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3.7 Bordered Determinants 47
and let B n denote the determinant of order (n + 1) obtained by bordering
A n by the column
T
X = x 1 x 2 x 3 ··· x n
on the right, the row
Y = y 1 y 2 y 3 ··· y n
at the bottom and the element z in position (n +1,n + 1). In some detail,
a 11 a 12 ··· a 1n x 1
a 21 a 22 ··· a 2n x 2
. (3.7.1)
B n = ........................
a n1 a n2 ··· a nn x n
z
y 1 y 2 ···
n+1
y n
Some authors border on the left and at the top but this method displaces
the element a ij to the position (i +1,j + 1), which is undesirable for both
practical and aesthetic reasons except in a few special cases.
In the theorems which follow, the notation is simplified by discarding the
suffix n.
Theorem 3.9.
n n
B = zA − A rs x r y s .
r=1 s=1
Proof. The coefficient of y s in B is (−1) n+s+1 F, where
F = C 1 ... C s−1 C s+1 ... C n X
n
=(−1) n+s G,
where
G = C 1 ... C s−1 XC s+1 ... C n .
n
The coefficient of x r in G is A rs . Hence, the coefficient of x r y s in B is
(−1) n+s+1+n+s A rs = −A rs .
The only term independent of the x’s and y’s is zA. The theorem
follows.
Let E ij denote the determinant obtained from A by
a. replacing a ij by z, i, j fixed,
b. replacing a rj by x r ,1 ≤ r ≤ n, r = i,
c. replacing a is by y s ,1 ≤ s ≤ n, s = j.
Theorem 3.10.
n n
B ij = zA ij − A ir,js x r y s = E ij .
r=1 s=1