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DEPARTAMENTODE CIENCIAS EXACTAS
ALGEBRALINEAL
PARCIALII
TALLERNO 3
TEMA: TRANSFORMACIONES LINEALES
INTEGRANTES:
- GUERRERO CÓRDOVA STEFANY PAOLA
- SUNTA BARZALLO JOSSELYN GABRIELA
- UNDA ROBLRES GRACE IVETTE
GRUPON 4
NRC:4264
FECHADE ENTREGA:4/3/2021
ÍNDICE
OBJETIVO…………………………………………………………3
 SOLUCIÓN EJERCICIO 1)………………………………………4
• SOLUCIÓN EJERCICIO 2) ……………………………………………..5
• SOLUCIÓN EJERCICIO 3)………………………………………………6
• SOLUCIÓN EJERCICIO 6)………………………………………………7
• SOLUCION EJERCICIO 6)………………………………………………8
• SOLUCIÓN EJERCICIO 7)………………………………………………9
• BIBLIOGRAFÍA
…………………………………………………………….10
OBJETIVOS:
• Analizar y calcular las transformaciones lineales.
• Desarrollar las destrezas obtenidas en hora clase.
DETERMINAR CUÁL DE LAS SIGUIENTES FUNCIONES, DEFINE UNA
TRANSFORMACIÓN LINEAL
𝟏. 𝒇 𝒙, 𝒚 = 𝟑(𝒙 − 𝒚, 𝒙 + 𝒚)
𝑢 =
𝑥1
𝑦2
𝑣 =
𝑥2
𝑦2
𝑇 𝛼𝑢 + 𝛽𝑣 = 𝑇 𝛼
𝑥1
𝑦1
+ 𝛽
𝑥2
𝑦2
= 𝑇
𝛼𝑥1 + 𝛽𝑦1
𝛼𝑥2 + 𝛽𝑦2
= 3
𝛼𝑥1 + 𝛽𝑥2 − (𝛼𝑦1 + 𝛽𝑦2)
𝛼𝑥1 + 𝛽𝑥2 + (𝛼𝑦1 + 𝛽𝑦2)
= 3
𝛼 𝑥1 − 𝑦1 + 𝛽(𝑥2 − 𝑦2)
𝛼 𝑥1 + 𝑦1 + 𝛽(𝑥2 + 𝑦2)
= 3𝛼
𝑥1 − 𝑦1
𝑥1 + 𝑦1
+ 3𝛽
𝑥2 − 𝑦2
𝑥2 + 𝑦2
= 3 𝛼𝑢 + 𝛽𝑣 = 3 𝛼𝑇 𝑢 + 𝛽𝑇 𝑣
→ 𝑺𝒊𝒆𝒔 𝒖𝒏𝒂 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒄𝒊ó𝒏 𝑳𝒊𝒏𝒆𝒂𝒍.
𝑆𝑖 𝑒𝑠 𝑢𝑛𝑎 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑎𝑐𝑖ó𝑛 𝐿𝑖𝑛𝑒𝑎𝑙.
𝟐. 𝒇 𝒙, 𝒚, 𝒛 = (𝒙, 𝒚, 𝒛𝟐)
𝑢 =
𝑥1
𝑦1
𝑧1
𝑣 =
𝑥2
𝑦2
𝑧2
𝑇 𝛼𝑢 + 𝛽𝑣 = 𝑇 𝛼
𝑥1
𝑦1
𝑧1
+ 𝛽
𝑥2
𝑦2
𝑧2
= 𝑇
𝛼𝑥1 + 𝛽𝑥2
𝛼𝑦1 + 𝛽𝑦2
𝛼𝑧1 + 𝛽𝑧2
=
𝛼𝑥1 + 𝛽𝑥2
𝛼𝑦1 + 𝛽𝑦2
(𝛼𝑧1 + 𝛽𝑧2)2
=
𝛼𝑥1 + 𝛽𝑥2
𝛼𝑦1 + 𝛽𝑦2
(𝛼𝑧1)2+2𝑧1𝑧2𝛼𝛽 + (𝛽 𝑧2 )2
≠
𝑥
𝑦
𝑧2
𝑁𝑜 𝑒𝑠 𝑢𝑛𝑎 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑎𝑐𝑖ó𝑛 𝐿𝑖𝑛𝑒𝑎𝑙.
