Regalo de Navidad para Antonio José Fernández Soriano...

Regalo de Navidad para Antonio José Fernández Soriano (ejercicio número 1)

Preguntas 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 TotalPuntuación obtenida

Observaciones

1. El ejercicio se debe entregar el primer día de clase después de las vacaciones de Navidad.

2. Además de los enunciados rellenos con las soluciones, debes entregar el desarrollo de laresolución de los problemas.

3. Entregad dos folios grapados.

4. Estos ejercicios no eliminan ninguna pregunta del examen, valdrán para complementar lanotas final del cuatrimestre.

5. En la corrección se atenderá principalmente al resultado final y el desarrollo complementarála valoración de cada uno de los ejercicios.

Sea f : R4 → R5 la aplicación lineal definida por

Mβ5c β5

c(f) =

1 2 3 10 1 5 20 0 1 30 0 0 00 0 0 0

Responde a las siguientes cuestiones:

1. Encuentra bases β y β′ tales que Mββ′ =

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

Plantilla para la solución:

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

2. Considera las bases

β4 = {(1, 1, 1, 1), (1, 2, 1, 0), (3, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 2, 3, 1, 0), (5, 2, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

y calcula Mβ4β5(f)

Mβ4β5(f) =

3. Calcula las matrices Mβ4β4

cy Mβ5β5

Mβ4β4c

Mβ5β5

4. Calcula Ker f e Im f dando ecuaciones cartesianas, dimensiones y bases (todo ello respecto

de las bases β4 y β5 respectivamente).

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

5. Sea g : R4 → R4 dada por g(x, y, z, t)β4 = (x + 2y + 3z + 5t, y, z + t, y + z + t)β4 , calculaMβ4β4(g)

Mβ4β4(g) =

6. Calcula Ker g (respecto a la base β4 todos los resultados que des).

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

7. Calcula Im g (respecto a la base β4 todos los resultados que des).

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

8. Calcula (Ker g + Ker f) + Im g

(Ker g + Ker f) + Im g

(Ker g + Ker f) + Im g =

βKer g+Im g =

dim[(Ker g + Ker f) + Im g] =

9. Calcula Mβ4β5(f ◦ g) (observa que Mβ4β5(f ◦ g) = Mβ4β5(f)Mβ4β4(g))

Mβ4β5(f ◦ g) =

10. Calcula la expresión analítica de f ◦ g

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

11. Sea ahora la matriz A formada por las tres primeras filas y columnas de Mβ4c β5

c(f) y calcula

(sin diagonalizar) la potencia 7 de A (tienes que seguir como ejemplo un ejercicio del tema 2que hicimos en clase donde se descomponía A como la suma de la matriz identidad con unamatriz nilpotente).

12. Encuentra (si existen) matrices P y P−1 tales que P−1AP es una matriz diagonal diciendo

quién es ésta. En este ejercicio

− 1 5 3 − 9− 3 7 3 − 9− 3 5 5 − 9− 2 2 6 − 8

; P−1 =

13. Calcula el siguiente determinante∣∣∣∣∣∣∣∣

4 11 15 224 5 8 92 1 1 12 4 7 13

∣∣∣∣∣∣∣∣ =

14. Calcula la primitiva que sigue

∫4x + 8

x2 − 4x + 13dx =

15. Dada A =

(1 20 1

)se pide calcular (usando el binomio de Newton)

16. Recordamos que el conjunto M2×2(R), formado por todas las matrices de tamaño 2 × 2sobre los números reales, es un espacio vectorial de dimensión 4 cuya base canónica es βc =

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(2 −22 2

)se pide justificar si el conjunto

de matrices β = {I2, A,A2, A3} es o no es una base de M2×2(R).

17. Factoriza (dejando claros todos los cálculos que hagas) el polinomio

p(x) = x4 + 24 =

18. Dados los espacios vectoriales U = {(x, y, z, t) : 5x + 2y + 3z = 2y + 5z = 5z = 0} yV =< (0, 1, 3, 0), (0, 0, 1, 0) > se pide calcular:

a) Una base de U .b) Las ecuaciones cartesianas de V respecto de la base β = {(1, 1, 1, 1), (1, 1, 1, 0), (1, 1, 0, 0), (1, 0, 0, 0)}.c) Una base de U ∩ V .d) Una base de U + V .e) Las ecuaciones de U + V respecto de la base canónica.

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

19. Discute (según el valor del parámetro α) el sistema de ecuaciones x + 2αy + z = 8x + αy + z = 103x + αy = 9

Hasta aquí el ejercicio de Antonio José Fernández Soriano.

Regalo de Navidad para Noelia Alcaraz Gómez (ejercicio número 2)

Observaciones

Mβ5c β5

c(f) =

1 2 6 10 1 3 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 2, 1, 0), (6, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 2, 6, 1, 0), (3, 2, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 10 − 12 6− 3 12 − 12 6− 3 10 − 10 6− 2 7 − 9 7

; P−1 =

4 20 11 194 8 6 82 1 1 12 7 5 11

∣∣∣∣∣∣∣∣ =

∫4x + 8

x2 − 4x + 40dx =

15. Dada A =

(1 20 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(2 −22 2

p(x) = x4 + 24 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Noelia Alcaraz Gómez.

Regalo de Navidad para Laura María Gómez Rocamora (ejercicio número 3)

Observaciones

Mβ5c β5

c(f) =

1 2 5 10 1 6 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 2, 1, 0), (5, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 2, 5, 1, 0), (6, 2, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 6 0 − 6− 3 8 0 − 6− 3 6 2 − 6− 2 3 3 − 5

; P−1 =

4 17 17 164 7 9 72 1 1 12 6 8 9

∣∣∣∣∣∣∣∣ =

∫4x + 8

x2 − 4x + 29dx =

15. Dada A =

(1 20 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(2 −22 2

p(x) = x4 + 24 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Laura María Gómez Rocamora.

Regalo de Navidad para Antonio Alfonso Giménez García (ejercicio número 4)

Observaciones

Mβ5c β5

c(f) =

1 2 4 10 1 5 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 2, 1, 0), (4, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 2, 4, 1, 0), (5, 2, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 6 0 − 6− 3 8 0 − 6− 3 6 2 − 6− 2 3 3 − 5

; P−1 =

4 14 15 174 6 8 62 1 1 12 5 7 11

∣∣∣∣∣∣∣∣ =

∫4x + 8

x2 − 4x + 20dx =

15. Dada A =

(1 20 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(2 −22 2

p(x) = x4 + 24 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Antonio Alfonso Giménez García.

