two equations in three unknowns

Two Homogeneous Linear Equations in 3 unknowns. PhL 12.9.04 This sounds completely trivial, but I have spent several hours unable to draw a clean set of conclusions! So let's start by writing down our two equations (where constants can take ANY values) a x + by + c z = 0 a'x + b'y + c'z = 0 You can think of (a,b,c) and (a',b',c') as the normals to the two planes implied by these equations. However, if (a,b,c) = (0,0,0), then there is no plane associated with the first equation, it just says 0=0 so we just throw it out and then we have 1 equation in 3 unknowns. So now we start our case analysis: (Or just read conclusions at the end! ) _________________________________________________________ Case 1. Assume that (a',b',c') = (0,0,0) so we have 1 equation in 3 unknowns. Case 1A: If (a,b,c) = (0,0,0) as well, then both equations say 0=0 and we then have NO equations at all, so we cannot even talk about what the "solutions" look like. I guess any (x,y,z) solves NO equations. Case 1B: If (a,b,c) = (a,0,0), then we have that x=0 as our only constraint. The solution is then (0,y,z) for any y,z, and we have geo = 2. Similarly if (a,b,c) = (0,b,0) or if (a,b,c) = (0,0,c). Case 1C. If (a,b,c) = (a,b,0), then we have ax+by=0 so we can solve for y = kx , and then the solution has the form (x,kx,z) for any x and any z. We can take our basis as (1,k,0) and (0,0,1) which are clearly linearly independent, so again we have geo = 2. Same for cases (0,b,c) and (a,0,c) Case 1D. If (a,b,c) = (a,b,c), then we have ax+by+cz = 0. This is a plane with n = (a,b,c) and any point on this plane satisfies our equation set. Our ray is then perhaps (x,y,-(ax+by)/c). One basis vector can then be (0,y,-by/c) and the other (x,0,-ax/c). These are lin indep and again we have geo = 2. Theorem: If you have 1 linear homo equation in 3 unknowns, and the equation is not just 0=0, then the solution of the equation has two independent basis vectors. ___________________________________________________________ Case 2. Assume that (a',b',c') = (a',0,0). This says that x=0. Case 2A. If (a,b,c) = (a,b,c) or (0,b,c), then we have by+cz = 0 and we solve for z = ky. Ray is then (0,y,ky) and our basis vector is perhaps (0,1,k) and that is all there is. So geo = 1 ! k = -b/c. Case 2B. If (a,b,c) = (a,0,c) or (0,0,c) then we have cz = 0 so z = 0. Solution is then (0,y,0) and geo = 1. Same idea for (a,b,c) = (a,b,0) or (0,b,0). Case 2C: If (a,b,c) = (a,0,0) or (0,0,0), then second equation 0 = 0 with x=0, so ray is (0,y,z) and geo = 2. Case 2D: If (a,b,c) = (0,b,0) then y = 0 and ray is (0,0,z) so geo = 1. _________________________________________________________ Case 3: Assume that (a',b',c') = (a',b',0) which says a'x + b'y = 0 so y = kx or x = Ky. Case 3A: If (a,b,c) = (a,b,c), then ax +b(kx) +cz = 0 so z = k'x and ray = (x,kx,k'x) or (1,k,k') and then we have geo = 1. Case 3B: If (a,b,c) = (a,b,0), then ax +b(kx) = 0 and x = 0, so ray is (0,0,z) and again geo = 1. Case 3C: If (a,b,c) = (0,b,c) then b(kx)+cz = 0 so z = k'x and ray is then (x,kx,k'x) and geo = 1. Same idea for (a,0,c). Case 3D: If (a,b,c) = (a,0,0) then x=0 so ray is (0,0,z) and geo = 1. Same idea for (0,b,0) Case 3E: If (a,b,c) = (0,0,c) then z=0 and ray is (x,kx,0) and geo = 1 _________________________________________________________ Case 4: Assume that (a',b',c') = (a',b',c'). Case 4A: If (a,b,c) = (a,b,c), then solve other equation for z = z(x,y) and plug that in here to get ax + by + cz(x,y) = 0 and find that y = kx, so resulting ray is (x,kx, k'x) so (1,k,k') and we have geo = 1. Case 4B: If (a,b,c) = (a,b,0), then solve other equation for y = y(x,z) and plug in here to get ax + by(x,z) = 0. Solve this for z = k'x. Solution is then (x, y(x,k'x), k'x) = (x,k"x,k'x), geo = 1. Case 4C: If (a,b,c) = (a,0,0), then get x=0 from this equation. Other then says b'y+c'z = 0 so get z = ky, and solution is then (0,y,ky) or (0,1,k) and geo = 1. _____________________________________________________ Table of Results: Gives the geometric multiplicity of the solution ray set. 1 2 3 4 (a'b'c') = 000 a'00 a'b'0 a'b'c' abc = 000 3 2 2 2 a00 2 2 1 1 ab0 2 1 1 1 abc 2 1 1 1 0bc 2 1 1 1 a0c 2 1 1 1 00c 2 1 1 1 0b0 2 1 1 1 Conclusions: Again write down our two equations, a x + by + c z = 0 a'x + b'y + c'z = 0 You can think of (a,b,c) and (a',b',c') as the normals to the two planes implied by these equations. However, if (a,b,c) = (0,0,0), then there is no plane associated with the first equation, it just says 0=0 so we just throw it out and then we have 1 equation in 3 unknowns. Case (a). If the second equation is a multiple of the first, then we really only have one equation. In this case, we are dealing with a sub-problem: 1 equation in 3 unknowns. Assuming that the first equation does not just say 0 = 0, then this is treated in Case 1 above and we found that geo = 2 in all subcases. This seems very reasonable: you have one plane and (x,y,z) must lie on that plane, so geo = 2. Case (b). If one of the equations says 0 = 0, then again, we have 1 equation in 3 unknowns, and as in case (a) above, we get geo = 2. There is one plane and it is 2D. Case (c). If neither equation is 0=0 and neither equation is a multiple of the other, then you are talking about the intersection of two planes which is a line, and you expect geo = 1. This is what we found in all the cases above. We can summarize these cases by saying: if the two equations are linearly independent, then geo = 1, otherwise geo = 2. If either equation is 0=0, we include that in the linearly dependent situation. If both equations are 0=0, then in fact we have geo = 3 because there are then no planes at all!