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• PARCIAL III
TALLER N° 1
TEMA.-
 TRANSFORMACIONES
LINEALES
 DEPARTAMENTO DE
CIENCIAS EXCATAS
ALGEBRA LINEAL
INTEGRANTES .-
GALARZA PILAGUANO ROSA LISBETH.
GUANOLUISA GUALOTUÑA RICHARD
IVAN.
QUINATOA ZAPATA WILSON PAUL.
NRC.-
4261
FECHA: 4/03/2021
PERIODO: NOVIEMBRE 2020_ABRIL
2021
 INDICE
 Objetivos…………………………...…...……….4
 Desarrollo…………….………………………….5
 Ejercicio 1.1………………………………….5
 Ejercicio 2.2………………………………...6
 Ejercicio 3.3…………………………………7
 Ejercicio 4.4…………………………………9
 Ejercicio 5.5………………………………..11
 OBJETIVOS.-
 Aprender sobre las definiciones de transformación
lineal y conocer los conceptos fundamentales, tales
como núcleo e imagen de una transformación lineal
otro de los aspectos importantes es saber sobre su
utilidad la cuál es muy importante dentro de las
matemáticas, geometría, análisis, etc.
 DESARROLLO
 EJERCICIO 1.1
 1.1 𝒇 𝒙, 𝒚 = 𝟑 (𝒙 − 𝒚, 𝒙 + 𝒚)

 𝑻 = ℝ𝟐
→ ℝ𝟐

 𝑢 = 𝑥1
𝑦1
𝑣 = 𝑥2
𝑦2
 𝑇 = 𝛼𝑢 + 𝛽𝑣 =
𝛼𝑥1 𝛽𝑥2
𝛼𝑦1 𝛽𝑦2
 =
𝛼𝑥1 + 𝛽𝑥2 − 𝛼𝑦1 − 𝛽𝑦2
𝛼𝑥1 + 𝛽𝑥2 + 𝛼𝑦1 + 𝛽𝑦2
 = 3
(𝛼𝑥1+𝛽𝑥2 − 𝛼𝑦1 − 𝛽𝑦2)
(𝛼𝑥1+𝛽𝑥2 + 𝛼𝑦1 + 𝛽𝑦2)
 =
3𝛼𝑥1 + 3𝛽𝑥2 − 3𝛼𝑦1 − 3𝛽𝑦2
3𝛼𝑥1 + 3𝛽𝑥2 + 3𝛼𝑦1 + 3𝛽𝑦2
 =
3𝛼(𝑥1−𝑦1) + 3𝛽(𝑥2 − 𝑦2)
