416 Algebra
which has the property that
Mn=(
Fn+ 1 Fn
Fn Fn− 1)
, forn≥ 1.Taking determinants, we have
Fn+ 1 Fn− 1 −Fn^2 =detMn=(detM)n=(− 1 )n,as desired.
(J.D. Cassini)
208.Subtract thepth row from the(p+ 1 )st, then the(p− 1 )st from thepth, and so on.
Using the identity
(n
k)
−
(n− 1
k)
=
(n− 1
k− 1)
, the determinant becomes
∣
∣
∣∣
∣∣
∣∣
∣∣1
(m
1)
···
(m
p)
0
(m
0)
···
(m
p− 1)
..
.
..
.
... ..
.
0
(m− 1 +p
0)
···
(m− 1 +p
p− 1)
∣
∣
∣∣
∣∣
∣∣
∣∣
.
Expanding by the first row, we obtain a determinant of the same form but withmreplaced
bym−1 andpreplaced byp−1. Forp=0 the determinant is obviously equal to 1,
and an induction onpproves that this is also true in the general case.
(C. Nast ̆ ̆asescu, C. Ni ̧ta, M. Brandiburu, D. Joi ̧ta, ̆ Exerci ̧tii ̧si Probleme de Algebra ̆
(Exercises and Problems in Algebra), Editura Didactica ̧ ̆si Pedagogica, Bucharest, 1983) ̆
209.The determinant
∣∣
∣
∣∣
∣∣
∣∣
(x 1
0)(x 2
0)
···
(xn
0)
(x 1
1)(x 2
1)
···
(xn
1)
..
.
..
.
... ..
(.
x 1
n− 1)(x 2
n− 1)
···
(xn
n− 1)
∣∣
∣
∣∣
∣∣
∣∣
is an integer. On the other hand, for some positive integer( mandk, the binomial coefficient
m
k
)
is a linear combination ofmk,(m
k− 1)
,...,
(m
0)
whose coefficients do not depend on
m. In this linear combination the coefficient ofmkis 1/k!. Hence by performing row
operations in the above determinant we can transform it into
∣∣
∣∣
∣∣
∣∣
∣
11 ··· 1
x 1 x 2 ··· xn
..
...
.
... ..
.
xn 1 −^1 x 2 n−^1 ···xnn−^1