Page 84 - MATINF Nr. 6

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84 PROBLEME DE MATEMATICA PENTRU CONCURSURI

Clasa a XI-a

„ Ž
2020 a −ab
M 111. Se consider˘a matricea X(a, b) = 0 2019 b , unde a, b ∈ C.
0 0 2020

„ Ž
1 x y
a) Ar˘atat ,i c˘a exist˘a o infinitate de matrice P ∈ M 3 (C) de forma P = 0 1 z astfel
0 0 1
ˆıncˆat P · X(a, b) = X(0, 0) · P.

n ∗
b) Calculat ,i (X(a, b)) , unde n ∈ N .
Stelian Corneliu Andronescu s , i Costel B˘alc˘au, Pites , ti

„ Ž
2020 a + 2019x −ab + bx + 2020y
Solut ,ie. a) Avem P · X(a, b) = X(0, 0) · P ⇔ 0 2019 b + 2020z =
0 0 2020
„ Ž  „ Ž
2020 2020x 2020y  x = a 1 a y
0 2019 2019z ⇔ z = −b ⇔ P = 0 1 −b , cu y ∈ C.
0 0 2020  y ∈ C 0 0 1

„ Ž
1 a 0
b) Conform a) avem X(a, b) = P −1 · X(0, 0) · P, cu P = 0 1 −b . Rezult˘a c˘a
0 0 1
„ Ž „ n Ž „ Ž
1 −a −ab 2020 0 0 1 a 0
n −1 n n
(X(a, b)) = P ·(X(0, 0)) ·P = 0 1 b · 0 2019 0 · 0 1 −b
0 0 1 0 0 2020 n 0 0 1
„ n n n n n Ž
2020 a (2020 − 2019 ) −ab (2020 − 2019 )
n
n
= 0 2019 n b (2020 − 2019 ) .
0 0 2020 n
M 112. Fie matricele A, B ∈ M 3 (C) astfel ˆıncˆat AB = BA, det(A + B) = det A + det B s , i
det(A − B) = det A − det B. Ar˘atat ,i c˘a:

3
3
3
a) det (A + B ) = (det A + det B) ;
3
3
3
b) det (A − B ) = (det A − det B) .
Daniel Jinga, Pites , ti
2
Solut ,ie (Daniel V˘acaru, Pites , ti). Fie polinomul P (x) = det (A + xB) = det A + ax + bx +
3
(det B)x . Observ˘am c˘a P (1) = det (A + B) = det A + a + b + det B, deci a + b = 0, iar
P (−1) = det (A − B) = det A − a + b − det B, deci a − b = 0. Rezult˘a c˘a a = b = 0, deci
3
P (x) = det (A + xB) = det A + x det B.
3
a) Consider˘am ecuat , ia x − 1 = 0, cu r˘ad˘acinile x 1 = 1, x 2 s , i x 3 . Folosind Formulele
lui Vi`ete x 1 + x 2 + x 3 = 0, x 1 x 2 + x 2 x 3 + x 3 x 1 = 0, x 1 x 2 x 3 = 1 s , i ipoteza AB = BA,
3
2
2
avem (A + x 1 B) (A + x 2 B) (A + x 3 B) = A +(x 1 + x 2 + x 3 ) A B +(x 1 x 2 + x 2 x 3 + x 3 x 1 ) AB +