3𝛼(𝑥1 + 𝑦1) + 3𝛽(𝑥2 + 𝑦2)
 = 3𝛼 𝑥1−𝑦1
𝑥1+𝑦1
+ 3𝛽 𝑥2−𝑦2
𝑥2+𝑦2
 𝑺𝒊 𝒅𝒆𝒇𝒊𝒏𝒆 𝒖𝒏 𝑻. 𝑳
 DESARROLLO
 EJERCICIO 2.2.-
 2.2 𝒇 𝒙, 𝒚 = (𝒙, 𝒚, 𝒛𝟐
)
 𝑻 = ℝ𝟐
→ ℝ𝟐

 𝑢 =
𝑥1
𝑦1
𝑧1
𝑣 =
𝑥2
𝑦2
𝑧2
 𝑇 = 𝛼𝑢 + 𝛽𝑣 =
𝛼𝑥1 𝛽𝑥2
𝛼𝑦1 𝛽𝑦2
𝛼𝑧1 𝛽𝑧2
 =
𝛼𝑥1 + 𝛽𝑥2
𝛼𝑥1 + 𝛽𝑥2
𝛼𝑥1 + 𝛽𝑧2
 =
𝛼𝑥1 + 𝛽𝑥2
𝛼𝑥1 + 𝛽𝑥2
(𝛼𝑥1 + 𝛽𝑧2)2
 =
𝛼𝑥1 + 𝛽𝑥2
𝛼𝑥1 + 𝛽𝑥2
(𝛼𝑧1)2
+2 𝛼𝑧1 𝛽𝑧2 + (𝛽𝑧2)2
 = 𝛼
𝑥1
𝑦1
𝑧1
+ 𝛼𝛽
𝑧1
𝑧2
𝛽
𝑥2
𝑦2
𝑧2
 𝑵𝒐 𝒅𝒆𝒇𝒊𝒏𝒆 𝒖𝒏𝒂 𝑻. 𝑳
 DESARROLLO
 EJERCICIO 3.3
 3.3 𝒇 𝒙, 𝒚, 𝒛 = 𝒙 + 𝟐𝒚 − 𝟑𝒛, 𝟑𝒙 − 𝒚 + 𝟓𝒛, 𝒙 − 𝒚 − 𝒛
 𝑇 =
𝑥1
𝑦1
𝑧1
+
𝑥2
𝑦2
𝑧2
 𝑢 =
𝑥1
𝑦1
𝑧1
𝑣 =
𝑥2
𝑦2
𝑧2
 𝑇 =
𝛼𝑥1 + 𝛽𝑥2
𝛼𝑦1 + 𝛽𝑦2
𝛼𝑧1 + 𝛽𝑧2
 𝒙 + 𝟐𝒚 − 𝟑𝒛
 (𝛼𝑥1 + 𝛽𝑥2 + 2 𝛼𝑦1 + 𝛽𝑦2 − 3(𝛼𝑧1 + 𝛽𝑧2)
 (𝛼𝑥1 + 𝛽𝑥2 + 2𝛼𝑦1 + 2𝛽𝑦2 − 3𝛼𝑧1 − 3𝛽𝑧2
 𝛼 𝑥1 + 2𝑦1 − 3𝑧1 + 𝛽 𝑥2 + 2𝑦2 − 3𝑧2
 𝟑𝒙 − 𝒚 + 𝟓𝒛
 3 𝛼𝑥1 + 𝛽𝑥2 − 𝛼𝑦1 + 𝛽𝑦2 + 5 𝛼𝑧1 + 𝛽𝑧2
 3 𝛼𝑥1 + 3𝛽𝑥2 − 𝛼𝑦1 − 𝛽𝑦2 + 5𝛼𝑧1 + 5𝛽𝑧2
 𝛼 3 𝑥1 − 𝑦1 + 5𝑧1 + 𝛽(3𝑥2 − 𝑦2 + 5𝑧2)
EJERCICIO 3.3
 𝒙 − 𝒚 − 𝒛
 𝛼𝑥1 + 𝛽𝑥2 − 𝛼𝑦1 + 𝛽𝑦2 − 𝛼𝑧1 + 𝛽𝑧2
 𝛼𝑥1 + 𝛽𝑥2 − 𝛼𝑦1 − 𝛽𝑦2 − 𝛼𝑧1 − 𝛽𝑧2
 𝛼 𝑥1 − 𝑦1 − 𝑧1 + 𝛽 𝑥2 − 𝑦2 − 𝑧2

 𝛼 𝑥1 + 2𝑦1 − 3𝑧1 + 𝛽 𝑥2 + 2𝑦2 − 3𝑧2
 𝛼 3 𝑥1 − 𝑦1 + 5𝑧1 + 𝛽(3𝑥2 − 𝑦2 + 5𝑧2)
 𝛼 𝑥1 − 𝑦1 − 𝑧1 + 𝛽 𝑥2 − 𝑦2 − 𝑧2

 𝛼
𝑥1 + 2𝑦1 − 3𝑧1
3 𝑥1 − 𝑦1 + 5𝑧1
𝑥1 − 𝑦1 − 𝑧1
+ 𝛽
𝑥2 + 2𝑦2 − 3𝑧2
3𝑥2 − 𝑦2 + 5𝑧2
𝑥2 − 𝑦2 − 𝑧2
 𝑆𝑖 𝑒𝑠 𝑇. 𝐿.
