3.7 Bordered Determinants 47
and letBndenote the determinant of order (n+ 1) obtained by bordering
Anby the column
X=
[
x 1 x 2 x 3 ···xn
]T
on the right, the row
Y=
[
y 1 y 2 y 3 ···yn
]
at the bottom and the elementzin position (n+1,n+ 1). In some detail,
Bn=
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
a 11 a 12 ··· a 1 n x 1
a 21 a 22 ··· a 2 n x 2
........................
an 1 an 2 ··· ann xn
y 1 y 2 ··· yn z
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
n+1
. (3.7.1)
Some authors border on the left and at the top but this method displaces
the elementaijto 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
suffixn.
Theorem 3.9.
B=zA−
n
∑
r=1
n
∑
s=1
Arsxrys.
Proof. The coefficient ofysinBis (−1)
n+s+1
F, where
F=
∣
∣C
1 ...Cs− 1 Cs+1...CnX
∣
∣
n
=(−1)
n+s
G,
where
G=
∣
∣
C 1 ...Cs− 1 XCs+1...Cn
∣
∣
n
.
The coefficient ofxrinGisArs. Hence, the coefficient ofxrysinBis
(−1)
n+s+1+n+s
Ars=−Ars.
The only term independent of the x’s and y’s is zA. The theorem
follows.
LetEijdenote the determinant obtained fromAby
a.replacingaijbyz,i, jfixed,
b.replacingarjbyxr,1≤r≤n,r=i,
c. replacingaisbyys,1≤s≤n,s=j.
Theorem 3.10.
Bij=zAij−
n
∑
r=1
n
∑
s=1
Air,jsxrys=Eij.