Advanced book on Mathematics Olympiad

Algebra 433

x 1

2

+

x 2

3

+···+

xn

n+ 1

= 0 ,

···

x 1

n

+

x 2

n+ 1

+···+

xn

2 n− 1

= 0

has only the trivial solution. For a solution(x 1 ,x 2 ,...,xn)consider the polynomial

P(x)=x 1 (x+ 1 )(x+ 2 )···(x+n− 1 )+x 2 x(x+ 2 )···(x+n− 1 )+···

+xnx(x+ 1 )···(x+n− 2 ).

Bringing to the common denominator each equation, we can rewrite the system in short

form asP( 1 )=P( 2 )= ··· =P (n)= 0 .The polynomialP(x)has degreen−1;

the only way it can havenzeros is if it is identically zero. Taking successivelyx =

0 ,− 1 ,− 2 ,...,−n, we deduce thatxi=0 for alli. Hence the system has only the trivial

solution, and the matrix is invertible.

For the second part, note that the sum of the entries of a matrixAis equal to the sum

of the coordinates of the vectorA1, where1 is the vector( 1 , 1 ,..., 1 ). Hence the sum

of the entries of the inverse matrix is equal tox 1 +x 2 +···+xn, where(x 1 ,x 2 ,...,xn)

is the unique solution to the system of linear equations

x 1

1

+

x 2

2

+···+

xn

n

= 1 ,

x 1

2

+

x 2

3

+···+

xn

n+ 1

= 1 ,

···

x 1

n

+

x 2

n+ 1

+···+

xn

2 n− 1

= 1.

This time, for a solution tothissystem, we consider the polynomial

Q(x)=x 1 (x+ 1 )(x+ 2 )···(x+n− 1 )+···+xnx(x+ 1 )···(x+n− 2 )

−x(x+ 1 )···(x+n− 1 ).

Again we observe thatQ( 1 )=Q( 2 )= ··· =Q(n)=0. BecauseQ(x)has degreen
and dominating coefficient−1, it follows thatQ(x)=−(x− 1 )(x− 2 )···(x−n).So

x 1

(x+ 1 )(x+ 2 )···(x+n− 1 )

xn−^1

+···+xn

x(x+ 1 )···(x+n− 2 )

xn−^1

=

x(x+ 1 )···(x+n− 1 )−(x− 1 )(x− 2 )···(x−n)

xn−^1

.

The reason for writing this complicated relation is that asx →∞, the left-hand side

becomesx 1 +x 2 +···+xn, while the right-hand side becomes the coefficient ofxn−^1 in

the numerator. And this coefficient is