Solución_Control_N_6_-_Sección_C

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TIPO A 
1. Halle el vector de coordenadas del polinomio 𝑝(𝑡) = 7 − 𝑡2 + 4𝑡3 respecto de la base 
𝐵 = {1 + 2𝑡 − 𝑡2 + 4𝑡3, 𝑡2, 3 − 𝑡, 1} (Aplique isomorfismo vectorial) 
 
Aplicando isomorfismo vectorial: 𝑃𝐵�̅�𝐵 = �̅� 
 
𝑃𝐵 = [
4 0
−1 1
0 0
0 0
2 0
1 0
−1 0
3 1
] 
 
�̅� = [
4
−1
0
7
] 
[
4 0
−1 1
0 0
0 0
2 0
1 0
−1 0
3 1
| 
4
−1
0
7
]𝐻1(
1
4
) [
1 0
−1 1
0 0
0 0
2 0
1 0
−1 0
3 1
| 
1
−1
0
7
] 
𝐻21(1)
𝐻31(−2)
𝐻41(−1)
[
1 0
0 1
0 0
0 0
0 0
0 0
−1 0
3 1
| 
1
0
−2
6
] 
 
𝐻3(−1) [
1 0
0 1
0 0
0 0
0 0
0 0
1 0
3 1
| 
1
0
2
6
]𝐻34(−3) [
1 0
0 1
0 0
0 0
0 0
0 0
1 0
0 1
| 
1
0
2
0
] 
 
 
�̅�𝐵 = [
1
0
2
0
] → 𝒕𝟑 + 𝟐𝒕 
 
 
2. Encontrar la base 𝐵 de 𝑅2 tal que el vector de coordenadas relativo a la base 𝐵 de 
𝑥̅ = (2, −1) es 𝑥̅ 𝐵 = (2,1) y el vector de coordenadas relativo a la base 𝐵 de �̅� = (1,−2) 
es �̅�𝐵 = (1,1). 
 
𝑃𝐵 = [
𝑎 𝑏
𝑐 𝑑
] 
 
𝑃𝐵�̅�𝐵 = �̅� [
𝑎 𝑏
𝑐 𝑑
] [
2
1
] = [
2
−1
] 
 
𝑃𝐵�̅�𝐵 = �̅� [
𝑎 𝑏
𝑐 𝑑
] [
1
1
] = [
1
−2
] 
 
 
𝑃𝐵 = [
1 0
1 −3
] 
 
 
 
 
 
 
 
3. Encontrar la matriz de cambio de base “M” de B a C. Si: 
 
𝐵 = {[
1
1
] , [
2
3
]} 
 
𝐶 = {[
1
0
] , [
2
5
]} 
 
𝑃𝐶 = [
1 2
0 5
] → 𝑃𝐶
−1 = [
1 −2/5
0 1/5
] 
 
𝑀 = 𝑃𝐶
−1𝑃𝐵 = [
1 −2/5
0 1/5
] [
1 2
1 3
] = [
𝟑/𝟓 𝟒/𝟓
𝟏/𝟓 𝟑/𝟓
] 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
TIPO B 
1. Encontrar la matriz de cambio de base “N” de C a B. Si: 
 
𝐵 = {[
1
1
] , [
0
2
]} 
 
𝐶 = {[
1
2
] , [
3
4
]} 
 
 
𝑃𝐵 = [
1 0
1 2
] → 𝑃𝐵
−1 = [
1 0
−1/2 1/2
] 
 
𝑁 = 𝑃𝐵
−1𝑃𝐶 = [
1 0
−1/2 1/2
] [
1 3
2 4
] = [
𝟏 𝟑
𝟏/𝟐 𝟏/𝟐
] 
 
 
2. Halle el vector de coordenadas del polinomio 𝑝(𝑡) = 9 + 5𝑡2 + 4𝑡3 respecto de la base 
𝐵 = {1 + 2𝑡 − 𝑡2 + 4𝑡3, 𝑡2, 3 − 𝑡, 1} (Aplique isomorfismo vectorial) 
 
𝑃𝐵 = [
4 0
−1 1
0 0
0 0
2 0
1 0
−1 0
3 1
] 
 
�̅� = [
4
5
0
9
] 
[
4 0
−1 1
0 0
0 0
2 0
1 0
−1 0
3 1
| 
4
5
0
9
]𝐻1(
1
4
) [
1 0
−1 1
0 0
0 0
2 0
1 0
−1 0
3 1
| 
1
5
0
9
] 
𝐻21(1)
𝐻31(−2)
𝐻41(−1)
[
1 0
0 1
0 0
0 0
0 0
0 0
−1 0
3 1
| 
1
6
−2
8
] 
 
𝐻3(−1) [
1 0
0 1
0 0
0 0
0 0
0 0
1 0
3 1
| 
1
6
2
8
]𝐻34(−3) [
1 0
0 1
0 0
0 0
0 0
0 0
1 0
0 1
| 
1
6
2
2
] 
 
 
�̅�𝐵 = [
1
6
2
2
] → 𝒕𝟑 + 𝟔𝒕𝟐 + 𝟐𝒕 + 𝟐 
 
 
 
 
 
 
 
 
 
 
3. Encontrar la base 𝐵 de 𝑅2 tal que el vector de coordenadas relativo a la base 𝐵 de 
𝑥̅ = (2, −1) es 𝑥̅ 𝐵 = (2,1) y el vector de coordenadas relativo a la base 𝐵 de 
�̅� = (1,−2) es �̅�𝐵 = (1,1). 
 
𝑃𝐵 = [
𝑎 𝑏
𝑐 𝑑
] 
 
𝑃𝐵�̅�𝐵 = �̅� [
𝑎 𝑏
𝑐 𝑑
] [
2
1
] = [
2
−1
] 
 
𝑃𝐵�̅�𝐵 = �̅� [
𝑎 𝑏
𝑐 𝑑
] [
1
1
] = [
1
−2
] 
 
 
𝑷𝑩 = [
𝟏 𝟎
𝟏 −𝟑
]