Advanced book on Mathematics Olympiad

410 Algebra ( 2 cos π 5 ,2 cos π 5 ,2 cos 3 π 5 ,2 cos 3 π 5 ,2 cosπ ) . (Romanian Mathematical Olympiad, 2002, proposed by T. A ...

Algebra 411 −( 2 n+ 1 )x dn dxn ( 1 −x^2 )n− (^12) −( 2 n+ 1 ) ( n 1 ) d dx x dn−^1 dxn−^1 ( 1 −x^2 )n− (^12) =−( 2 n+ 1 )x dn d ...

412 Algebra Remark.This decomposition plays a special role, especially for linear operators on infinite-dimensional spaces. IfAa ...

Algebra 413 (p+q)A^2 B^2 =ABand(p+q)B^2 A^2 =BA. We have seen thatA^2 andB^2 commute, and so we find thatAandBcommute as well, w ...

414 Algebra 204.First solution: Computed by hand, the second, third, and fourth powers ofJ 4 (λ)are ⎛ ⎜ ⎜ ⎝ λ^22 λ 10 0 λ^22 λ 1 ...

Algebra 415 we see that we could almost use the equality from the statement, but the factors in two terms come in the wrong orde ...

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 =det ...

Algebra 417 The Vandermonde determinant has the value ∏ i>j(xi−xj). It follows that our determinant is equal to ∏ i>j(xi−x ...

418 Algebra det(A)= ∣∣ ∣∣ ∣ ∣∣ ∣∣ ∣∣ 100 ··· 00 010 ··· 00 .. . .. . .. . ... .. . .. . 000 ··· 10 ± 1 ∓ 2 ± 3 ··· −n+ 1 n+ 1 ∣∣ ...

Algebra 419 214.We reduce the problem to a computation with 4×4 determinants. Expanding according to the rule of Laplace, we see ...

420 Algebra and the latter is nonnegative, as seen in the introduction. Taking the limit withap- proaching zero, we obtain det( ...

Algebra 421 withak,bk∈R. It follows that detQ(X)= ∏m k= 1 det [ (X+ak)^2 +b^2 kIn ] ≥ 0 , and the latter is positive, since for ...

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 prov ...

Algebra 423 223.We know thatAA∗=A∗A=(detA)I 3 ,soifAis invertible then so isA∗, and A=detA(A∗)−^1. Also, detAdetA∗=(detA)^3 ; he ...

424 Algebra Let α=max i ⎛ ⎝ ∑n j= 1 |aij| ⎞ ⎠< 1. Then ∑ k ∣ ∣∣ ∣∣ ∣ ∑ j aijajk ∣ ∣∣ ∣∣ ∣ ≤ ∑ j,k |aijajk|= ∑ j ( |aij| ∑ k | ...

Algebra 425 ∑n k= 1 cos(m±j)kα= (− 1 )m+j+^1 2 − 1 2 . It follows that form =j, the(m, j )entry of the matrixA^2 is zero. Hen ...

426 Algebra Second solution: Simply write In=kAk+^1 −(k+ 1 )Ak+In=(A−In)(kAk−Ak−^1 −···−A−In), which gives the inverse written i ...

Algebra 427 3 (x 1 +x 2 +x 3 )+ 3 (x 4 +x 5 +x 6 )+···+ 3 (x 97 +x 98 +x 99 )+ 3 x 100 = 0. The terms in the parentheses are all ...

428 Algebra = ∣ ∣∣ ∣∣ ∣∣ ∣ −y− 10 0 − 1 0 −y 1 − 1 − 11 −y− 1 − 1 −10 0−y ∣ ∣∣ ∣∣ ∣∣ ∣ , which, after expanding with the rule of ...

Algebra 429 (a 11 +a 12 +a 13 )x 1 +(a 12 +a 13 )m+a 13 n= 0 , (a 31 +a 32 +a 33 )x 1 +(a 32 +a 33 )m+a 33 n= 0. The hypothesesa ...