Regalo de Navidad para Ángel Tudela Mayol (ejercicio número 5)

Observaciones

Mβ5c β5

c(f) =

1 3 3 10 1 6 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 3, 1, 0), (3, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 3, 3, 1, 0), (6, 3, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 4 6 − 12− 3 6 6 − 12− 3 4 8 − 12− 2 1 9 − 11

; P−1 =

6 11 17 226 5 9 93 1 1 13 4 8 13

∣∣∣∣∣∣∣∣ =

∫4x + 12

x2 − 6x + 18dx =

15. Dada A =

(1 30 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(3 −33 3

p(x) = x4 + 34 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Ángel Tudela Mayol.

Regalo de Navidad para Matías López Bastida (ejercicio número 6)

Observaciones

Mβ5c β5

c(f) =

1 3 6 10 1 2 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 3, 1, 0), (6, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 3, 6, 1, 0), (2, 3, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 11 − 15 9− 3 13 − 15 9− 3 11 − 13 9− 2 8 − 12 10

; P−1 =

6 20 9 196 8 5 83 1 1 13 7 4 11

∣∣∣∣∣∣∣∣ =

∫4x + 12

x2 − 6x + 45dx =

15. Dada A =

(1 30 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(3 −33 3

p(x) = x4 + 34 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Matías López Bastida.

Regalo de Navidad para Javier Bustillos Cuéllar (ejercicio número 7)

Observaciones

Mβ5c β5

c(f) =

1 3 5 10 1 7 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 3, 1, 0), (5, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 3, 5, 1, 0), (7, 3, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 5 3 − 9− 3 7 3 − 9− 3 5 5 − 9− 2 2 6 − 8

; P−1 =

6 17 19 166 7 10 73 1 1 13 6 9 9

∣∣∣∣∣∣∣∣ =

∫4x + 12

x2 − 6x + 34dx =

15. Dada A =

(1 30 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(3 −33 3

p(x) = x4 + 34 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Javier Bustillos Cuéllar.

Regalo de Navidad para José Miguel Sánchez Zapata (ejercicio número 8)

Observaciones

Mβ5c β5

c(f) =

1 3 4 10 1 1 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 3, 1, 0), (4, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 3, 4, 1, 0), (1, 3, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 10 − 12 6− 3 12 − 12 6− 3 10 − 10 6− 2 7 − 9 7

; P−1 =

6 14 7 176 6 4 63 1 1 13 5 3 11

∣∣∣∣∣∣∣∣ =

∫4x + 12

x2 − 6x + 25dx =

15. Dada A =

(1 30 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(3 −33 3

p(x) = x4 + 34 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de José Miguel Sánchez Zapata.

Regalo de Navidad para Juan Francisco García Pedreño (ejercicio número 9)

Observaciones

Mβ5c β5

c(f) =

1 4 3 10 1 4 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 4, 1, 0), (3, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 4, 3, 1, 0), (4, 4, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 6 0 − 6− 3 8 0 − 6− 3 6 2 − 6− 2 3 3 − 5

; P−1 =

8 11 13 228 5 7 94 1 1 14 4 6 13

∣∣∣∣∣∣∣∣ =

∫4x + 16

x2 − 8x + 25dx =

15. Dada A =

(1 40 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(4 −44 4

p(x) = x4 + 44 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Juan Francisco García Pedreño.

Regalo de Navidad para Daniel Guillén Buendía (ejercicio número 10)

Observaciones

Mβ5c β5

c(f) =

1 4 6 10 1 1 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 4, 1, 0), (6, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 4, 6, 1, 0), (1, 4, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 12 − 18 12− 3 14 − 18 12− 3 12 − 16 12− 2 9 − 15 13

; P−1 =

8 20 7 198 8 4 84 1 1 14 7 3 11

∣∣∣∣∣∣∣∣ =

∫4x + 16

x2 − 8x + 52dx =

15. Dada A =

(1 40 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(4 −44 4

p(x) = x4 + 44 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Daniel Guillén Buendía.

Regalo de Navidad para José Antonio Madrid Font (ejercicio número 11)

Observaciones

Mβ5c β5

c(f) =

1 4 5 10 1 8 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 4, 1, 0), (5, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 4, 5, 1, 0), (8, 4, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 4 6 − 12− 3 6 6 − 12− 3 4 8 − 12− 2 1 9 − 11

; P−1 =

8 17 21 168 7 11 74 1 1 14 6 10 9

∣∣∣∣∣∣∣∣ =

∫4x + 16

x2 − 8x + 41dx =

15. Dada A =

(1 40 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(4 −44 4

p(x) = x4 + 44 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de José Antonio Madrid Font.

Regalo de Navidad para Alberto Corbalán Bernad (ejercicio número 12)

Observaciones

Mβ5c β5

c(f) =

1 4 4 10 1 6 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 4, 1, 0), (4, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 4, 4, 1, 0), (6, 4, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 5 3 − 9− 3 7 3 − 9− 3 5 5 − 9− 2 2 6 − 8

; P−1 =

8 14 17 178 6 9 64 1 1 14 5 8 11

∣∣∣∣∣∣∣∣ =

∫4x + 16

x2 − 8x + 32dx =

15. Dada A =

(1 40 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(4 −44 4

p(x) = x4 + 44 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Alberto Corbalán Bernad.

Regalo de Navidad para Francisco Fernández Sánchez (ejercicio número 13)

Observaciones

Mβ5c β5

c(f) =

1 5 1 10 1 2 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 5, 1, 0), (1, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 5, 1, 1, 0), (2, 5, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 6 0 − 6− 3 8 0 − 6− 3 6 2 − 6− 2 3 3 − 5

; P−1 =

10 5 9 2210 3 5 95 1 1 15 2 4 13

∣∣∣∣∣∣∣∣ =

∫4x + 20

x2 − 10x + 26dx =

15. Dada A =

(1 50 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(5 −55 5

p(x) = x4 + 54 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Francisco Fernández Sánchez.

Regalo de Navidad para Pablo González Alcaraz (ejercicio número 14)

Observaciones

Mβ5c β5

c(f) =

1 5 1 10 1 3 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 5, 1, 0), (1, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 5, 1, 1, 0), (3, 5, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 5 3 − 9− 3 7 3 − 9− 3 5 5 − 9− 2 2 6 − 8

; P−1 =

10 5 11 1910 3 6 85 1 1 15 2 5 11

∣∣∣∣∣∣∣∣ =

∫4x + 20

x2 − 10x + 26dx =

15. Dada A =

(1 50 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(5 −55 5

p(x) = x4 + 54 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Pablo González Alcaraz.

