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5.6 Distinct Matrices with Nondistinct Determinants 211

1 (n+1)
r
= c r−t B .
n+1,n+1−t
t=0
B n
Hence, from the fourth equation in (5.5.11) and applying Lemma (b) and
(5.5.31),
n r
(n+1)
B n Q 2n = x r c r−t B
n+1,n+1−t
r=0 t=0
n j
(n+1)
= B c r x n+r−j
n+1,j+1
j=0 r=0
n
(n+1)

= ψ j x n−j B n+1,j+1 .
j=0
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
h n,2n+j =0, j ≥ 2,
k n,2n+j =0, j ≥ 1,
lead respectively to

S n+2 =0, all n, (X)

T n+1 =0, all n, (Y)
where S n+2 denotes the determinant obtained from A n+2 by replacing its
last row by the row

n+2
c n+j−1 c n+j c n+j+1 ··· c 2n+j
and T n+1 denotes the determinant obtained from B n+1 by replacing its last
row by the row

.
n+1
c n+j c n+j+1 c n+j+2 ··· c 2n+j
Regarding (X) and (Y) as conditions, what is their significance?

5.6 Distinct Matrices with Nondistinct
Determinants

5.6.1 Introduction
Two matrices [a ij ] m and [b ij ] n are equal if and only if m = n and a ij =
b ij ,1 ≤ i, j ≤ n. No such restriction applies to determinants. Consider