MathAction #178 Linear Agebra problem

Topics

FrontPage
- IssueTracker
  - #178 Linear Agebra problem <-- You are here.

This doesn't work:

fricas

(1) -> L := [ A = 2*P1+P2, B = 2*P2+P1, C = 2*Q1+Q2, D = 2*Q2+Q1]

$\label{eq1}\begin{array}{@{}l} \displaystyle \left[{A ={P 2 +{2 \ P 1}}}, \:{B ={{2 \ P 2}+ P 1}}, \:{C ={Q 2 +{2 \ Q 1}}}, \: \right. \ \ \displaystyle \left.{D ={{2 \ Q 2}+ Q 1}}\right]$

(1)

Type: List(Equation(Polynomial(Integer)))

fricas

solve(L, [P1,P2])

$\label{eq2}\left[ \right]$

(2)

Type: List(List(Equation(Fraction(Polynomial(Integer)))))

But it should, observe this:

fricas

SA:=solve([L.1,L.2],[P1,P2])

$\label{eq3}\left[{\left[{P 1 ={\frac{- B +{2 \ A}}{3}}}, \:{P 2 ={\frac{{2 \ B}- A}{3}}}\right]}\right]$

(3)

Type: List(List(Equation(Fraction(Polynomial(Integer)))))

fricas

SB:=solve([L.3,L.4],[Q1,Q2])

$\label{eq4}\left[{\left[{Q 1 ={\frac{- D +{2 \ C}}{3}}}, \:{Q 2 ={\frac{{2 \ D}- C}{3}}}\right]}\right]$

(4)

Type: List(List(Equation(Fraction(Polynomial(Integer)))))

First two equationa do not depend on , the later two don't depend on . Now check it again the initial set:

fricas

S:=[SA.1.1,SA.1.2,SB.1.1,SB.1.2]

$\label{eq5}\begin{array}{@{}l} \displaystyle \left[{P 1 ={\frac{- B +{2 \ A}}{3}}}, \:{P 2 ={\frac{{2 \ B}- A}{3}}}, \: \right. \ \ \displaystyle \left.{Q 1 ={\frac{- D +{2 \ C}}{3}}}, \:{Q 2 ={\frac{{2 \ D}- C}{3}}}\right]$

(5)

Type: List(Equation(Fraction(Polynomial(Integer))))

fricas

map(x+->eval(x,S),L)

$\label{eq6}\left[{A = A}, \:{B = B}, \:{C = C}, \:{D = D}\right]$

(6)

Type: List(Equation(Expression(Integer)))

Same as bug #170 --kratt6, Tue, 21 Jun 2005 02:20:21 -0500 reply

Status: open => duplicate

Subject: Be Bold !!

	( 15 subscribers )