SlideShare a Scribd company logo
1 of 14
• 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

More Related Content

What's hot

What's hot (18)

Tugas 1 Matematika 2
Tugas 1 Matematika 2Tugas 1 Matematika 2
Tugas 1 Matematika 2
 
Calculo
CalculoCalculo
Calculo
 
Factoring common monomial
Factoring common monomialFactoring common monomial
Factoring common monomial
 
Διαγώνισμα Α Γυμνασίου 13-01-2016
Διαγώνισμα Α Γυμνασίου 13-01-2016Διαγώνισμα Α Γυμνασίου 13-01-2016
Διαγώνισμα Α Γυμνασίου 13-01-2016
 
TABLA CENTROIDE .pdf
TABLA CENTROIDE .pdfTABLA CENTROIDE .pdf
TABLA CENTROIDE .pdf
 
Tugas matematika-1-semester-2
Tugas matematika-1-semester-2Tugas matematika-1-semester-2
Tugas matematika-1-semester-2
 
師父與徒弟的列式
師父與徒弟的列式師父與徒弟的列式
師父與徒弟的列式
 
師父與徒弟的列式
師父與徒弟的列式師父與徒弟的列式
師父與徒弟的列式
 
Tugas Calculus : Limit (Hal. 8-14)
Tugas Calculus : Limit (Hal. 8-14)Tugas Calculus : Limit (Hal. 8-14)
Tugas Calculus : Limit (Hal. 8-14)
 
Calculus revision card
Calculus  revision cardCalculus  revision card
Calculus revision card
 
Calculus revision card
Calculus  revision cardCalculus  revision card
Calculus revision card
 
ضرب وحيدات الحد
ضرب وحيدات الحدضرب وحيدات الحد
ضرب وحيدات الحد
 
Practica 1 numeros naturales y enteros
Practica 1 numeros naturales y enterosPractica 1 numeros naturales y enteros
Practica 1 numeros naturales y enteros
 
Tugas mtk 1
Tugas mtk 1Tugas mtk 1
Tugas mtk 1
 
Tugas matematika 1 (semester 2)
Tugas matematika 1 (semester 2)Tugas matematika 1 (semester 2)
Tugas matematika 1 (semester 2)
 
Mohr circle
Mohr circleMohr circle
Mohr circle
 
Reducible equation to quadratic form
Reducible equation to quadratic formReducible equation to quadratic form
Reducible equation to quadratic form
 
Ejercicios: Círculo y parábola
Ejercicios: Círculo y parábolaEjercicios: Círculo y parábola
Ejercicios: Círculo y parábola
 

Similar to TRANSFORMACIONES LINEALES

Similar to TRANSFORMACIONES LINEALES (20)

Algebra
AlgebraAlgebra
Algebra
 
Integral calculus
Integral calculusIntegral calculus
Integral calculus
 
Taller 1 parcial 3
Taller 1 parcial 3Taller 1 parcial 3
Taller 1 parcial 3
 
taller transformaciones lineales
taller transformaciones linealestaller transformaciones lineales
taller transformaciones lineales
 
SUEC 高中 Adv Maths (Locus) (Part 2).pptx
SUEC 高中 Adv Maths (Locus) (Part 2).pptxSUEC 高中 Adv Maths (Locus) (Part 2).pptx
SUEC 高中 Adv Maths (Locus) (Part 2).pptx
 
Tugas blog-matematika-kelompok-3
Tugas blog-matematika-kelompok-3Tugas blog-matematika-kelompok-3
Tugas blog-matematika-kelompok-3
 
Tugas blog-matematika-kelompok-3
Tugas blog-matematika-kelompok-3Tugas blog-matematika-kelompok-3
Tugas blog-matematika-kelompok-3
 
SUEC 高中 Adv Maths (Locus) (Part 1).pptx
SUEC 高中 Adv Maths (Locus) (Part 1).pptxSUEC 高中 Adv Maths (Locus) (Part 1).pptx
SUEC 高中 Adv Maths (Locus) (Part 1).pptx
 
Latihan 8.3 Thomas (Kalkulus Integral)
Latihan 8.3 Thomas (Kalkulus Integral)Latihan 8.3 Thomas (Kalkulus Integral)
Latihan 8.3 Thomas (Kalkulus Integral)
 
Hsc maths formulae for board exam
Hsc maths formulae for board examHsc maths formulae for board exam
Hsc maths formulae for board exam
 
Hsc maths formulae for board exam
Hsc maths formulae for board examHsc maths formulae for board exam
Hsc maths formulae for board exam
 
Derivación 1.
Derivación 1.Derivación 1.
Derivación 1.
 
