5.6 Distinct Matrices with Nondistinct Determinants 211
=
1
Bn
r
∑
t=0
cr−tB
(n+1)
n+1,n+1−t.
Hence, from the fourth equation in (5.5.11) and applying Lemma (b) and
(5.5.31),
BnQ 2 n=
n
∑
r=0
x
r
r
∑
t=0
cr−tB
(n+1)
n+1,n+1−t
=
n
∑
j=0
B
(n+1)
n+1,j+1
j
∑
r=0
crx
n+r−j
=
n
∑
j=0
ψjx
n−j
B
(n+1)
n+1,j+1
This sum is an expansion of the determinant in part (b) of the theorem.
This completes the proofs of both parts of the theorem.
Exercise.Show that the equations
hn, 2 n+j=0,j≥ 2 ,
kn, 2 n+j=0,j≥ 1 ,
lead respectively to
Sn+2=0, alln, (X)
Tn+1=0, alln, (Y)
whereSn+2denotes the determinant obtained fromAn+2by replacing its
last row by the row
[
cn+j− 1 cn+jcn+j+1···c 2 n+j
]
n+2
andTn+1denotes the determinant obtained fromBn+1by replacing its last
row by the row
[
cn+jcn+j+1cn+j+2···c 2 n+j
]
n+1
Regarding (X) and (Y) as conditions, what is their significance?
5.6 Distinct Matrices with Nondistinct Determinants
5.6.1 Introduction
Two matrices [aij]mand [bij]nare equal if and only ifm=nandaij=
bij,1≤i, j≤n. No such restriction applies to determinants. Consider