 Ejercicio 4.4
 4.4 Sea f una transformación lineal de ℝ𝟑
→ ℝ𝟑
, suponga que
𝒇 𝟏, 𝟎, 𝟏 = 𝟏, −𝟏, 𝟑 𝒚 𝒇 𝟐, 𝟏, 𝟎 =
𝟎, 𝟐, 𝟏 ; 𝑫𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒆 𝒇 −𝟏, −𝟐, 𝟑

 𝑇
1
0
1
=
1
−1
3
; 𝑇
2
1
0
=
0
2
1

𝑥
𝑦
𝑧
= 𝛼
1
0
1
+ 𝛽
2
1
0
 𝑇(𝛼𝑢 + 𝛽𝑣) = 𝛼𝑇(𝑢) + 𝛽𝑇(𝑣)
 𝑥 = 𝛼 + 2𝛽 𝑦 = 𝛽 𝑧 = 𝛼
 𝛼 = 𝑥 − 2𝑦
 𝛽 =
𝑥 − 𝑧
2
 Ejercicio 4.4
 𝑇
𝑥
𝑦
𝑧
= 𝑇 𝛼
1
0
1
+ 𝛽
2
1
0
= 𝛼𝑇
1
0
1
+ 𝛽𝑇
2
1
0
 𝑇
𝑥
𝑦
𝑧
=
𝑥 − 𝑧
2
1
0
1
+ 𝑥 − 2𝑦
2
1
0
 𝑇
𝑥
𝑦
𝑧
=
5𝑥 − 8𝑦 − 𝑧
2
𝑥 − 2𝑦
𝑥 − 𝑧
2
 𝑇
−1
−2
3
=
5𝑥 − 8𝑦 − 𝑧
2
𝑥 − 2𝑦
𝑥 − 𝑧
2
 𝑇
−1
−2
3
=
5(−1) − 8(−2) − (3)
2
(−1) − 2(−2)
(−1) − (3)
2
𝑇
−1
−2
3
==
4
3
−2
 EJERCICIO 5.5
• Sea 𝒇 𝒖𝒏𝒂 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒄𝒊ó𝒏 𝒍𝒊𝒏𝒆𝒂𝒍 𝒅𝒆 ℝ𝟑
𝒆𝒏 𝑷𝟐 𝒕𝒂𝒍 𝒒𝒖𝒆 𝒇((𝟏, 𝟏, 𝟏)) = 𝟏 – 𝟐𝒕 + 𝒕𝟐
,
𝒇((𝟐, 𝟎, 𝟎)) = 𝟑 + 𝒕 – 𝒕𝟐, 𝒇((𝟎, 𝟒, 𝟓)) = 𝟐 + 𝟑𝒕 +
𝟑𝒕𝟐; 𝑫𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒆 𝒇((𝟐, −𝟑, 𝟏)).
 𝑇: ℝ3 → 𝑃2
 𝑇𝑢
1
1
1
=
1
−2
1
𝑇𝑣
2
0
0
=
3
1
−1
𝑇𝑤
0
4
5
=
2
3
3

𝑥
𝑦
𝑧
= 𝛼
1
1
1
+ 𝛽
2
0
0
+ 𝛿
0
4
5
 𝛼 + 2𝛽 = 𝑥
 𝛼 + 4𝛿 = 𝑦
 𝛼 + 5𝛿 = 𝑧
 Ejercicio 5.5
 𝐴 =
1
1
1
2 0
0 4
0 5
𝑥
𝑦
𝑧
𝐹2−𝐹1→ 𝐹2 𝑦 𝐹3−𝐹1→ 𝐹3
 𝐴 =
1
0
0
2 0
−2 4
−2 5
𝑥
𝑦 − 𝑥
𝑧 − 𝑥
𝐹3−𝐹2→ 𝐹3
 𝐴 =
1
0
0
2 0
−2 4
0 1
𝑥
𝑦 − 𝑥
𝑧 − 𝑦
 𝛿 = 𝑧 − 𝑦
 −2𝛽 + 4𝛿 = 𝑦 − 𝑥
 −2𝛽 = 𝑦 − 𝑥 − 4(𝑧 − 𝑦)
 −2𝛽 = 𝑦 − 𝑥 − 4𝑧 + 4𝑦
 𝛽 =
𝑥−5𝑦+4𝑧
2
 𝛼 + 2𝛽 = 𝑥 𝛼 = 𝑥 − 2
𝑥−5𝑦+4𝑧
2
 𝛼 + 2𝛽 = 𝑥 𝛼 = 𝑥 − 2
𝑥−5𝑦+4𝑧
2
 Ejercicio 5.5
 𝛼 = 5𝑦 − 4𝑧
 𝑇
𝑥
𝑦
𝑧
= 𝛼𝑢 + 𝛽𝑣 + 𝛿𝑤 = 𝛼𝑇 𝑢 + 𝛽𝑇 𝑣 + 𝛿𝑇 𝑤
 𝑇
𝑥
𝑦
𝑧
= 5𝑦 − 4𝑧 𝑇 𝑢 +
𝑥−5𝑦+4𝑧
2
𝑇 𝑣 + 𝑧 − 𝑦 𝑇 𝑤
