From: Fred lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Sat, February 19, 2011 6:53:26 PM Subject: [math-fun] (yawn!) Simultaneous Linear Equations with Literal Coefficients Everybody can solve these, can't they --- just like quadratic equations ... Here's a simple example: solve for x,y,z, b z - c y = f , c x - a z = g , a y - b x = h . Oops --- the determinant vanishes: | c 0 -a | | -b a 0 | = 0 . | 0 -c b | No solution, says our CAS equation solver, firmly. Except of course, if the RHS also vanishes there is obviously a 1-dimensional homogeneous solution; but simply substituting f,g,h -> 0 should encourage our CAS to disgorge some version of that too. End of story? Not quite: you see, adding a equn1 + b equn2 + c equn3 eliminates all variables, leaving a f + b g + c h = 0 ; and in my particular application, this constraint happens to be satisfied by the coefficients. Now some head-scratching unearths a 3-dimensional (in general) solution spanned by the symmetric linear basis [ x -> -h/b, y -> 0, z -> +f/b ] , [ x -> +g/c, y -> -f/c, z -> 0 ] , [ x -> 0, y -> +h/a, z -> -g/a ] . In particular, summing eventually delivers the symmetric solution x -> (g/c - h/b)/3 , y -> (h/a - f/c)/3 , z -> (f/b - g/a)/3 , which I knew ought to exist by the symmetry of the equations. I have encountered this type of problem --- solving SLE's on a manifold --- on a number of occasions, without ever finding an effective solution algorithm. Surely it must have been considered before --- any pointers, anyone? Is it reasonable to expect that the next generation of CAS solvers might inform us about constraints on our coefficients under which such equations might possess extra solutions? Fred Lunnon _______________________________________________ We have vectors A = (a,b,c), Z = (x,y,z), and F = (f,g,h). The linear equations become a vector cross product AxZ = F. A solution exists only when A.F = 0, and then the most general solution consists of the sum of any one particular solution P with an arbitrary solution of the homogeneous equation AxZ = 0. The latter has the 1-dimensional solution space consisting of multiples of A. So the general solution is Z = P + kA. If A' is any vector in the plane normal to F, and linearly independent of A, then AxA' lies along F, and so we have a solution Z = ((F.F)/[F,A,A']) A', [...] denoting the scalar triple product. Any such A' is of the form FxB with B arbitrary. Then [F,A,FxB] = ((FxB)xF).A = ((F.F)B - (B.F)F).A = (F.F)(A.B), and Z = (FxB)/(A.B) is a general solution. If A' = FxB is linearly dependent on A, we have 0 = Ax(FxB) = (A.B)F - (A.F)B = (A.B)F, so B is restricted to A.B nonzero. The solution found in the original post is of this form with B = (1/a, 1/b, 1/c) and A.B = 3. -- Gene