Regalo de Navidad para Emilio Gil Estade (ejercicio número 15)

Observaciones

Mβ5c β5

c(f) =

1 5 1 10 1 4 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 5, 1, 0), (1, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 5, 1, 1, 0), (4, 5, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 4 6 − 12− 3 6 6 − 12− 3 4 8 − 12− 2 1 9 − 11

; P−1 =

10 5 13 1610 3 7 75 1 1 15 2 6 9

∣∣∣∣∣∣∣∣ =

∫4x + 20

x2 − 10x + 26dx =

15. Dada A =

(1 50 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(5 −55 5

p(x) = x4 + 54 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Emilio Gil Estade.

Regalo de Navidad para Celia Ruiz Zamora (ejercicio número 16)

Observaciones

Mβ5c β5

c(f) =

1 5 1 10 1 5 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 5, 1, 0), (1, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 5, 1, 1, 0), (5, 5, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 3 9 − 15− 3 5 9 − 15− 3 3 11 − 15− 2 0 12 − 14

; P−1 =

10 5 15 1710 3 8 65 1 1 15 2 7 11

∣∣∣∣∣∣∣∣ =

∫4x + 20

x2 − 10x + 26dx =

15. Dada A =

(1 50 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(5 −55 5

p(x) = x4 + 54 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Celia Ruiz Zamora.

Regalo de Navidad para Sebastián García Abellán (ejercicio número 17)

Observaciones

Mβ5c β5

c(f) =

1 6 2 10 1 2 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 6, 1, 0), (2, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 6, 2, 1, 0), (2, 6, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 7 − 3 − 3− 3 9 − 3 − 3− 3 7 − 1 − 3− 2 4 0 − 2

; P−1 =

12 8 9 2212 4 5 96 1 1 16 3 4 13

∣∣∣∣∣∣∣∣ =

∫4x + 24

x2 − 12x + 40dx =

15. Dada A =

(1 60 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(6 −66 6

p(x) = x4 + 64 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Sebastián García Abellán.

Regalo de Navidad para José Ángel Rubio López (ejercicio número 18)

Observaciones

Mβ5c β5

c(f) =

1 6 2 10 1 3 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 6, 1, 0), (2, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 6, 2, 1, 0), (3, 6, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 6 0 − 6− 3 8 0 − 6− 3 6 2 − 6− 2 3 3 − 5

; P−1 =

12 8 11 1912 4 6 86 1 1 16 3 5 11

∣∣∣∣∣∣∣∣ =

∫4x + 24

x2 − 12x + 40dx =

15. Dada A =

(1 60 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(6 −66 6

p(x) = x4 + 64 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de José Ángel Rubio López.

Regalo de Navidad para Lourdes Fábregas (ejercicio número 19)

Observaciones

Mβ5c β5

c(f) =

1 6 2 10 1 4 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 6, 1, 0), (2, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 6, 2, 1, 0), (4, 6, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 5 3 − 9− 3 7 3 − 9− 3 5 5 − 9− 2 2 6 − 8

; P−1 =

12 8 13 1612 4 7 76 1 1 16 3 6 9

∣∣∣∣∣∣∣∣ =

∫4x + 24

x2 − 12x + 40dx =

15. Dada A =

(1 60 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(6 −66 6

p(x) = x4 + 64 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Lourdes Fábregas.

Regalo de Navidad para Ana María Antolí Gil (ejercicio número 20)

Observaciones

Mβ5c β5

c(f) =

1 6 2 10 1 5 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 6, 1, 0), (2, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 6, 2, 1, 0), (5, 6, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 4 6 − 12− 3 6 6 − 12− 3 4 8 − 12− 2 1 9 − 11

; P−1 =

12 8 15 1712 4 8 66 1 1 16 3 7 11

∣∣∣∣∣∣∣∣ =

∫4x + 24

x2 − 12x + 40dx =

15. Dada A =

(1 60 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(6 −66 6

p(x) = x4 + 64 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Ana María Antolí Gil.

Regalo de Navidad para Álvaro Conesa Aparicio (ejercicio número 21)

Observaciones

Mβ5c β5

c(f) =

1 7 2 10 1 6 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 7, 1, 0), (2, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 7, 2, 1, 0), (6, 7, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 3 9 − 15− 3 5 9 − 15− 3 3 11 − 15− 2 0 12 − 14

; P−1 =

14 8 17 2214 4 9 97 1 1 17 3 8 13

∣∣∣∣∣∣∣∣ =

∫4x + 28

x2 − 14x + 53dx =

15. Dada A =

(1 70 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(7 −77 7

p(x) = x4 + 74 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Álvaro Conesa Aparicio.

Regalo de Navidad para Enrique López López (ejercicio número 22)

Observaciones

Mβ5c β5

c(f) =

1 7 2 10 1 7 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 7, 1, 0), (2, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 7, 2, 1, 0), (7, 7, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 2 12 − 18− 3 4 12 − 18− 3 2 14 − 18− 2 − 1 15 − 17

; P−1 =

14 8 19 1914 4 10 87 1 1 17 3 9 11

∣∣∣∣∣∣∣∣ =

∫4x + 28

x2 − 14x + 53dx =

15. Dada A =

(1 70 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(7 −77 7

p(x) = x4 + 74 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Enrique López López.

Regalo de Navidad para Miguel Ángel Belmonte Ortiz (ejercicio número 23)

Observaciones

Mβ5c β5

c(f) =

1 7 3 10 1 3 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 7, 1, 0), (3, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 7, 3, 1, 0), (3, 7, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 7 − 3 − 3− 3 9 − 3 − 3− 3 7 − 1 − 3− 2 4 0 − 2

; P−1 =

14 11 11 1614 5 6 77 1 1 17 4 5 9

∣∣∣∣∣∣∣∣ =

∫4x + 28

x2 − 14x + 58dx =

15. Dada A =

(1 70 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(7 −77 7

p(x) = x4 + 74 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Miguel Ángel Belmonte Ortiz.

