Page 109 - MATINF Nr. 9-10

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PROBLEME DE MATEMATICA PENTRU CONCURSURI 109

 1010
 −b + c = 2
adic˘a −b + c + i(2a − b) = −2 1010 (−1 − i). Obt , inem sistemul 2a − b = 2 1010 , avˆand
4a + 2b + c = 2
 2021
2 2020 + 2 1010 2 · 2 2020 − 3 · 2 1010 2 · 2 2020 + 2 · 2 1010
solut , ia a = , b = , c = .
5 5 5
Ñ é
0 −2 −6
2
2
Avem A 2021 = aA + bA + cI 3 , iar A = 4 4 4 , deci
−2 −2 0
Ñ 2020 1010 2020 1010 2020 1010 é
4 · 2 − 2 2 · 2 − 8 · 2 −4 · 2 − 9 · 2
1
A 2021 = 4 · 2 2020 + 4 · 2 1010 2 · 2 2020 + 12 · 2 1010 −4 · 2 2020 + 16 · 2 1010 .
5 2020 1010 2020 1010 2020 1010
−4 · 2 + 2 −2 · 2 − 2 · 2 4 · 2 − 2
à í
a b c d 0
k l m 0 −d
M 176. Fie matricea A = n p 0 −m −c ∈ M 5 (Z).
q 0 −p −l −b
0 −q −n −k −a
a) Ar˘atat ,i c˘ dac˘ produsul abcdklmnpq nu se divide cu 4, atunci rang (A) = 4.
a
a
b) R˘amˆane afirmat ,ia adev˘arat˘ dac˘ produsul abcdklmnpq nu se divide cu 8?
a
a
Stelian Corneliu Andronescu s , i Costel B˘alc˘au, Pites , ti

Solut ,ie. a) Avem

a k n q 0 −a −k −n −q 0

b l p 0 −q −b −l −p 0 q

t

det(A) = det(A ) = c m 0 −p −n = (−1) 5 −c −m 0 p n .

d 0 −m −l −k −d 0 m l k

0 −d −c −b −a 0 d c b a

0 d c b a

−d 0 m l k

a
a
Interschimbˆand L 1 cu L 5 s , i L 2 cu L 4 rezult˘ c˘ det(A) = − −c −m 0 p n .

−b −l −p 0 q

−a −k −n −q 0

a b c d 0

k l m 0 −d

a
a
Interschimbˆand acum C 1 cu C 5 s , i C 2 cu C 4 rezult˘ c˘ det(A) = − n p 0 −m −c .

q 0 −p −l −b

0 −q −n −k −a
Am obt , inut c˘a det(A) = − det(A), deci det(A) = 0 s , i astfel rezult˘ c˘a rang (A) ≤ 4.
a

k l m 0

n p 0 −m
2

Pe de alt˘ parte, minorul (1, 5) al matricei A este δ 1,5 = = k(kp +pmq−
a
q 0 −p −l

0 −q −n −k

2
2
2
2
pnl)−l(knp+mnq−ln )+m(mq −lnq+kpq) = (kp−ln+mq) . Analog, δ 2,4 = (ap−bn+cq) ,
2
2
2
δ 3,3 = (al − bk + dq) , δ 4,2 = (am − ck + dn) , δ 5,1 = (bm − cl + dp) .