 𝑇
𝑥
𝑦
𝑧
=
1
2
2 5𝑦 − 4𝑧
1
−2
1
+ 𝑥 − 5𝑦 + 4𝑧
3
1
−1
+ 2 𝑧 − 𝑦
2
3
3
 𝐹1 = 10𝑦 − 8𝑧 + 3𝑥 − 15𝑦 + 12𝑧 ∓ 4𝑧 − 4𝑦
 𝐹1 = 3𝑥 − 9𝑦 + 8𝑧
 𝐹2 = −20𝑦 + 16𝑧 + 𝑥 − 5𝑦 + 4𝑧 + 6𝑧 − 6𝑦
 𝐹2 = 𝑥 − 31𝑦 + 26𝑧
 𝐹3 = 10𝑦 − 8𝑧 − 𝑥 + 5𝑦 − 4𝑧 + 6𝑧 − 6𝑦
 𝐹3 = −𝑥 + 9𝑦 − 6𝑧
 𝑇
𝑥
𝑦
𝑧
=
1
2
3𝑥 − 9𝑦 + 8𝑧
𝑥 − 31𝑦 + 26𝑧
−𝑥 + 9𝑦 − 6𝑧
 Ejercicio 5.5
 𝑫𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒆 𝒇 𝟐, −𝟑, 𝟏
 𝑇
𝑥
𝑦
𝑧
=
1
2
3 2 − 9 −3 + 8 1
2 − 31 −3 + 26 1
− 2 + 9 −3 − 6 1
 𝑇
𝑥
𝑦
𝑧
=
1
2
3 + 27 + 8
2 + 93 + 26
−2 − 27 − 6
 𝑇
𝑥
𝑦
𝑧
=
1
2
38
121
−35

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TRANSFORMACIONES LINEALES

  • 1. • PARCIAL III TALLER N° 1 TEMA.-  TRANSFORMACIONES LINEALES  DEPARTAMENTO DE CIENCIAS EXCATAS ALGEBRA LINEAL
  • 2. INTEGRANTES .- GALARZA PILAGUANO ROSA LISBETH. GUANOLUISA GUALOTUÑA RICHARD IVAN. QUINATOA ZAPATA WILSON PAUL. NRC.- 4261 FECHA: 4/03/2021 PERIODO: NOVIEMBRE 2020_ABRIL 2021
  • 3.  INDICE  Objetivos…………………………...…...……….4  Desarrollo…………….………………………….5  Ejercicio 1.1………………………………….5  Ejercicio 2.2………………………………...6  Ejercicio 3.3…………………………………7  Ejercicio 4.4…………………………………9  Ejercicio 5.5………………………………..11
  • 4.  OBJETIVOS.-  Aprender sobre las definiciones de transformación lineal y conocer los conceptos fundamentales, tales como núcleo e imagen de una transformación lineal otro de los aspectos importantes es saber sobre su utilidad la cuál es muy importante dentro de las matemáticas, geometría, análisis, etc.