Regalo de Navidad para Mario González Ortín (ejercicio número 24)

Observaciones

Mβ5c β5

c(f) =

1 7 3 10 1 4 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 7, 1, 0), (3, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 7, 3, 1, 0), (4, 7, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 6 0 − 6− 3 8 0 − 6− 3 6 2 − 6− 2 3 3 − 5

; P−1 =

14 11 13 1714 5 7 67 1 1 17 4 6 11

∣∣∣∣∣∣∣∣ =

∫4x + 28

x2 − 14x + 58dx =

15. Dada A =

(1 70 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(7 −77 7

p(x) = x4 + 74 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Mario González Ortín.

Regalo de Navidad para Pedro Martínez López (ejercicio número 25)

Observaciones

Mβ5c β5

c(f) =

1 3 3 10 1 5 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 3, 1, 0), (3, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 3, 3, 1, 0), (5, 3, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 5 3 − 9− 3 7 3 − 9− 3 5 5 − 9− 2 2 6 − 8

; P−1 =

6 11 15 226 5 8 93 1 1 13 4 7 13

∣∣∣∣∣∣∣∣ =

∫4x + 12

x2 − 6x + 18dx =

15. Dada A =

(1 30 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(3 −33 3

p(x) = x4 + 34 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Pedro Martínez López.

Regalo de Navidad para Elena Pérez Gómez (ejercicio número 26)

Observaciones

Mβ5c β5

c(f) =

1 4 3 10 1 4 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 4, 1, 0), (3, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 4, 3, 1, 0), (4, 4, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 6 0 − 6− 3 8 0 − 6− 3 6 2 − 6− 2 3 3 − 5

; P−1 =

8 11 13 198 5 7 84 1 1 14 4 6 11

∣∣∣∣∣∣∣∣ =

∫4x + 16

x2 − 8x + 25dx =

15. Dada A =

(1 40 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(4 −44 4

p(x) = x4 + 44 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Elena Pérez Gómez.

Regalo de Navidad para Carlos Gómez-Gil de la Cerda (ejercicio número 27)

Observaciones

Mβ5c β5

c(f) =

1 5 3 10 1 7 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 5, 1, 0), (3, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 5, 3, 1, 0), (7, 5, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 3 9 − 15− 3 5 9 − 15− 3 3 11 − 15− 2 0 12 − 14

; P−1 =

10 11 19 1610 5 10 75 1 1 15 4 9 9

∣∣∣∣∣∣∣∣ =

∫4x + 20

x2 − 10x + 34dx =

15. Dada A =

(1 50 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(5 −55 5

p(x) = x4 + 54 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Carlos Gómez-Gil de la Cerda.

Regalo de Navidad para Ana Marina Carralero Hernández (ejercicio número 28)

Observaciones

Mβ5c β5

c(f) =

1 6 3 10 1 8 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 6, 1, 0), (3, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 6, 3, 1, 0), (8, 6, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 2 12 − 18− 3 4 12 − 18− 3 2 14 − 18− 2 − 1 15 − 17

; P−1 =

12 11 21 1712 5 11 66 1 1 16 4 10 11

∣∣∣∣∣∣∣∣ =

∫4x + 24

x2 − 12x + 45dx =

15. Dada A =

(1 60 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(6 −66 6

p(x) = x4 + 64 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Ana Marina Carralero Hernández.

Regalo de Navidad para Diego Barberán Verdú (ejercicio número 29)

Observaciones

Mβ5c β5

c(f) =

1 2 8 10 1 5 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 2, 1, 0), (8, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 2, 8, 1, 0), (5, 2, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 10 − 12 6− 3 12 − 12 6− 3 10 − 10 6− 2 7 − 9 7

; P−1 =

4 26 15 184 10 8 92 1 1 12 9 7 9

∣∣∣∣∣∣∣∣ =

∫4x + 8

x2 − 4x + 68dx =

15. Dada A =

(1 20 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(2 −22 2

p(x) = x4 + 24 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Diego Barberán Verdú.

Regalo de Navidad para José María Navarro Alarcón (ejercicio número 30)

Observaciones

Mβ5c β5

c(f) =

1 2 8 10 1 4 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 2, 1, 0), (8, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 2, 8, 1, 0), (4, 2, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 11 − 15 9− 3 13 − 15 9− 3 11 − 13 9− 2 8 − 12 10

; P−1 =

4 26 13 174 10 7 82 1 1 12 9 6 9

∣∣∣∣∣∣∣∣ =

∫4x + 8

x2 − 4x + 68dx =

15. Dada A =

(1 20 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(2 −22 2

p(x) = x4 + 24 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de José María Navarro Alarcón.

Regalo de Navidad para Jesús Aguirre Cárcel (ejercicio número 31)

Observaciones

Mβ5c β5

c(f) =

1 2 8 10 1 5 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 2, 1, 0), (8, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 2, 8, 1, 0), (5, 2, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 10 − 12 6− 3 12 − 12 6− 3 10 − 10 6− 2 7 − 9 7

; P−1 =

4 26 15 164 10 8 72 1 1 12 9 7 9

∣∣∣∣∣∣∣∣ =

∫4x + 8

x2 − 4x + 68dx =

15. Dada A =

(1 20 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(2 −22 2

p(x) = x4 + 24 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Jesús Aguirre Cárcel.

Regalo de Navidad para Álvaro Alejandro Martínez Millán (ejercicio número 32)

Observaciones

Mβ5c β5

c(f) =

1 2 8 10 1 2 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 2, 1, 0), (8, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 2, 8, 1, 0), (2, 2, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 13 − 21 15− 3 15 − 21 15− 3 13 − 19 15− 2 10 − 18 16

; P−1 =

4 26 9 154 10 5 62 1 1 12 9 4 9

∣∣∣∣∣∣∣∣ =

∫4x + 8

x2 − 4x + 68dx =

15. Dada A =

(1 20 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(2 −22 2

p(x) = x4 + 24 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Álvaro Alejandro Martínez Millán.

Regalo de Navidad para Adrián Ibánez del Toro (ejercicio número 33)

Observaciones

Mβ5c β5

c(f) =

1 3 8 10 1 1 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 3, 1, 0), (8, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 3, 8, 1, 0), (1, 3, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 14 − 24 18− 3 16 − 24 18− 3 14 − 22 18− 2 11 − 21 19

; P−1 =

6 26 7 226 10 4 93 1 1 13 9 3 13

∣∣∣∣∣∣∣∣ =

∫4x + 12

x2 − 6x + 73dx =

15. Dada A =

(1 30 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(3 −33 3

p(x) = x4 + 34 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Adrián Ibánez del Toro.

