Malet's Proof that every Equation has Boots. 
225 
Thus 
(D5 shown screened). 
Treat D, thus: (8), (6), (5), (7), (.) in brackets denoting the minors of 
the last column o£ D; (viz. 8657.), multiply the 1st row of D, by (8), the 
2nd by (6), and the last three by (5), (7), (.) as shown below. 
0 
•x(8) 
■x(6) 
... X (5) 
...x(7) 
...x(.) 
and add columns for a new last row. 
The elements of this new row will be seven 5-line determinants, of 
which the first four columns are those of D^, the last one varying and 
being shown below between verticals. 
Thus 
0 
2 
4 
6 
0 
2 
4 
6 
8 
0 
2 
4 
0 
2 
4 
6 
1 
3 
1 
3 
5 
1 
3 
5 
1 
3 
5 
7 
1— I 
3 
5 
7 
1 
3 
5 
7 
so that the seven values are D3, D^, . 
Noticing that the original last row was multiplied by (.) which 
0 2 4: 
= 0 . 1 3 =0 . D3 (say), and reading off from the new last row,. 
1 3 51 
. 0 . -D; . 0 . D; + D5 . 0 . D;' say, 
Therefore when D5 = 0, D^, and D3 have opposite signs. 
