[186 
186] ON THE DETERMINATION OF THE VALUE OF A CERTAIN DETERMINANT. 121 
IN 
And of course if x denote the number of lines or columns of the determinant, then 
£4 is the value of the determinant. This theorem, or a particular case of it, is due 
to Prof. Sylvester: I have not been able to find an easier demonstration than the 
following one, which, it must be admitted, is somewhat complicated. I observe that 
JJ X satisfies the equation 
U x — eu x _ x + (x — 1) (n — x + 2) £4- 2 = 0. 
Hence writing x — 1 and x — 2 for x, we have the system 
U x —6U x - 1 -\-(x—l)(n — x + 2) U x ~ 2 =0, 
£4-i ~ 0£4-2 + (x — 2)(n — x + 3) £4-s = 0, 
£4—2 - 0U x -, + (x-3) (n -x + 4<) £/4-4 = 0, 
or, eliminating £4-i and £4_ 3 , 
£4 + {{® — 1) (n — x + 2) + (x — 2) (n — x + 3) — 6 2 } £4-2 
+ (x — 2)(x — 3) (n — x + 3) (n — x + 4) £4-4 = 0. 
Suppose, for shortness, 
(6 + x — 1) (6 + x — 3) (6 + x — 5)...(0 — x + 5) (6 — x + 3) {6 — x + 1) = H x , 
and assume 
£4 — ~Ax,o H x d-c! H a; _ 2 ...+ ( yA. xs -£4—2*)'• •., 
where A x g is independent of 6, then 
£4 contains the term {—) S A X S H X - 2S , 
£4-2 contains the term (—) S A X _ 2 S H x _ 2S _ 2 , 
which is to be multiplied by 
(x — 1) (n — x + 2) + (x —2){n — x + 3) — 6-. 
This multiplier may be written under the form 
(x — 1) (n — x + 2) + (x - 2) (n — x + 3) — (x — 2s — l) 2 — {6~ — (x — 2s — l) 2 } 
= M x<8 — [6- — (x — 2s — l) 2 }, 
if, for shortness, 
= (x — 1) (n — x + 2) + (x — 2) (n — x + 3) — (x — 2s — l) 2 . 
Now 
multiplied into 
C. III. 
M Xt s - {6 2 -{x-2s- l) 2 } 
( — yA X —2,3 /4-23-2 
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