end{array}\right)

X=\left(\begin{array}{rrr}
2 & -2 & 0 \\
-1 & 4 & -1 \\
-5 & 4 & 4
\end{array}\right)


or 

Y=\left(\begin{array}{r}
0 \\
0 \\
0 \\
0
\end{array}\right)

X=\left(\begin{array}{rrrr}
1 & 2 & 2 & 2 \\
2 & 4 & 6 & 8 \\
3 & 6 & 8 & 10 \\
0 & 0 & 0 & 0
\end{array}\right)
ortollj gravatar imageortollj ( 2019-05-25 09:19:45 +0100 )edit

what is headers for Latex code in this forum ? in the post above A=N and Y=M (now I cant edit my comment !!, too late)

ortollj gravatar imageortollj ( 2019-05-25 09:30:08 +0100 )edit

Emmanuel this does not give a solution too.

nEC=4
A=matrix(QQ,[[ 1 , 2 ,  2 , 2 ],[ 2 , 4 , 6 , 8 ],[ 3, 6 , 8 , 10 ],[ 0, 0 , 0 , 0 ]])
Y=matrix(QQ,[0,0,0,0]).transpose()
#A=matrix([[var("a_{}_{}".format(u,v), latex_name="a_{{{},{}}}".format(u,v)) for v in (1..nEC)] for u in (1..nEC)])
X=vector([var("x_{}".format(u)) for u in (1..nEC)])
#Y=vector([var("y_{}".format(u)) for u in (1..nEC)])
show(" A : ",A," X : ",X," Y : ",Y)
Sol=solve(map(lambda u,v:u==v, A*X, Y),[u for u in X]);Sol
ortollj gravatar imageortollj ( 2019-05-25 10:27:18 +0100 )edit

Emmanuel, your code does not work either with these two matrices:

A=matrix(QQ,[[ 2 , 4, -3],[-3 ,-2 , 4],[ 2 ,-5 ,-2]])
Y=matrix(QQ,[[-5, -2],[ 3 , 1],[ 0 , 3]])
ortollj gravatar imageortollj ( 2019-05-25 10:42:16 +0100 )edit

I forgot to precise, that my code is OK with these matrices

ortollj gravatar imageortollj ( 2019-05-25 12:19:28 +0100 )edit

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

2 followers

Stats

Asked: 2019-05-23 11:11:22 +0100

Seen: 4,264 times

Last updated: Jun 20 '19