Tugas mtk
Tugas mtkTugas mtk
Tugas mtk
 
Tugas mtk
Tugas mtkTugas mtk
Tugas mtk
 
Tugas mtk
Tugas mtkTugas mtk
Tugas mtk
 
Tugas mtk
Tugas mtkTugas mtk
Tugas mtk
 
MAT 230 CH 7 Notes 7.4 (1).pptx
MAT 230 CH 7 Notes 7.4 (1).pptxMAT 230 CH 7 Notes 7.4 (1).pptx
MAT 230 CH 7 Notes 7.4 (1).pptx
 
Tugas mtk
Tugas mtkTugas mtk
Tugas mtk
 
Tugas mtk
Tugas mtkTugas mtk
Tugas mtk
 
Factorising
FactorisingFactorising
Factorising
 

Recently uploaded

VIP Call Girls Ankleshwar 7001035870 Whatsapp Number, 24/07 Booking
VIP Call Girls Ankleshwar 7001035870 Whatsapp Number, 24/07 BookingVIP Call Girls Ankleshwar 7001035870 Whatsapp Number, 24/07 Booking
VIP Call Girls Ankleshwar 7001035870 Whatsapp Number, 24/07 Booking
dharasingh5698
 
AKTU Computer Networks notes --- Unit 3.pdf
AKTU Computer Networks notes ---  Unit 3.pdfAKTU Computer Networks notes ---  Unit 3.pdf
AKTU Computer Networks notes --- Unit 3.pdf
ankushspencer015
 
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
ssuser89054b
 
notes on Evolution Of Analytic Scalability.ppt
notes on Evolution Of Analytic Scalability.pptnotes on Evolution Of Analytic Scalability.ppt
notes on Evolution Of Analytic Scalability.ppt
MsecMca
 
Call for Papers - African Journal of Biological Sciences, E-ISSN: 2663-2187, ...
Call for Papers - African Journal of Biological Sciences, E-ISSN: 2663-2187, ...Call for Papers - African Journal of Biological Sciences, E-ISSN: 2663-2187, ...
Call for Papers - African Journal of Biological Sciences, E-ISSN: 2663-2187, ...
Christo Ananth
 
Call Girls in Ramesh Nagar Delhi 💯 Call Us 🔝9953056974 🔝 Escort Service
Call Girls in Ramesh Nagar Delhi 💯 Call Us 🔝9953056974 🔝 Escort ServiceCall Girls in Ramesh Nagar Delhi 💯 Call Us 🔝9953056974 🔝 Escort Service
Call Girls in Ramesh Nagar Delhi 💯 Call Us 🔝9953056974 🔝 Escort Service
9953056974 Low Rate Call Girls In Saket, Delhi NCR
 

Recently uploaded (20)

VIP Call Girls Ankleshwar 7001035870 Whatsapp Number, 24/07 Booking
VIP Call Girls Ankleshwar 7001035870 Whatsapp Number, 24/07 BookingVIP Call Girls Ankleshwar 7001035870 Whatsapp Number, 24/07 Booking
VIP Call Girls Ankleshwar 7001035870 Whatsapp Number, 24/07 Booking
 
The Most Attractive Pune Call Girls Manchar 8250192130 Will You Miss This Cha...
The Most Attractive Pune Call Girls Manchar 8250192130 Will You Miss This Cha...The Most Attractive Pune Call Girls Manchar 8250192130 Will You Miss This Cha...
The Most Attractive Pune Call Girls Manchar 8250192130 Will You Miss This Cha...
 
