Prova 1 + resolução de Algebra Linear

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Álgebra Linear – P1 
 
 
Favor não compartilharem com nerds tetudos e similares. 
 
 
 
 
 
 
 
 
 
 
 
1) V = { (x,y,z) ∈ R³ : x + y – z = 0 } 
W = { (x,y,z) ∈ R³ : x = y } 
 
Para V: 
 X= -y+z 
 
 V = (-y+z,y,z) 
 
 V = y(-1,1,0) + z(1,0,1)  duas dimensões 
 
β V = { (-1,1,0) , (1,0,1) } 
 
 
Para W: 
 W = (x,y,z) 
 
 W = y(1,1,0) + z(1,0,1) 
 
 β W = { (1,1,0) , (0,0,1) } 
 
Para V+W: 
 
 
 
-1 1 0 
 
-1 1 0 L1+L2 0 1 0 
 
0 1 0 
1 0 1 L2-L4 1 0 0 
 
1 0 0 
 
1 0 0 
1 1 0 
 
1 1 0 
 
1 1 0 L3-L4-L1 0 0 0 
0 0 1 
 
0 0 1 
 
0 0 1 
 
0 0 1 
 
β V+W = { (1,0,0),(0,0,1) } possuí 3 dimensões. 
 
 
Para V ∩ W: 
 
V ∩ W = { (x,y,z) PERTENCE R³ : x+y-z = 0, x = y } 
 
| x = y 
| x+x-z = 0  z = 2x 
 
V ∩ W = { (x,x,2x) PERTENCE R³ : x PERTENCE R }  { x ( 1,1,2) }  
 
β V ∩ W = (1,1,2) 
 
 
Adicionar a dimensão, e os vetores resultantes no fim do ex
 
2) 
a) 
1 0 0 0 
 
1 
0 1 0 0 P = Vetor (1,2,0,1) = 2 
0 0 1 0 
 
0 
0 0 0 1 
 
1 
 
 
 
 
 
 
b) 
 
1 0 0 0 
1 1 0 0 
1 0 -1 0 
1 0 0 -1 
 
β P = (1,2,0,1) = a * (1,0,0,0) + b * (1,1,0,0) + c * (1,0,-1,0) + d * (1,0,0,-1) 
 
a + b + c + d = 
1 
 
a + 2 + 0 + (-1) = 1 
b = 2 
 
a = 1 - 2 + 1 
-c = 0 -> c = 0 
 
a = 0 
-d = 1 -> d = -1 
 
V = ( 0 , 2 , 0 , -1 ) 
 
1 0 0 0 
 
0 
1 1 0 0 P = Vetor (1,2,0,1) = 2 
1 0 -1 0 
 
0 
1 0 0 -1 
 
-1 
 
 
 
 
 
 
 
 
 
 
3) 
B1 = { (2,-1) , (3,4) } 
B2 = { (1,0) , (0,1) }  bases de R² 
 
 [ ]
 
 
 = 
 
(2,-1) = α(2,-1) + β(3,4) 
(3,1) = γ(2,-1) + δ(3,4) 
 
 | 2 = 2α + 3β  | 2 = 2α + 3β  α = 1 
-2 * |-1 = =α + 4β  |-2 = 2α + 8β  β = 0 
 
 | 3 = 2γ + 3δ  3 = 2γ + 3δ  γ = 0 
2 * | 4 = -γ + 4δ  8 = -2γ + 8δ  δ = 1 
 
[ ]
 
 
 = | α β | = | 1 0 | 
 | γ δ | | 0 1 | 
 
 
 
 [ ]
 
 
 = 
 
(1,0) = α(2,-1) + β(3,4) 
(0,1) = γ(2,-1) + δ(3,4) 
 
 | 1 = 2α + 3β  1 = 2α + 3β  α = 
 
 
 
 | 
2 * | 0 = -α + 4β  0 = -2α + 8β  β = 
 
 
 
 
 
 | 0 = 2γ + 3δ  0 = 2γ + 3δ  γ = 
 
 
 
 | 
2 * | 1 = -γ + 4δ  2 = -2γ + 8δ  δ = 
 
 
 
 
 
[ ]
 
 
 = | 
 
 | = | 
 
 
 
 
 
 
 
 
 | 
 
 
 
 [ ]
 
 
 = 
 
(2,-1) = α(1,0) + β(0,1) 
(3,4) = γ(1,0) + δ(0,1) 
 
| 2 = α 
| -1 = β 
 
| 3 = γ 
| 4 = δ 
 
 [ ]
 
 
 = | 
 
 
 | 
 
 
 
 
 
 
 
 
 
 
 
 
4) 
β = { V1 , V2 , .. Vn } ∈ V 
 
VETOR U ∈ V  ∃ α1, α2...,αn ∈ R 
VETOR V ∈ V  ∃ n , β1,β2,...,βn ∈ R 
 
VETOR U = α1V1 + α2V2 + ... + αnVn 
VETOR V = β1V1 + β2V2 + ... + βnVn 
 
β é LI, portanto é β, então VETOR U e VETOR V podem ser escritos como combinação 
linear ÚNICA de { V1, V2, ... Vn} 
 
.:. VETOR u = VETOR v => VETOR u = VETOR v = VETOR NULO 
 
VETOR1(α1-β1)+ VETOR2 (α2-β2) +...+ VETORn (αn-βn) = VETOR NULO 
 
α1-β1 = α2- β2 = α3 – β3 = VETOR NULO = combinação linear unica 
 
5) 
V = R³ U ∩ V = {0} (deve ser provado) 
U = [ (1,0,0) ] 
W = [ (1,1,0) , (0,1,1) ] 
 
VETOR V ∈ U ∩ W < = > VETOR e U e VETOR V ∈ W < = >α, β, γ ∈ R 
 
U ∩ W => U = W => α(1,0,0) = β (1,1,0) + γ(0,1,1) 
 α(1,0,0) – β(1,1,0) , γ(0,1,1) = 0 
 
α – β = 0 => α = 0 
β – γ = 0 => β = 0 => .:. α = β = γ = 0 => U ∩ W = { 0 } 
γ = 0