Exam (elaborations)
Ejercicios resueltos Álgebra
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Apuntes sobre Matemáticas
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Control 1 [1AT1-1AT2] AiG/EETAC/UPC 11-10-181. Considerem els polinomis p(z) = z 5 + 7z 4 + 11z 3 − 9z 2 + 54 i q(z) = 4z 6 − 9z 4 + 4z 2 − 9.
(a){20pt} Troba totes les arrels complexes de p(z) i expresseu-les tant en forma binòmica com en forma exponencial.
√ π √ π
p(z) = (z + 3)3 · (z 2 − z + 2) = 0 ⇒ z ∈ {−3, 1 + j, 1 − j} ⇔ z ∈ {3eπj , 2e 4 j , 2e− 4 j }
(b){20pt} Dona la factorització de q(z) en factors irreductibles a C[x] i a R[x].
πj 5πj
[pista: e 4 , e 4 i − 32 són tres de les seves arrels.]
q(z) = (2z + 3) · (2z − 3) · (z 4 + 1)
√ √ √ √ √ √ √ √
2 2 2 2 2 2 2 2
q(z) = (2z + 3) · (2z − 3) · (z − 2 − 2 j) · (z − 2 + 2 j) · (z + 2 + 2 j) · (z + 2 − 2 j)
√ √
2 2 1 2 2 1
q(z) = (2z + 3) · (2z − 3) · ((z − 2 ) + 2 ) · (z + 2 ) + 2)
√ √
q(z) = (2z + 3) · (2z − 3) · (z 2 − 2z + 1) · (z 2 + 2z + 1)
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2. Considerem la familia uniparamètrica de sistemes d’equacions lineals homogènies A · x = o d’ordre 4x4 i paràmetre λ on:
0 −4 0 −1 x 0
0 2 0 0 y
, o= 0
C= , A = C − λ · I4 , x=
0 0 0 0 z 0
4 8 −12 4 t 0
(a){20pt} Quan λ = 2, classifica i resol aquest s.e.l.h.
−2 −4 0 −1
0 0 0 0
λ = 2 ⇒ A=
0 0 −2 0
4 8 −12 2
r(A) = 2 ⇒ Compatible Indeterminat, amb dos graus de llibertat.
Solució general: (x, y, z, t) = (λ, µ, 0, −2λ − 4µ)
Subespai vectorial de solucions: W = h{(1, 0, 0, −2), (0, 1, 0, −4)}i
(b){20pt} Esbrina per quins valors del paràmetre λ aquest s.e.l.h. és compatible indeterminat.
−λ −4 0 −1
0 2−λ 0 0
= λ4 − 6λ3 + 12λ2 − 8λ = λ(λ − 2)3
0 0 −λ 0
4 8 −12 4 − λ
Per tant, aquest s.e.l.h. és compatible indeterminat si i només si λ ∈ {0, 2}.
——————————————————————————————————————————————————
3. Considerem els subespais vectorials de R3 :
U = {(x, y, z) ∈ R3 : x + y − 2z = 0, 2x + y − 3z = 0} ,
V = h{~v1 , ~v2 }i, on: ~v1 = (2, 4, 3), ~v2 = (3, −1, 1), i
W = {(x, y, z) ∈ R3 : 2x + y − z = 0} ,
(a){20pt} Calcula la dimensió, les equacions implı́cites i una base dels subespais vectorials U , V , U + V i U ∩ V .
U = h(1, 1, 1)i V = {(x, y, z) ∈ R3 : x + y − 2z = 0 Per tant, U ⊂ V .
Per tant, U ∩ V = U , U + V = V .
(b){20pt} Esbrina si el subespais vectorials U i W són suplementaris.
W = h{(1, 0, 2), (0, 1, 1)}i
1 1 1
1 0 2 6= 0 ⇒ dim(U + W ) = 3 ⇒ dim(U ∩ W ) = 0 ⇒ R3 = U ⊕ V
0 1 1
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