of Edinburgh, Sessio7i 1871-72. 
I verify this by the former solution, as follows:—We have 
* = E^ r , = thati > = g> = 
to ° CD 3 X 1 
681 
The equation thus becomes 
a -f q, b , , c 
cl , e + q , / 
0 
7i , i + q 
or 
(pq - r) 3 
that is 
to 2 
q 3 + po? + 2 q + r - ^ = 0 . 
But we have 
— w = q 3 -j- 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 ). 
0, 1, 0 
0, 0, 1 
We have here p — 3, q = 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=3orq=-l. Taking, as we 
may do, r= +1, we have the two solutions (p= 3, q= 3, r= 1) and 
(p= -1, q= “1, r = 1). 
For the first of these 6 = - 6, <p = 21, od — -64, pq - r = 8, and 
thence 
L = - g-(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, to = 0, pq-r=0, and the 
solution takes the form \/M = 2 (M - l) 2 - 1. There is, in 