𝟑. 𝒇((𝒙, 𝒚, 𝒛)) = (𝒙 + 𝟐𝒚 – 𝟑𝒛, 𝟑𝒙 − 𝒚 + 𝟓𝒛, 𝒙 – 𝒚 – 𝒛)
U=
𝑥1
𝑦1
𝑧1
V=
𝑥2
𝑦2
𝑧2
T(𝛼𝑢 + 𝛽𝑣)=T 𝛼
𝑥1
𝑦1
𝑧1
+ 𝛽
𝑥2
𝑦2
𝑧2
T
𝛼𝑥1 + 𝛽𝑥2
𝛼𝑦1 + 𝛽𝑦2
𝛼𝑧1 + 𝛽𝑧2
=
𝛼𝑥1 + 𝛽𝑥2 + 2 𝛼𝑦1 + 𝛽𝑦2 − 3(𝛼𝑧1 + 𝛽𝑧2)
3 𝛼𝑥1 + 𝛽𝑥2 − 𝛼𝑦1 + 𝛽𝑦2 + 5(𝛼𝑧1 + 𝛽𝑧2)
𝛼𝑥1 + 𝛽𝑥2 − 𝛼𝑦1 + 𝛽𝑦2 − (𝛼𝑧1 + 𝛽𝑧2)
𝛼
𝑥1 + 2𝑦1 − 3𝑧1
3𝑧1 − 𝑧1 + 5𝑧1
𝑥1 − 𝑦1 − 𝑧1
+ 𝛽
𝑥2 + 2𝑦2 − 3𝑧2
3𝑥2 − 𝑦2 + 5𝑧2
𝑥2 − 𝑦2 − 𝑧2
T(𝛼𝑢 + 𝛽𝑣) =𝛼𝑇 𝑢 + 𝛽𝑇 𝑣 → 𝑺𝒊 𝒆𝒔 𝒖𝒏𝒂 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒄𝒊ó𝒏 𝑳𝒊𝒏𝒆𝒂𝒍.
6.- Sea f una transformación lineal de 𝑹𝟑
en 𝑹𝟑
, suponga que 𝒇((𝟏, 𝟎, 𝟏)) = (𝟏, −𝟏, 𝟑) 𝒚 𝒇((𝟐, 𝟏, 𝟎)) =
(𝟎, 𝟐, 𝟏); Determine 𝒇((−𝟏, −𝟐, 𝟑)).
𝑥
𝑦
𝑧
= 𝛼
1
0
1
+ β
2
1
0
+ δ
−1
−2
3
1
0
1
2
1
0
−1
−2
3
𝑥
𝑦
𝑧
F3-F1 F3
1
0
0
2
1
−2
−1
−2
4
𝑥
𝑦
𝑧 − 𝑥
𝐹3 + 2𝐹2 → 𝐹3
1
0
0
2
1
0
−1
−2
0
𝑥
𝑦
𝑧 − 𝑥 + 2𝑦
𝑇
𝑥
𝑦
𝑧
= 𝑇
1
0
1
+ β
2
1
0
+ δ
−1
−2
3
𝑇
𝑥
𝑦
𝑧
= 𝑇 𝛼
1
0
1
+ Tβ
2
1
0
+ Tδ
−1
−2
3
δ = 0
δ = 0
β - 2 δ = y
β – 0 = y
β = y
𝛼 + δ − δ = x
𝛼 + 2𝑦 − 0 = x
𝛼 = 𝑥 − 2𝑦
𝑇
𝑥
𝑦
𝑧
= 𝑥 − 2𝑦
1
0
1
+ y
2
1
0
+ 0
−1
−2
3
𝑇
𝑥
𝑦
𝑧
= 𝑥 − 2𝑦
1
−1
3
+ y
0
2
1
+ 0
𝑁1
𝑁2
𝑁3
𝑇
𝑥
𝑦
𝑧
= 𝑥 − 2𝑦
1
−1
3
+ y
0
2
1
𝑇
𝑥
𝑦
𝑧
=
𝑥 − 2𝑦 + 0
−𝑥 + 2𝑦 + 2𝑦
3𝑥 − 6𝑦 + 𝑦
=
𝑥 − 2𝑦
−𝑥 + 4𝑦
3𝑥 − 5𝑦
𝑇
−1
−2
3
=
𝟑
−𝟕
𝟕
7.Sea una transformación lineal 𝑹𝟑
en 𝑹𝟑
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
f 𝛼𝑢 + 𝛽𝑣 + 𝛿𝑤 = 𝛼𝑓 𝑢 + 𝛽 𝑣 + 𝛿𝑓(𝑤)
2
−3
1
