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5.8 Some Applications of Algebraic Computing 233
and denote the result by Z n+1 . Verify the formula
2 2 2
n −2
2
2
Z n+1 = −n K n (x − 1) (x − n )
for several values of n.
Both formulas have been proved analytically, but the details are
complicated and it has been decided to omit them.
Exercise. Show that
a 11 a 12 ··· a 1n x
2
a 21 a 22 ··· a 2n x
1 1
··· ··· ··· ··· ··· = − K n F n, −n; ; −x ,
2 2
n
a n1 a n2 ··· x
n−1
a nn
1 x ··· x −
1
3 2n−1 2
where
(1 + x) i+j−1 − x i+j−1
a ij =
i + j − 1
and where F(a, b; c; x) is the hypergeometric function.
5.8.7 Determinantal Identities Related to Matrix Identities
If M r ,1 ≤ r ≤ s, denote matrices of order n and
s
M r = 0, s > 2,
r=1
then, in general,
s
|M r | =0, s > 2,
r=1
that is, the corresponding determinantal identity is not valid. However,
there are nontrivial exceptions to this rule.
Let P and Q denote arbitrary matrices of order n. Then
1. a. (PQ + QP)+(PQ − QP) − 2PQ = 0, all n,
b. |PQ + QP| + |PQ − QP|−|2PQ| =0, n =2.
2
2
2. a. (P − Q)(P + Q) − (P − Q ) − (PQ − QP)= 0, all n,
2
2
b. |(P − Q)(P + Q)|−|P − Q |−|PQ − QP| =0, n =2.
2
2
3. a. (P − Q)(P + Q) − (P − Q )+(PQ + QP) − 2PQ = 0, all n,
2
2
b. |(P − Q)(P + Q)|−|P − Q | + |PQ + QP|−|2PQ| =0, n =2.
The matrix identities 1(a), 2(a), and 3(a) are obvious. The corresponding
determinantal identities 1(b), 2(b), and 3(b) are not obvious and no neat
proofs have been found, but they can be verified manually or on a computer.
Identity 3(b) can be obtained from 1(b) and 2(b) by eliminating |PQ−QP|.