Regalo de Navidad para Daviz Pérez Lozano (ejercicio número 34)

Observaciones

Mβ5c β5

c(f) =

1 3 9 10 1 6 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 3, 1, 0), (9, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 3, 9, 1, 0), (6, 3, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 10 − 12 6− 3 12 − 12 6− 3 10 − 10 6− 2 7 − 9 7

; P−1 =

6 29 17 216 11 9 83 1 1 13 10 8 13

∣∣∣∣∣∣∣∣ =

∫4x + 12

x2 − 6x + 90dx =

15. Dada A =

(1 30 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(3 −33 3

p(x) = x4 + 34 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Daviz Pérez Lozano.

Regalo de Navidad para Francisco David Morales Fernández (ejercicio número 35)

Observaciones

Mβ5c β5

c(f) =

1 3 9 10 1 5 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 3, 1, 0), (9, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 3, 9, 1, 0), (5, 3, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 11 − 15 9− 3 13 − 15 9− 3 11 − 13 9− 2 8 − 12 10

; P−1 =

6 29 15 206 11 8 73 1 1 13 10 7 13

∣∣∣∣∣∣∣∣ =

∫4x + 12

x2 − 6x + 90dx =

15. Dada A =

(1 30 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(3 −33 3

p(x) = x4 + 34 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Francisco David Morales Fernández.

Regalo de Navidad para Elina Krassimirova Anguelova (ejercicio número 36)

Observaciones

Mβ5c β5

c(f) =

1 3 9 10 1 4 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 3, 1, 0), (9, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 3, 9, 1, 0), (4, 3, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 12 − 18 12− 3 14 − 18 12− 3 12 − 16 12− 2 9 − 15 13

; P−1 =

6 29 13 196 11 7 63 1 1 13 10 6 13

∣∣∣∣∣∣∣∣ =

∫4x + 12

x2 − 6x + 90dx =

15. Dada A =

(1 30 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(3 −33 3

p(x) = x4 + 34 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Elina Krassimirova Anguelova.

Regalo de Navidad para Carlos Andrés Santiago Civantos (ejercicio número 37)

Observaciones

Mβ5c β5

c(f) =

1 4 9 10 1 3 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 4, 1, 0), (9, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 4, 9, 1, 0), (3, 4, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 13 − 21 15− 3 15 − 21 15− 3 13 − 19 15− 2 10 − 18 16

; P−1 =

8 29 11 268 11 6 94 1 1 14 10 5 17

∣∣∣∣∣∣∣∣ =

∫4x + 16

x2 − 8x + 97dx =

15. Dada A =

(1 40 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(4 −44 4

p(x) = x4 + 44 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Carlos Andrés Santiago Civantos.

Regalo de Navidad para Fernando Monteagudo López (ejercicio número 38)

Observaciones

Mβ5c β5

c(f) =

1 4 9 10 1 2 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 4, 1, 0), (9, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 4, 9, 1, 0), (2, 4, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 14 − 24 18− 3 16 − 24 18− 3 14 − 22 18− 2 11 − 21 19

; P−1 =

8 29 9 168 11 5 84 1 1 14 10 4 8

∣∣∣∣∣∣∣∣ =

∫4x + 16

x2 − 8x + 97dx =

15. Dada A =

(1 40 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(4 −44 4

p(x) = x4 + 44 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Fernando Monteagudo López.

Regalo de Navidad para Fernando Costa Hernández (ejercicio número 39)

Observaciones

Mβ5c β5

c(f) =

1 4 9 10 1 1 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 4, 1, 0), (9, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 4, 9, 1, 0), (1, 4, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 15 − 27 21− 3 17 − 27 21− 3 15 − 25 21− 2 12 − 24 22

; P−1 =

8 29 7 158 11 4 74 1 1 14 10 3 8

∣∣∣∣∣∣∣∣ =

∫4x + 16

x2 − 8x + 97dx =

15. Dada A =

(1 40 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(4 −44 4

p(x) = x4 + 44 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Fernando Costa Hernández.

Regalo de Navidad para Carmen María Hernández Jara (ejercicio número 40)

Observaciones

Mβ5c β5

c(f) =

1 4 7 10 1 4 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 4, 1, 0), (7, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 4, 7, 1, 0), (4, 4, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 10 − 12 6− 3 12 − 12 6− 3 10 − 10 6− 2 7 − 9 7

; P−1 =

8 23 13 148 9 7 64 1 1 14 8 6 8

∣∣∣∣∣∣∣∣ =

∫4x + 16

x2 − 8x + 65dx =

15. Dada A =

(1 40 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(4 −44 4

p(x) = x4 + 44 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Carmen María Hernández Jara.

Regalo de Navidad para Pedro Muñoz Díaz (ejercicio número 41)

Observaciones

Mβ5c β5

c(f) =

1 5 7 10 1 3 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 5, 1, 0), (7, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 5, 7, 1, 0), (3, 5, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 11 − 15 9− 3 13 − 15 9− 3 11 − 13 9− 2 8 − 12 10

; P−1 =

10 23 11 2110 9 6 95 1 1 15 8 5 12

∣∣∣∣∣∣∣∣ =

∫4x + 20

x2 − 10x + 74dx =

15. Dada A =

(1 50 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(5 −55 5

p(x) = x4 + 54 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Pedro Muñoz Díaz.

Regalo de Navidad para Gloria Motos Cascales (ejercicio número 42)

Observaciones

Mβ5c β5

c(f) =

1 3 3 10 1 4 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 3, 1, 0), (3, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 3, 3, 1, 0), (4, 3, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 6 0 − 6− 3 8 0 − 6− 3 6 2 − 6− 2 3 3 − 5

; P−1 =

6 11 13 196 5 7 63 1 1 13 4 6 13

∣∣∣∣∣∣∣∣ =

∫4x + 12

x2 − 6x + 18dx =

15. Dada A =

(1 30 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(3 −33 3

p(x) = x4 + 34 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Gloria Motos Cascales.

Regalo de Navidad para Patricio David Tripiana García (ejercicio número 43)

Observaciones

Mβ5c β5

c(f) =

1 4 3 10 1 5 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 4, 1, 0), (3, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 4, 3, 1, 0), (5, 4, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 5 3 − 9− 3 7 3 − 9− 3 5 5 − 9− 2 2 6 − 8

; P−1 =

8 11 15 268 5 8 94 1 1 14 4 7 17

∣∣∣∣∣∣∣∣ =

∫4x + 16

x2 − 8x + 25dx =

15. Dada A =

(1 40 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(4 −44 4

p(x) = x4 + 44 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Patricio David Tripiana García.