= 𝛼
11
1
1
+ 𝛽
2
0
0
+ 𝛿
0
4
5
1 2 0 2
1 0 4 −3
1 0 5 1
𝐹2 − 𝐹1 → 𝐹2
𝐹3 − 𝐹1 → 𝐹3
1 2 0 2
0 −2 4 −5
0 −2 5 −1
𝐹3 − 𝐹2 → 𝐹3
1 2 0 2
0 −2 4 −5
0 0 1 4
-
𝛿 = 4
−2𝛽 + 4𝛿 = −5
𝛼 + 2𝛽 = 2
δ=4
β =
−5−4(4)
−2
=
21
2
α = −2
21
2
+ 2 = −19
= -19 1 − 2𝑡 + 𝑡2 +
21
2
3 + 𝑡 − 𝑡2 + 4 2 + 3𝑡 + 3𝑡2
=−19+38t−19𝑡2
+
63
2
+
21
2
𝑡 −
21
2
𝑡2
+ 8 + 12𝑡 + 12𝑡2
𝒇 𝟐, −𝟑, 𝟏 =-
𝟑𝟓
𝟐
𝒕𝟐 +
𝟏𝟐𝟏
𝟐
𝒕 +
𝟒𝟏
𝟐
BIBLIOGRAFÍA:
• https://www.youtube.com/watch?v=7DcJDjOyGVM&feature=youtu.be
• https://youtu.be/Ha7dbOUYvmo

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Algebra

  • 1. DEPARTAMENTODE CIENCIAS EXACTAS ALGEBRALINEAL PARCIALII TALLERNO 3 TEMA: TRANSFORMACIONES LINEALES INTEGRANTES: - GUERRERO CÓRDOVA STEFANY PAOLA - SUNTA BARZALLO JOSSELYN GABRIELA - UNDA ROBLRES GRACE IVETTE GRUPON 4 NRC:4264 FECHADE ENTREGA:4/3/2021
  • 2. ÍNDICE OBJETIVO…………………………………………………………3  SOLUCIÓN EJERCICIO 1)………………………………………4 • SOLUCIÓN EJERCICIO 2) ……………………………………………..5 • SOLUCIÓN EJERCICIO 3)………………………………………………6 • SOLUCIÓN EJERCICIO 6)………………………………………………7 • SOLUCION EJERCICIO 6)………………………………………………8 • SOLUCIÓN EJERCICIO 7)………………………………………………9 • BIBLIOGRAFÍA …………………………………………………………….10
  • 3. OBJETIVOS: • Analizar y calcular las transformaciones lineales. • Desarrollar las destrezas obtenidas en hora clase.