University management System project report..pdf
University management System project report..pdfUniversity management System project report..pdf
University management System project report..pdf
 
Vivazz, Mieres Social Housing Design Spain
Vivazz, Mieres Social Housing Design SpainVivazz, Mieres Social Housing Design Spain
Vivazz, Mieres Social Housing Design Spain
 
(INDIRA) Call Girl Meerut Call Now 8617697112 Meerut Escorts 24x7
(INDIRA) Call Girl Meerut Call Now 8617697112 Meerut Escorts 24x7(INDIRA) Call Girl Meerut Call Now 8617697112 Meerut Escorts 24x7
(INDIRA) Call Girl Meerut Call Now 8617697112 Meerut Escorts 24x7
 
Intze Overhead Water Tank Design by Working Stress - IS Method.pdf
Intze Overhead Water Tank  Design by Working Stress - IS Method.pdfIntze Overhead Water Tank  Design by Working Stress - IS Method.pdf
Intze Overhead Water Tank Design by Working Stress - IS Method.pdf
 
chapter 5.pptx: drainage and irrigation engineering
chapter 5.pptx: drainage and irrigation engineeringchapter 5.pptx: drainage and irrigation engineering
chapter 5.pptx: drainage and irrigation engineering
 
Generative AI or GenAI technology based PPT
Generative AI or GenAI technology based PPTGenerative AI or GenAI technology based PPT
Generative AI or GenAI technology based PPT
 
Double Revolving field theory-how the rotor develops torque
Double Revolving field theory-how the rotor develops torqueDouble Revolving field theory-how the rotor develops torque
Double Revolving field theory-how the rotor develops torque
 
AKTU Computer Networks notes --- Unit 3.pdf
AKTU Computer Networks notes ---  Unit 3.pdfAKTU Computer Networks notes ---  Unit 3.pdf
AKTU Computer Networks notes --- Unit 3.pdf
 
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
 
Call for Papers - International Journal of Intelligent Systems and Applicatio...
Call for Papers - International Journal of Intelligent Systems and Applicatio...Call for Papers - International Journal of Intelligent Systems and Applicatio...
Call for Papers - International Journal of Intelligent Systems and Applicatio...
 
notes on Evolution Of Analytic Scalability.ppt
notes on Evolution Of Analytic Scalability.pptnotes on Evolution Of Analytic Scalability.ppt
notes on Evolution Of Analytic Scalability.ppt
 
Call Girls Wakad Call Me 7737669865 Budget Friendly No Advance Booking
Call Girls Wakad Call Me 7737669865 Budget Friendly No Advance BookingCall Girls Wakad Call Me 7737669865 Budget Friendly No Advance Booking
Call Girls Wakad Call Me 7737669865 Budget Friendly No Advance Booking
 
Call for Papers - African Journal of Biological Sciences, E-ISSN: 2663-2187, ...
Call for Papers - African Journal of Biological Sciences, E-ISSN: 2663-2187, ...Call for Papers - African Journal of Biological Sciences, E-ISSN: 2663-2187, ...
Call for Papers - African Journal of Biological Sciences, E-ISSN: 2663-2187, ...
 
The Most Attractive Pune Call Girls Budhwar Peth 8250192130 Will You Miss Thi...
The Most Attractive Pune Call Girls Budhwar Peth 8250192130 Will You Miss Thi...The Most Attractive Pune Call Girls Budhwar Peth 8250192130 Will You Miss Thi...
The Most Attractive Pune Call Girls Budhwar Peth 8250192130 Will You Miss Thi...
 
Call Girls in Ramesh Nagar Delhi 💯 Call Us 🔝9953056974 🔝 Escort Service
Call Girls in Ramesh Nagar Delhi 💯 Call Us 🔝9953056974 🔝 Escort ServiceCall Girls in Ramesh Nagar Delhi 💯 Call Us 🔝9953056974 🔝 Escort Service
Call Girls in Ramesh Nagar Delhi 💯 Call Us 🔝9953056974 🔝 Escort Service
 
Java Programming :Event Handling(Types of Events)
Java Programming :Event Handling(Types of Events)Java Programming :Event Handling(Types of Events)
Java Programming :Event Handling(Types of Events)
 
Call Girls Walvekar Nagar Call Me 7737669865 Budget Friendly No Advance Booking
Call Girls Walvekar Nagar Call Me 7737669865 Budget Friendly No Advance BookingCall Girls Walvekar Nagar Call Me 7737669865 Budget Friendly No Advance Booking
Call Girls Walvekar Nagar Call Me 7737669865 Budget Friendly No Advance Booking
 
PVC VS. FIBERGLASS (FRP) GRAVITY SEWER - UNI BELL
PVC VS. FIBERGLASS (FRP) GRAVITY SEWER - UNI BELLPVC VS. FIBERGLASS (FRP) GRAVITY SEWER - UNI BELL
PVC VS. FIBERGLASS (FRP) GRAVITY SEWER - UNI BELL
 

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