  • 5.  DESARROLLO  EJERCICIO 1.1  1.1 𝒇 𝒙, 𝒚 = 𝟑 (𝒙 − 𝒚, 𝒙 + 𝒚)   𝑻 = ℝ𝟐 → ℝ𝟐   𝑢 = 𝑥1 𝑦1 𝑣 = 𝑥2 𝑦2  𝑇 = 𝛼𝑢 + 𝛽𝑣 = 𝛼𝑥1 𝛽𝑥2 𝛼𝑦1 𝛽𝑦2  = 𝛼𝑥1 + 𝛽𝑥2 − 𝛼𝑦1 − 𝛽𝑦2 𝛼𝑥1 + 𝛽𝑥2 + 𝛼𝑦1 + 𝛽𝑦2  = 3 (𝛼𝑥1+𝛽𝑥2 − 𝛼𝑦1 − 𝛽𝑦2) (𝛼𝑥1+𝛽𝑥2 + 𝛼𝑦1 + 𝛽𝑦2)  = 3𝛼𝑥1 + 3𝛽𝑥2 − 3𝛼𝑦1 − 3𝛽𝑦2 3𝛼𝑥1 + 3𝛽𝑥2 + 3𝛼𝑦1 + 3𝛽𝑦2  = 3𝛼(𝑥1−𝑦1) + 3𝛽(𝑥2 − 𝑦2) 3𝛼(𝑥1 + 𝑦1) + 3𝛽(𝑥2 + 𝑦2)  = 3𝛼 𝑥1−𝑦1 𝑥1+𝑦1 + 3𝛽 𝑥2−𝑦2 𝑥2+𝑦2  𝑺𝒊 𝒅𝒆𝒇𝒊𝒏𝒆 𝒖𝒏 𝑻. 𝑳
  • 6.  DESARROLLO  EJERCICIO 2.2.-  2.2 𝒇 𝒙, 𝒚 = (𝒙, 𝒚, 𝒛𝟐 )  𝑻 = ℝ𝟐 → ℝ𝟐   𝑢 = 𝑥1 𝑦1 𝑧1 𝑣 = 𝑥2 𝑦2 𝑧2  𝑇 = 𝛼𝑢 + 𝛽𝑣 = 𝛼𝑥1 𝛽𝑥2 𝛼𝑦1 𝛽𝑦2 𝛼𝑧1 𝛽𝑧2  = 𝛼𝑥1 + 𝛽𝑥2 𝛼𝑥1 + 𝛽𝑥2 𝛼𝑥1 + 𝛽𝑧2  = 𝛼𝑥1 + 𝛽𝑥2 𝛼𝑥1 + 𝛽𝑥2 (𝛼𝑥1 + 𝛽𝑧2)2  = 𝛼𝑥1 + 𝛽𝑥2 𝛼𝑥1 + 𝛽𝑥2 (𝛼𝑧1)2 +2 𝛼𝑧1 𝛽𝑧2 + (𝛽𝑧2)2  = 𝛼 𝑥1 𝑦1 𝑧1 + 𝛼𝛽 𝑧1 𝑧2 𝛽 𝑥2 𝑦2 𝑧2  𝑵𝒐 𝒅𝒆𝒇𝒊𝒏𝒆 𝒖𝒏𝒂 𝑻. 𝑳
  • 7.  DESARROLLO  EJERCICIO 3.3  3.3 𝒇 𝒙, 𝒚, 𝒛 = 𝒙 + 𝟐𝒚 − 𝟑𝒛, 𝟑𝒙 − 𝒚 + 𝟓𝒛, 𝒙 − 𝒚 − 𝒛  𝑇 = 𝑥1 𝑦1 𝑧1 + 𝑥2 𝑦2 𝑧2  𝑢 = 𝑥1 𝑦1 𝑧1 𝑣 = 𝑥2 𝑦2 𝑧2  𝑇 = 𝛼𝑥1 + 𝛽𝑥2 𝛼𝑦1 + 𝛽𝑦2 𝛼𝑧1 + 𝛽𝑧2  𝒙 + 𝟐𝒚 − 𝟑𝒛  (𝛼𝑥1 + 𝛽𝑥2 + 2 𝛼𝑦1 + 𝛽𝑦2 − 3(𝛼𝑧1 + 𝛽𝑧2)  (𝛼𝑥1 + 𝛽𝑥2 + 2𝛼𝑦1 + 2𝛽𝑦2 − 3𝛼𝑧1 − 3𝛽𝑧2  𝛼 𝑥1 + 2𝑦1 − 3𝑧1 + 𝛽 𝑥2 + 2𝑦2 − 3𝑧2  𝟑𝒙 − 𝒚 + 𝟓𝒛  3 𝛼𝑥1 + 𝛽𝑥2 − 𝛼𝑦1 + 𝛽𝑦2 + 5 𝛼𝑧1 + 𝛽𝑧2  3 𝛼𝑥1 + 3𝛽𝑥2 − 𝛼𝑦1 − 𝛽𝑦2 + 5𝛼𝑧1 + 5𝛽𝑧2  𝛼 3 𝑥1 − 𝑦1 + 5𝑧1 + 𝛽(3𝑥2 − 𝑦2 + 5𝑧2)