  • 4. DETERMINAR CUÁL DE LAS SIGUIENTES FUNCIONES, DEFINE UNA TRANSFORMACIÓN LINEAL 𝟏. 𝒇 𝒙, 𝒚 = 𝟑(𝒙 − 𝒚, 𝒙 + 𝒚) 𝑢 = 𝑥1 𝑦2 𝑣 = 𝑥2 𝑦2 𝑇 𝛼𝑢 + 𝛽𝑣 = 𝑇 𝛼 𝑥1 𝑦1 + 𝛽 𝑥2 𝑦2 = 𝑇 𝛼𝑥1 + 𝛽𝑦1 𝛼𝑥2 + 𝛽𝑦2 = 3 𝛼𝑥1 + 𝛽𝑥2 − (𝛼𝑦1 + 𝛽𝑦2) 𝛼𝑥1 + 𝛽𝑥2 + (𝛼𝑦1 + 𝛽𝑦2) = 3 𝛼 𝑥1 − 𝑦1 + 𝛽(𝑥2 − 𝑦2) 𝛼 𝑥1 + 𝑦1 + 𝛽(𝑥2 + 𝑦2) = 3𝛼 𝑥1 − 𝑦1 𝑥1 + 𝑦1 + 3𝛽 𝑥2 − 𝑦2 𝑥2 + 𝑦2 = 3 𝛼𝑢 + 𝛽𝑣 = 3 𝛼𝑇 𝑢 + 𝛽𝑇 𝑣 → 𝑺𝒊𝒆𝒔 𝒖𝒏𝒂 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒄𝒊ó𝒏 𝑳𝒊𝒏𝒆𝒂𝒍. 𝑆𝑖 𝑒𝑠 𝑢𝑛𝑎 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑎𝑐𝑖ó𝑛 𝐿𝑖𝑛𝑒𝑎𝑙.
  • 5. 𝟐. 𝒇 𝒙, 𝒚, 𝒛 = (𝒙, 𝒚, 𝒛𝟐) 𝑢 = 𝑥1 𝑦1 𝑧1 𝑣 = 𝑥2 𝑦2 𝑧2 𝑇 𝛼𝑢 + 𝛽𝑣 = 𝑇 𝛼 𝑥1 𝑦1 𝑧1 + 𝛽 𝑥2 𝑦2 𝑧2 = 𝑇 𝛼𝑥1 + 𝛽𝑥2 𝛼𝑦1 + 𝛽𝑦2 𝛼𝑧1 + 𝛽𝑧2 = 𝛼𝑥1 + 𝛽𝑥2 𝛼𝑦1 + 𝛽𝑦2 (𝛼𝑧1 + 𝛽𝑧2)2 = 𝛼𝑥1 + 𝛽𝑥2 𝛼𝑦1 + 𝛽𝑦2 (𝛼𝑧1)2+2𝑧1𝑧2𝛼𝛽 + (𝛽 𝑧2 )2 ≠ 𝑥 𝑦 𝑧2 𝑁𝑜 𝑒𝑠 𝑢𝑛𝑎 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑎𝑐𝑖ó𝑛 𝐿𝑖𝑛𝑒𝑎𝑙.
  • 6. 𝟑. 𝒇((𝒙, 𝒚, 𝒛)) = (𝒙 + 𝟐𝒚 – 𝟑𝒛, 𝟑𝒙 − 𝒚 + 𝟓𝒛, 𝒙 – 𝒚 – 𝒛) U= 𝑥1 𝑦1 𝑧1 V= 𝑥2 𝑦2 𝑧2 T(𝛼𝑢 + 𝛽𝑣)=T 𝛼 𝑥1 𝑦1 𝑧1 + 𝛽 𝑥2 𝑦2 𝑧2 T 𝛼𝑥1 + 𝛽𝑥2 𝛼𝑦1 + 𝛽𝑦2 𝛼𝑧1 + 𝛽𝑧2 = 𝛼𝑥1 + 𝛽𝑥2 + 2 𝛼𝑦1 + 𝛽𝑦2 − 3(𝛼𝑧1 + 𝛽𝑧2) 3 𝛼𝑥1 + 𝛽𝑥2 − 𝛼𝑦1 + 𝛽𝑦2 + 5(𝛼𝑧1 + 𝛽𝑧2) 𝛼𝑥1 + 𝛽𝑥2 − 𝛼𝑦1 + 𝛽𝑦2 − (𝛼𝑧1 + 𝛽𝑧2) 𝛼 𝑥1 + 2𝑦1 − 3𝑧1 3𝑧1 − 𝑧1 + 5𝑧1 𝑥1 − 𝑦1 − 𝑧1 + 𝛽 𝑥2 + 2𝑦2 − 3𝑧2 3𝑥2 − 𝑦2 + 5𝑧2 𝑥2 − 𝑦2 − 𝑧2 T(𝛼𝑢 + 𝛽𝑣) =𝛼𝑇 𝑢 + 𝛽𝑇 𝑣 → 𝑺𝒊 𝒆𝒔 𝒖𝒏𝒂 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒄𝒊ó𝒏 𝑳𝒊𝒏𝒆𝒂𝒍.
