348 HITCHCOCK. 



12. Interpretation of the Equations (58). 



It is manifest that the equations (58) are biHnear: and that the set 

 of six b's are determined by the choice of three axes /3i, (82 63, while 

 the set of nine A's govern the other four axes. I shall first show how 

 various simple solutions of these eciuations may be written down, 

 and shall then discuss the general solution as a linear function of these 

 simple solutions. 



One solution is seen by inspection to be: ^u, /I22, ^33, proportional 

 to hii, &22, and 633, with the other six ^4's all zero. The resulting value 

 of Fp, which we may call F\p, is, by (31), 



Fip = ^ihnX-iXi + ^ihiX-zXi + jSafess-TLTo (59i)= 



which, of course, may be multiplied by a scalar constant. It is easy 

 to check the result by operating with V, using the values of the &'s. 

 from (57i), and showing that the expression is of the form Vpb. 



A second solution is equally obvious: let ^23 + ^32, Azi + An, and 

 Aio + Ail be proportional to 623, 631 and 612 with Aw = A22 =Asz = 

 and A23 = A32, Az\ = An, An = A21, giving the solution 



F2P = 0i(bi2X3Xi + 613X1.T2) + i32(fe2i3:'2.r3 + &23a*ia-2) + |S3(&3i.r2.T3 + bsoXsXi} 



(592)' 



We note that this solution is null if /jos, &31, and &12 are all zero, that is, 

 if the vectors /3i, ^2, and /Ss form a rectangular system. Let us suppose- 

 for the moment that one of the b's, as 693, is not zero. 



These two solutions are linear in the b's, and are not altered by 

 advancing subscripts. A third solution, is found by assigning arbi- 

 trarily An = A22 = A2Z + ^32 = and solving for the ^'s in terms 

 of the b's. The result is 



^33 == —2631623, ^31 = b2dbn, Au — —623611, Ai2 = 622611, 



A21 = 622611, -^23 = — 622631, 7132 = 'T"622631>. 



which may be checked by substituting in (58). Whence by (31) 



Fsp = |3i(6226ii.T3;ri — 623611.T1.r2) + i32(6226iia-2.T3 — 62263ia;ia;2) 



+ i33(6236n.r2.T3 + 62263ia-3.Ti — 263i623a-i.r2) (SQs) 



quadratic and unsymmetrical in the 6's. 



A fourth solution may now be obtained by advancing subscripts, as 

 also, of course, a fifth, which might be used in place of F2P in case 

 all 6's with unequal subscripts vanish, but a simpler treatment of this- 