  • 8. EJERCICIO 3.3  𝒙 − 𝒚 − 𝒛  𝛼𝑥1 + 𝛽𝑥2 − 𝛼𝑦1 + 𝛽𝑦2 − 𝛼𝑧1 + 𝛽𝑧2  𝛼𝑥1 + 𝛽𝑥2 − 𝛼𝑦1 − 𝛽𝑦2 − 𝛼𝑧1 − 𝛽𝑧2  𝛼 𝑥1 − 𝑦1 − 𝑧1 + 𝛽 𝑥2 − 𝑦2 − 𝑧2   𝛼 𝑥1 + 2𝑦1 − 3𝑧1 + 𝛽 𝑥2 + 2𝑦2 − 3𝑧2  𝛼 3 𝑥1 − 𝑦1 + 5𝑧1 + 𝛽(3𝑥2 − 𝑦2 + 5𝑧2)  𝛼 𝑥1 − 𝑦1 − 𝑧1 + 𝛽 𝑥2 − 𝑦2 − 𝑧2   𝛼 𝑥1 + 2𝑦1 − 3𝑧1 3 𝑥1 − 𝑦1 + 5𝑧1 𝑥1 − 𝑦1 − 𝑧1 + 𝛽 𝑥2 + 2𝑦2 − 3𝑧2 3𝑥2 − 𝑦2 + 5𝑧2 𝑥2 − 𝑦2 − 𝑧2  𝑆𝑖 𝑒𝑠 𝑇. 𝐿.
  • 9.  Ejercicio 4.4  4.4 Sea f una transformación lineal de ℝ𝟑 → ℝ𝟑 , suponga que 𝒇 𝟏, 𝟎, 𝟏 = 𝟏, −𝟏, 𝟑 𝒚 𝒇 𝟐, 𝟏, 𝟎 = 𝟎, 𝟐, 𝟏 ; 𝑫𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒆 𝒇 −𝟏, −𝟐, 𝟑   𝑇 1 0 1 = 1 −1 3 ; 𝑇 2 1 0 = 0 2 1  𝑥 𝑦 𝑧 = 𝛼 1 0 1 + 𝛽 2 1 0  𝑇(𝛼𝑢 + 𝛽𝑣) = 𝛼𝑇(𝑢) + 𝛽𝑇(𝑣)  𝑥 = 𝛼 + 2𝛽 𝑦 = 𝛽 𝑧 = 𝛼  𝛼 = 𝑥 − 2𝑦  𝛽 = 𝑥 − 𝑧 2
  • 10.  Ejercicio 4.4  𝑇 𝑥 𝑦 𝑧 = 𝑇 𝛼 1 0 1 + 𝛽 2 1 0 = 𝛼𝑇 1 0 1 + 𝛽𝑇 2 1 0  𝑇 𝑥 𝑦 𝑧 = 𝑥 − 𝑧 2 1 0 1 + 𝑥 − 2𝑦 2 1 0  𝑇 𝑥 𝑦 𝑧 = 5𝑥 − 8𝑦 − 𝑧 2 𝑥 − 2𝑦 𝑥 − 𝑧 2  𝑇 −1 −2 3 = 5𝑥 − 8𝑦 − 𝑧 2 𝑥 − 2𝑦 𝑥 − 𝑧 2  𝑇 −1 −2 3 = 5(−1) − 8(−2) − (3) 2 (−1) − 2(−2) (−1) − (3) 2 𝑇 −1 −2 3 == 4 3 −2
  • 11.  EJERCICIO 5.5 • Sea 𝒇 𝒖𝒏𝒂 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒄𝒊ó𝒏 𝒍𝒊𝒏𝒆𝒂𝒍 𝒅𝒆 ℝ𝟑 𝒆𝒏 𝑷𝟐 𝒕𝒂𝒍 𝒒𝒖𝒆 𝒇((𝟏, 𝟏, 𝟏)) = 𝟏 – 𝟐𝒕 + 𝒕𝟐 , 𝒇((𝟐, 𝟎, 𝟎)) = 𝟑 + 𝒕 – 𝒕𝟐, 𝒇((𝟎, 𝟒, 𝟓)) = 𝟐 + 𝟑𝒕 + 𝟑𝒕𝟐; 𝑫𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒆 𝒇((𝟐, −𝟑, 𝟏)).  