Advanced book on Mathematics Olympiad

422 Algebra

=

(

−

1

2

)n

det(In− 2 A+In)=

(

−

1

2

)n

det(A^2 − 2 A+In)

=

(

−

1

2

)n

det(A−In)^2 ≤ 0 ,

and the inequality is proved in this case, too.
(Romanian mathematics competition, 1987)

221.All the information about the inverse ofAis contained in its determinant. If we
compute the determinant ofAby expanding along thekth column, we obtain a polynomial
inxk, and the coefficient ofxmk−^1 is exactly the minor used for computing the entrybkm
of the adjoint matrix multiplied by(− 1 )k+m. Viewing the product

∏

i>j(xi−xj)as a
polynomial inxk, we have
∏

i>j

(xi−xj)=    (x 1 ,...,xk− 1 ,xk+ 1 ,...,xn)×(xk−x 1 )···(xk−xk− 1 )

×(xk+ 1 −xk)...(xn−xk)

=(− 1 )n−k  (x 1 ,...,xk− 1 ,xk+ 1 ,...,xn)×

∏

j   =k

(xk−xj).

In the product

∏

j   =k(xk−xj)the coefficient ofx

m− 1

k is

(− 1 )n−mSn−m(x 1 ,...,xk− 1 ,xk+ 1 ,...,xn).

Combining all these facts, we obtain

bkm=(− 1 )k+m   (x 1 ,x 2 ,...,xn)−^1 (− 1 )k+m(− 1 )n−k(− 1 )n−m

×   (x 1 ,...,xk− 1 ,xk+ 1 ,...,xn)Sm(x 1 ,...,xk− 1 ,xk+ 1 ,...,xn)

=(− 1 )k+m  (x 1 ,x 2 ,...,xn)−^1   (x 1 ,...,xk− 1 ,xk+ 1 ,...,xn)

×Sm(x 1 ,...,xk− 1 ,xk+ 1 ,...,xn),

as desired.

222.The inverse of a 2×2 matrixC=(cij)i,jwith integer entries is a matrix with integer
entries if and only if detC=±1 (one direction of this double implication follows from
the formula for the inverse, and the other from detC−^1 = 1 /detC).
With this in mind, let us consider the polynomialP(x)∈Z[x],P(x)=det(A+xB).
The hypothesis of the problem implies thatP( 0 ), P ( 1 ), P ( 2 ), P ( 3 ), P ( 4 )∈{− 1 , 1 }.By
the pigeonhole principle, three of these numbers are equal, and becauseP(x)has degree
at most 2, it must be constant. Therefore, det(A+xB)=±1 for allx, and in particular
forx=5 the matrixA+ 5 Bis invertible and has determinant equal to±1. Consequently,
the inverse of this matrix has integer entries.
(55th W.L. Putnam Mathematical Competition, 1994)