THE AXES OF A QUADRATIC VECTOE. 



349 



<;ase will be given below. (It is evident that we cannot in general have 

 more than four linearly independent solutions for the A's when the 6's 

 are assigned.) Thus 



Fip = /3i(— 2611631X2.1:3 + 631/J22X3.T1 + 633612.T1.T2) 



+ i32(— 631&22X2.T3 + 633622^1X2) + ^3{— 633612X2X3 + 633622X3X1) (594) 



It is furthermore clear, because the equations (58) are linear in the 

 A's, that any linear function of the solutions already obtained will be a 

 solution, the 6's being taken as known constants, and, we will thus 

 obtain the most general solution. For from the four solutions already 

 written we may pick out the matrix of the four sets of values of An, 

 A22, Az3, namely 



611, 

 0, 

 0, 



-2612631, 



622, 

 0, 

 0, 

 0, 



&33 







—2631623 

 



"which by inspection are linearly independent sets. Hence the four 

 solutions J^i, F2, F3, Fi are linearly independent. 



Consider now the vector Fip. Remembering that the axes are the 

 vector solutions of the equation VpFp = 0, w^e expand p as /SiXi + 182X2 

 + 183X3 and have these three equations to determine the axes of Fip, 



633X1X2^ — 622X1X3^ = 0, 611X2X3^ — 633X2X1^ = 0, 622X3X1^ — 6iiX3.T2^ = 



Writing for brevity Ci = '^611, C2 = "^622, and C3 = ^633 we find 

 easily the matrix of the coefficients of /Si, ^2, ?z for the seven axes to be 



that is, the fourth axis is given by 184 = Ci/3i + €2^2 + C3i83 etc. By 

 inspection of this matrix we see the following relations of coplanarity 

 among the axes: (145) = (167) = (246) = (257) = (356) = (347) 

 = 0. The vector Fyp therefore is of the type mentioned in Art. 8 

 having four central axes, namely 184, (85, (86, 187. 