𝑇: ℝ3 → 𝑃2  𝑇𝑢 1 1 1 = 1 −2 1 𝑇𝑣 2 0 0 = 3 1 −1 𝑇𝑤 0 4 5 = 2 3 3  𝑥 𝑦 𝑧 = 𝛼 1 1 1 + 𝛽 2 0 0 + 𝛿 0 4 5  𝛼 + 2𝛽 = 𝑥  𝛼 + 4𝛿 = 𝑦  𝛼 + 5𝛿 = 𝑧
  • 12.  Ejercicio 5.5  𝐴 = 1 1 1 2 0 0 4 0 5 𝑥 𝑦 𝑧 𝐹2−𝐹1→ 𝐹2 𝑦 𝐹3−𝐹1→ 𝐹3  𝐴 = 1 0 0 2 0 −2 4 −2 5 𝑥 𝑦 − 𝑥 𝑧 − 𝑥 𝐹3−𝐹2→ 𝐹3  𝐴 = 1 0 0 2 0 −2 4 0 1 𝑥 𝑦 − 𝑥 𝑧 − 𝑦  𝛿 = 𝑧 − 𝑦  −2𝛽 + 4𝛿 = 𝑦 − 𝑥  −2𝛽 = 𝑦 − 𝑥 − 4(𝑧 − 𝑦)  −2𝛽 = 𝑦 − 𝑥 − 4𝑧 + 4𝑦  𝛽 = 𝑥−5𝑦+4𝑧 2  𝛼 + 2𝛽 = 𝑥 𝛼 = 𝑥 − 2 𝑥−5𝑦+4𝑧 2  𝛼 + 2𝛽 = 𝑥 𝛼 = 𝑥 − 2 𝑥−5𝑦+4𝑧 2
  • 13.  Ejercicio 5.5  𝛼 = 5𝑦 − 4𝑧  𝑇 𝑥 𝑦 𝑧 = 𝛼𝑢 + 𝛽𝑣 + 𝛿𝑤 = 𝛼𝑇 𝑢 + 𝛽𝑇 𝑣 + 𝛿𝑇 𝑤  𝑇 𝑥 𝑦 𝑧 = 5𝑦 − 4𝑧 𝑇 𝑢 + 𝑥−5𝑦+4𝑧 2 𝑇 𝑣 + 𝑧 − 𝑦 𝑇 𝑤  𝑇 𝑥 𝑦 𝑧 = 1 2 2 5𝑦 − 4𝑧 1 −2 1 + 𝑥 − 5𝑦 + 4𝑧 3 1 −1 + 2 𝑧 − 𝑦 2 3 3  𝐹1 = 10𝑦 − 8𝑧 + 3𝑥 − 15𝑦 + 12𝑧 ∓ 4𝑧 − 4𝑦  𝐹1 = 3𝑥 − 9𝑦 + 8𝑧  𝐹2 = −20𝑦 + 16𝑧 + 𝑥 − 5𝑦 + 4𝑧 + 6𝑧 − 6𝑦  𝐹2 = 𝑥 − 31𝑦 + 26𝑧  𝐹3 = 10𝑦 − 8𝑧 − 𝑥 + 5𝑦 − 4𝑧 + 6𝑧 − 6𝑦  𝐹3 = −𝑥 + 9𝑦 − 6𝑧  𝑇 𝑥 𝑦 𝑧 = 1 2 3𝑥 − 9𝑦 + 8𝑧 𝑥 − 31𝑦 + 26𝑧 −𝑥 + 9𝑦 − 6𝑧
  • 14.  Ejercicio 5.5  𝑫𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒆 𝒇 𝟐, −𝟑, 𝟏  𝑇 𝑥 𝑦 𝑧 = 1 2 3 2 − 9 −3 + 8 1 2 − 31 −3 + 26 1 − 2 + 9 −3 − 6 1  𝑇 𝑥 𝑦 𝑧 = 1 2 3 + 27 + 8 2 + 93 + 26 −2 − 27 − 6  𝑇 𝑥 𝑦 𝑧 = 1 2 38 121 −35