Advanced book on Mathematics Olympiad

418 Algebra

det(A)=

∣∣

∣∣

∣

∣∣

∣∣

∣∣

100 ··· 00

010 ··· 00

..

.

..

.

..

.

... ..

.

..

.

000 ··· 10

± 1 ∓ 2 ± 3 ··· −n+ 1 n+ 1

∣∣

∣∣

∣

∣∣

∣∣

∣∣

=n+ 1.

(9th International Mathematics Competition for University Students, 2002)

212.View the determinant as a polynomial in the independent variablesx 1 ,x 2 ,..., xn.

Because wheneverxi =xjthe determinant vanishes, it follows that the determinant is

divisible byxi−xj, and therefore by the product

∏

1 ≤i<j≤n(xj−xi). Because theki’s are

positive, the determinant is also divisible byx 1 x 2 ···xn. To solve the problem, it suffices

to show that for any positive integersx 1 ,x 2 ,...,xn, the product

x 1 x 2 ···xn

∏

1 ≤i<j≤n

(xj−xi)

is divisible byn!. This can be proved by induction onn. A parity check proves the case

n=2. Assume that the property is true for anyn−1 integers and let us prove it forn.

Either one of the numbersx 1 ,x 2 ,...,xnis divisible byn, or, by the pigeonhole principle,

the difference of two of them is divisible byn. In the first case we may assume thatxn

is divisible byn, in the latter thatxn−x 1 is divisible byn. In either case,

x 1 x 2 ···xn− 1

∏

1 ≤i<j≤n− 1

(xj−xi)

is divisible by(n− 1 )!, by the induction hypothesis. It follows that the whole product is

divisible byn×(n− 1 )!=n!as desired. We are done.

(proposed for the Romanian Mathematical Olympiad by N. Chichirim)

213.Expand the determinant as

det(xA+yB)=a 0 (x)y^3 +a 1 (x)y^2 +a 2 (x)y+a 3 (x),

whereai(x)are polynomials of degree at mosti,i= 0 , 1 , 2 ,3. Fory =0 this gives
det(xA)=x^3 detA=0, and hencea 3 (x)=0 for allx. Similarly, settingy =xwe
obtain det(xA+xB)=x^3 det(A+B)=0, and thusa 0 (x)x^3 +a 1 (x)x^2 +a 2 (x)x=0.
Also, fory =−xwe obtain det(xA−xB)=x^3 det(A−B)=0; thus−a 0 (x)x^3 +
a 1 (x)x^2 −ax(x)x=0. Adding these two relations givesa 1 (x) =0 for allx. For
x =0 we find that det(yB)=y^3 detB =0, and hencea 0 ( 0 )y^3 +ax( 0 )y =0 for
ally. Therefore,a 0 ( 0 )=0. Buta 0 (x)is a constant, soa 0 (x)=0. This implies that
a 2 (x)x=0 for allx, and soa 2 (x)=0 for allx. We conclude that det(xA+yB)is
identically equal to zero, and the problem is solved.
(Romanian mathematics competition, 1979, M. Martin)