Regalo de Navidad para María Ángeles García Martínez (ejercicio número 44)

Observaciones

Mβ5c β5

c(f) =

1 4 3 10 1 6 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 4, 1, 0), (3, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 4, 3, 1, 0), (6, 4, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 4 6 − 12− 3 6 6 − 12− 3 4 8 − 12− 2 1 9 − 11

; P−1 =

8 11 17 168 5 9 84 1 1 14 4 8 8

∣∣∣∣∣∣∣∣ =

∫4x + 16

x2 − 8x + 25dx =

15. Dada A =

(1 40 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(4 −44 4

p(x) = x4 + 44 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de María Ángeles García Martínez.

Regalo de Navidad para Alejandro Ottenwalder Rivas (ejercicio número 45)

Observaciones

Mβ5c β5

c(f) =

1 4 3 10 1 7 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 4, 1, 0), (3, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 4, 3, 1, 0), (7, 4, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 3 9 − 15− 3 5 9 − 15− 3 3 11 − 15− 2 0 12 − 14

; P−1 =

8 11 19 158 5 10 74 1 1 14 4 9 8

∣∣∣∣∣∣∣∣ =

∫4x + 16

x2 − 8x + 25dx =

15. Dada A =

(1 40 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(4 −44 4

p(x) = x4 + 44 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Alejandro Ottenwalder Rivas.

Regalo de Navidad para Juan Miguel Rodríguez Soler (ejercicio número 46)

Observaciones

Mβ5c β5

c(f) =

1 5 7 10 1 2 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 5, 1, 0), (7, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 5, 7, 1, 0), (2, 5, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 12 − 18 12− 3 14 − 18 12− 3 12 − 16 12− 2 9 − 15 13

; P−1 =

10 23 9 2010 9 5 85 1 1 15 8 4 12

∣∣∣∣∣∣∣∣ =

∫4x + 20

x2 − 10x + 74dx =

15. Dada A =

(1 50 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(5 −55 5

p(x) = x4 + 54 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Juan Miguel Rodríguez Soler.

Regalo de Navidad para Aránzazu Ros Liarte (ejercicio número 47)

Observaciones

Mβ5c β5

c(f) =

1 5 7 10 1 1 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 5, 1, 0), (7, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 5, 7, 1, 0), (1, 5, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 13 − 21 15− 3 15 − 21 15− 3 13 − 19 15− 2 10 − 18 16

; P−1 =

10 23 7 1910 9 4 75 1 1 15 8 3 12

∣∣∣∣∣∣∣∣ =

∫4x + 20

x2 − 10x + 74dx =

15. Dada A =

(1 50 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(5 −55 5

p(x) = x4 + 54 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Aránzazu Ros Liarte.

Regalo de Navidad para Alfonso Dávila Abellán (ejercicio número 48)

Observaciones

Mβ5c β5

c(f) =

1 5 6 10 1 3 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 5, 1, 0), (6, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 5, 6, 1, 0), (3, 5, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 10 − 12 6− 3 12 − 12 6− 3 10 − 10 6− 2 7 − 9 7

; P−1 =

10 20 11 1810 8 6 65 1 1 15 7 5 12

∣∣∣∣∣∣∣∣ =

∫4x + 20

x2 − 10x + 61dx =

15. Dada A =

(1 50 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(5 −55 5

p(x) = x4 + 54 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Alfonso Dávila Abellán.

Regalo de Navidad para Javier Olmedo Romero (ejercicio número 49)

Observaciones

Mβ5c β5

c(f) =

1 6 6 10 1 2 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 6, 1, 0), (6, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 6, 6, 1, 0), (2, 6, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 11 − 15 9− 3 13 − 15 9− 3 11 − 13 9− 2 8 − 12 10

; P−1 =

12 20 9 2512 8 5 96 1 1 16 7 4 16

∣∣∣∣∣∣∣∣ =

∫4x + 24

x2 − 12x + 72dx =

15. Dada A =

(1 60 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(6 −66 6

p(x) = x4 + 64 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Javier Olmedo Romero.

Regalo de Navidad para Julio José Jiménez Cava (ejercicio número 50)

Observaciones

Mβ5c β5

c(f) =

1 6 6 10 1 1 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 6, 1, 0), (6, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 6, 6, 1, 0), (1, 6, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 12 − 18 12− 3 14 − 18 12− 3 12 − 16 12− 2 9 − 15 13

; P−1 =

12 20 7 2412 8 4 86 1 1 16 7 3 16

∣∣∣∣∣∣∣∣ =

∫4x + 24

x2 − 12x + 72dx =

15. Dada A =

(1 60 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(6 −66 6

p(x) = x4 + 64 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Julio José Jiménez Cava.

Regalo de Navidad para Juan Manuel Girón Sola (ejercicio número 51)

Observaciones

Mβ5c β5

c(f) =

1 6 5 10 1 2 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 6, 1, 0), (5, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 6, 5, 1, 0), (2, 6, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 10 − 12 6− 3 12 − 12 6− 3 10 − 10 6− 2 7 − 9 7

; P−1 =

12 17 9 1412 7 5 76 1 1 16 6 4 7

∣∣∣∣∣∣∣∣ =

∫4x + 24

x2 − 12x + 61dx =

15. Dada A =

(1 60 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(6 −66 6

p(x) = x4 + 64 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Juan Manuel Girón Sola.

Regalo de Navidad para Francisco José Pérez Losella (ejercicio número 52)

Observaciones

Mβ5c β5

c(f) =

1 6 5 10 1 1 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 6, 1, 0), (5, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 6, 5, 1, 0), (1, 6, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 11 − 15 9− 3 13 − 15 9− 3 11 − 13 9− 2 8 − 12 10

; P−1 =

12 17 7 1312 7 4 66 1 1 16 6 3 7

∣∣∣∣∣∣∣∣ =

∫4x + 24

x2 − 12x + 61dx =

15. Dada A =

(1 60 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(6 −66 6

p(x) = x4 + 64 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Francisco José Pérez Losella.

Regalo de Navidad para Miguel García Alcobas (ejercicio número 53)

Observaciones

Mβ5c β5

c(f) =

1 7 1 10 1 2 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 7, 1, 0), (1, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 7, 1, 1, 0), (2, 7, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 6 0 − 6− 3 8 0 − 6− 3 6 2 − 6− 2 3 3 − 5

; P−1 =

14 5 9 2014 3 5 97 1 1 17 2 4 11

∣∣∣∣∣∣∣∣ =

∫4x + 28

x2 − 14x + 50dx =

15. Dada A =

(1 70 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(7 −77 7

p(x) = x4 + 74 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Miguel García Alcobas.