  • 7. 6.- Sea f una transformación lineal de 𝑹𝟑 en 𝑹𝟑 , suponga que 𝒇((𝟏, 𝟎, 𝟏)) = (𝟏, −𝟏, 𝟑) 𝒚 𝒇((𝟐, 𝟏, 𝟎)) = (𝟎, 𝟐, 𝟏); Determine 𝒇((−𝟏, −𝟐, 𝟑)). 𝑥 𝑦 𝑧 = 𝛼 1 0 1 + β 2 1 0 + δ −1 −2 3 1 0 1 2 1 0 −1 −2 3 𝑥 𝑦 𝑧 F3-F1 F3 1 0 0 2 1 −2 −1 −2 4 𝑥 𝑦 𝑧 − 𝑥 𝐹3 + 2𝐹2 → 𝐹3 1 0 0 2 1 0 −1 −2 0 𝑥 𝑦 𝑧 − 𝑥 + 2𝑦 𝑇 𝑥 𝑦 𝑧 = 𝑇 1 0 1 + β 2 1 0 + δ −1 −2 3 𝑇 𝑥 𝑦 𝑧 = 𝑇 𝛼 1 0 1 + Tβ 2 1 0 + Tδ −1 −2 3 δ = 0 δ = 0 β - 2 δ = y β – 0 = y β = y 𝛼 + δ − δ = x 𝛼 + 2𝑦 − 0 = x 𝛼 = 𝑥 − 2𝑦
  • 8. 𝑇 𝑥 𝑦 𝑧 = 𝑥 − 2𝑦 1 0 1 + y 2 1 0 + 0 −1 −2 3 𝑇 𝑥 𝑦 𝑧 = 𝑥 − 2𝑦 1 −1 3 + y 0 2 1 + 0 𝑁1 𝑁2 𝑁3 𝑇 𝑥 𝑦 𝑧 = 𝑥 − 2𝑦 1 −1 3 + y 0 2 1 𝑇 𝑥 𝑦 𝑧 = 𝑥 − 2𝑦 + 0 −𝑥 + 2𝑦 + 2𝑦 3𝑥 − 6𝑦 + 𝑦 = 𝑥 − 2𝑦 −𝑥 + 4𝑦 3𝑥 − 5𝑦 𝑇 −1 −2 3 = 𝟑 −𝟕 𝟕
  • 9. 7.Sea una transformación lineal 𝑹𝟑 en 𝑹𝟑 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 f 𝛼𝑢 + 𝛽𝑣 + 𝛿𝑤 = 𝛼𝑓 𝑢 + 𝛽 𝑣 + 𝛿𝑓(𝑤) 2 −3 1 = 𝛼 11 1 1 + 𝛽 2 0 0 + 𝛿 0 4 5 1 2 0 2 1 0 4 −3 1 0 5 1 𝐹2 − 𝐹1 → 𝐹2 𝐹3 − 𝐹1 → 𝐹3 1 2 0 2 0 −2 4 −5 0 −2 5 −1 𝐹3 − 𝐹2 → 𝐹3 1 2 0 2 0 −2 4 −5 0 0 1 4 - 𝛿 = 4 −2𝛽 + 4𝛿 = −5 𝛼 + 2𝛽 = 2 δ=4 β = −5−4(4) −2 = 21 2 α = −2 21 2 + 2 = −19 = -19 1 − 2𝑡 + 𝑡2 + 21 2 3 + 𝑡 − 𝑡2 + 4 2 + 3𝑡 + 3𝑡2 =−19+38t−19𝑡2 + 63 2 + 21 2 𝑡 − 21 2 𝑡2 + 8 + 12𝑡 + 12𝑡2 𝒇 𝟐, −𝟑, 𝟏 =- 𝟑𝟓 𝟐 𝒕𝟐 + 𝟏𝟐𝟏 𝟐 𝒕 + 𝟒𝟏 𝟐