of Edinburgh, Session J 871-72. 681 
I verify this by the former solution, as follows : — We have 
pq - r pq-r 
y 
The equation thus becomes 
? ; that is, - - 0 , = - - q . 
a + q, b, , c 
d ,e + q, / 
g , h ,i + q 
(pq - r) : 
-v, = 0, 
r) 2 
that is 
q3 + p0 * + n + r _ = 0 , 
But we have 
— o) = q 3 + pq 2 + </q + r , 
and the equation thus becomes 
q 3 + pq 2 + qq + r - (pq - r) 2 = 0 ; 
viz., substituting for p , q, r their values in terms of p, q, r, this is 
the identity, 
q 3 + q 2 (p 2 - 2q) + q (q 2 - 2pr) + r 2 - (pq - r) 2 = 0 . 
An interesting case is where the given matrix M is unity ; that is 
M = ( 1, 0, 0 ) . 
I 0, 1, 0 I 
I 0, 0, 1 | 
We have here p= 3, 9 = 3, r- 1 ; the equation in q is 
q 4 - 6q 2 - 8q -3 = 0; 
that is (q- 3) (q+l) 3 = 0; viz., q = 3or q = -1. Taking, as we 
may do, r = q-l, we have the two solutions (p = 3, q= 3, r = 1) and 
(p= -1. q| -1, r=l). 
For the first of these 0= -6, <p = 21, w= -64, pq-r = 8, and 
thence 
L = - ^-(M 2 - 6M + 21) + 3, = 1, on writing therein M = 1 ; 
viz., we have L the matrix unity, a self-evident solution. 
But for the second, 6= -2, <p- 1, w=0, pq-r = 0, and the 
solution takes the form \/M = ^ (M - l) 2 - 1. There is, in 