Regalo de Navidad para Oscar Amorós Ferri (ejercicio número 54)

Observaciones

Mβ5c β5

c(f) =

1 7 1 10 1 3 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 7, 1, 0), (1, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 7, 1, 1, 0), (3, 7, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 5 3 − 9− 3 7 3 − 9− 3 5 5 − 9− 2 2 6 − 8

; P−1 =

14 5 11 1914 3 6 87 1 1 17 2 5 11

∣∣∣∣∣∣∣∣ =

∫4x + 28

x2 − 14x + 50dx =

15. Dada A =

(1 70 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(7 −77 7

p(x) = x4 + 74 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Oscar Amorós Ferri.

Regalo de Navidad para Rosa Ana Hurtado Latorre (ejercicio número 55)

Observaciones

Mβ5c β5

c(f) =

1 7 1 10 1 4 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 7, 1, 0), (1, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 7, 1, 1, 0), (4, 7, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 4 6 − 12− 3 6 6 − 12− 3 4 8 − 12− 2 1 9 − 11

; P−1 =

14 5 13 1814 3 7 77 1 1 17 2 6 11

∣∣∣∣∣∣∣∣ =

∫4x + 28

x2 − 14x + 50dx =

15. Dada A =

(1 70 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(7 −77 7

p(x) = x4 + 74 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Rosa Ana Hurtado Latorre.

Regalo de Navidad para Abraham Lorente Vela (ejercicio número 56)

Observaciones

Mβ5c β5

c(f) =

1 7 1 10 1 5 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 7, 1, 0), (1, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 7, 1, 1, 0), (5, 7, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 3 9 − 15− 3 5 9 − 15− 3 3 11 − 15− 2 0 12 − 14

; P−1 =

14 5 15 1714 3 8 67 1 1 17 2 7 11

∣∣∣∣∣∣∣∣ =

∫4x + 28

x2 − 14x + 50dx =

15. Dada A =

(1 70 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(7 −77 7

p(x) = x4 + 74 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Abraham Lorente Vela.

Regalo de Navidad para Eduardo Cuevas Tortosa (ejercicio número 57)

Observaciones

Mβ5c β5

c(f) =

1 3 1 10 1 6 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 3, 1, 0), (1, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 3, 1, 1, 0), (6, 3, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 2 12 − 18− 3 4 12 − 18− 3 2 14 − 18− 2 − 1 15 − 17

; P−1 =

6 5 17 246 3 9 93 1 1 13 2 8 15

∣∣∣∣∣∣∣∣ =

∫4x + 12

x2 − 6x + 10dx =

15. Dada A =

(1 30 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(3 −33 3

p(x) = x4 + 34 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Eduardo Cuevas Tortosa.

Regalo de Navidad para Andrés García (ejercicio número 58)

Observaciones

Mβ5c β5

c(f) =

1 4 1 10 1 7 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 4, 1, 0), (1, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 4, 1, 1, 0), (7, 4, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 1 15 − 21− 3 3 15 − 21− 3 1 17 − 21− 2 − 2 18 − 20

; P−1 =

8 5 19 238 3 10 84 1 1 14 2 9 15

∣∣∣∣∣∣∣∣ =

∫4x + 16

x2 − 8x + 17dx =

15. Dada A =

(1 40 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(4 −44 4

p(x) = x4 + 44 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Andrés García.

Regalo de Navidad para Ángel Reverte (ejercicio número 59)

Observaciones

Mβ5c β5

c(f) =

1 4 3 10 1 6 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 4, 1, 0), (3, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 4, 3, 1, 0), (6, 4, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 4 6 − 12− 3 6 6 − 12− 3 4 8 − 12− 2 1 9 − 11

; P−1 =

8 11 17 168 5 9 84 1 1 14 4 8 8

∣∣∣∣∣∣∣∣ =

∫4x + 16

x2 − 8x + 25dx =

15. Dada A =

(1 40 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(4 −44 4

p(x) = x4 + 44 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Ángel Reverte.

Regalo de Navidad para María Angeles Moya Navarro (ejercicio número 60)

Observaciones

Mβ5c β5

c(f) =

1 5 1 10 1 8 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 5, 1, 0), (1, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 5, 1, 1, 0), (8, 5, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 0 18 − 24− 3 2 18 − 24− 3 0 20 − 24− 2 − 3 21 − 23

; P−1 =

10 5 21 2210 3 11 75 1 1 15 2 10 15

∣∣∣∣∣∣∣∣ =

∫4x + 20

x2 − 10x + 26dx =

15. Dada A =

(1 50 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(5 −55 5

p(x) = x4 + 54 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de María Angeles Moya Navarro.

Regalo de Navidad para Francisco Antonio Campoy Aznar (ejercicio número 61)

Observaciones

Mβ5c β5

c(f) =

1 6 1 10 1 9 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 6, 1, 0), (1, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 6, 1, 1, 0), (9, 6, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 − 1 21 − 27− 3 1 21 − 27− 3 − 1 23 − 27− 2 − 4 24 − 26

; P−1 =

12 5 23 1212 3 12 66 1 1 16 2 11 6

∣∣∣∣∣∣∣∣ =

∫4x + 24

x2 − 12x + 37dx =

15. Dada A =

(1 60 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(6 −66 6

p(x) = x4 + 64 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Francisco Antonio Campoy Aznar.

Regalo de Navidad para Antonio Ángel Escudero Lidón (ejercicio número 62)

Observaciones

Mβ5c β5

c(f) =

1 2 2 10 1 3 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 2, 1, 0), (2, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 2, 2, 1, 0), (3, 2, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 6 0 − 6− 3 8 0 − 6− 3 6 2 − 6− 2 3 3 − 5

; P−1 =

4 8 11 184 4 6 92 1 1 12 3 5 9

∣∣∣∣∣∣∣∣ =

∫4x + 8

x2 − 4x + 8dx =

15. Dada A =

(1 20 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(2 −22 2

p(x) = x4 + 24 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Antonio Ángel Escudero Lidón.

Regalo de Navidad para Juan Antonio Núñez Reolid (ejercicio número 63)

Observaciones

Mβ5c β5

c(f) =

1 2 2 10 1 4 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 2, 1, 0), (2, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 2, 2, 1, 0), (4, 2, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 5 3 − 9− 3 7 3 − 9− 3 5 5 − 9− 2 2 6 − 8

; P−1 =

4 8 13 174 4 7 82 1 1 12 3 6 9

∣∣∣∣∣∣∣∣ =

∫4x + 8

x2 − 4x + 8dx =

15. Dada A =

(1 20 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(2 −22 2

p(x) = x4 + 24 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Juan Antonio Núñez Reolid.

Regalo de Navidad para María Isabel León Sánchez (ejercicio número 64)

Observaciones

Mβ5c β5

c(f) =

1 2 2 10 1 5 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 2, 1, 0), (2, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 2, 2, 1, 0), (5, 2, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 4 6 − 12− 3 6 6 − 12− 3 4 8 − 12− 2 1 9 − 11

; P−1 =

4 8 15 164 4 8 72 1 1 12 3 7 9

∣∣∣∣∣∣∣∣ =

∫4x + 8

x2 − 4x + 8dx =

15. Dada A =

(1 20 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(2 −22 2

p(x) = x4 + 24 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de María Isabel León Sánchez.

Regalo de Navidad para María Cerezo Martínez (ejercicio número 65)

Observaciones

Mβ5c β5

c(f) =

1 2 2 10 1 6 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 2, 1, 0), (2, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 2, 2, 1, 0), (6, 2, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 3 9 − 15− 3 5 9 − 15− 3 3 11 − 15− 2 0 12 − 14

; P−1 =

4 8 17 154 4 9 62 1 1 12 3 8 9

∣∣∣∣∣∣∣∣ =

∫4x + 8

x2 − 4x + 8dx =

15. Dada A =

(1 20 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(2 −22 2

p(x) = x4 + 24 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de María Cerezo Martínez.

Regalo de Navidad para Cristóbal Marín Fernández (ejercicio número 66)

Observaciones

Mβ5c β5

c(f) =

1 3 2 10 1 7 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 3, 1, 0), (2, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 3, 2, 1, 0), (7, 3, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 2 12 − 18− 3 4 12 − 18− 3 2 14 − 18− 2 − 1 15 − 17

; P−1 =

6 8 19 226 4 10 93 1 1 13 3 9 13

∣∣∣∣∣∣∣∣ =

∫4x + 12

x2 − 6x + 13dx =

15. Dada A =

(1 30 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(3 −33 3

p(x) = x4 + 34 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Cristóbal Marín Fernández.

Regalo de Navidad para Elena Clos Pérez (ejercicio número 67)

Observaciones

Mβ5c β5

c(f) =

1 3 2 10 1 8 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 3, 1, 0), (2, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 3, 2, 1, 0), (8, 3, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 1 15 − 21− 3 3 15 − 21− 3 1 17 − 21− 2 − 2 18 − 20

; P−1 =

6 8 21 216 4 11 83 1 1 13 3 10 13

∣∣∣∣∣∣∣∣ =

∫4x + 12

x2 − 6x + 13dx =

15. Dada A =

(1 30 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(3 −33 3

p(x) = x4 + 34 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Elena Clos Pérez.

Regalo de Navidad para Alejandro Acosta León (ejercicio número 68)

Observaciones

Mβ5c β5

c(f) =

1 3 2 10 1 9 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 3, 1, 0), (2, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 3, 2, 1, 0), (9, 3, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 0 18 − 24− 3 2 18 − 24− 3 0 20 − 24− 2 − 3 21 − 23

; P−1 =

6 8 23 206 4 12 73 1 1 13 3 11 13

∣∣∣∣∣∣∣∣ =

∫4x + 12

x2 − 6x + 13dx =

15. Dada A =

(1 30 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(3 −33 3

p(x) = x4 + 34 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Alejandro Acosta León.

Regalo de Navidad para Alumnno 04 (ejercicio número 69)

Observaciones

Mβ5c β5

c(f) =

1 4 3 10 1 7 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 4, 1, 0), (3, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 4, 3, 1, 0), (7, 4, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 3 9 − 15− 3 5 9 − 15− 3 3 11 − 15− 2 0 12 − 14

; P−1 =

8 11 19 158 5 10 74 1 1 14 4 9 8

∣∣∣∣∣∣∣∣ =

∫4x + 16

x2 − 8x + 25dx =

15. Dada A =

(1 40 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(4 −44 4

p(x) = x4 + 44 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Hasta aquí el ejercicio de Alumnno 04.

Observaciones

Mβ5c β5

c(f) =

1 4 3 10 1 8 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 4, 1, 0), (3, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 4, 3, 1, 0), (8, 4, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 2 12 − 18− 3 4 12 − 18− 3 2 14 − 18− 2 − 1 15 − 17

; P−1 =

8 11 21 148 5 11 64 1 1 14 4 10 8

∣∣∣∣∣∣∣∣ =

∫4x + 16

x2 − 8x + 25dx =

15. Dada A =

(1 40 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(4 −44 4

p(x) = x4 + 44 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Observaciones

Mβ5c β5

c(f) =

1 5 3 10 1 9 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 5, 1, 0), (3, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 5, 3, 1, 0), (9, 5, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 1 15 − 21− 3 3 15 − 21− 3 1 17 − 21− 2 − 2 18 − 20

; P−1 =

10 11 23 2110 5 12 95 1 1 15 4 11 12

∣∣∣∣∣∣∣∣ =

∫4x + 20

x2 − 10x + 34dx =

15. Dada A =

(1 50 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(5 −55 5

p(x) = x4 + 54 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Observaciones

Mβ5c β5

c(f) =

1 5 4 10 1 5 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 5, 1, 0), (4, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 5, 4, 1, 0), (5, 5, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 6 0 − 6− 3 8 0 − 6− 3 6 2 − 6− 2 3 3 − 5

; P−1 =

10 14 15 2010 6 8 85 1 1 15 5 7 12

∣∣∣∣∣∣∣∣ =

∫4x + 20

x2 − 10x + 41dx =

15. Dada A =

(1 50 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(5 −55 5

p(x) = x4 + 54 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Observaciones

Mβ5c β5

c(f) =

1 5 4 10 1 6 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 5, 1, 0), (4, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 5, 4, 1, 0), (6, 5, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 5 3 − 9− 3 7 3 − 9− 3 5 5 − 9− 2 2 6 − 8

; P−1 =

10 14 17 1910 6 9 75 1 1 15 5 8 12

∣∣∣∣∣∣∣∣ =

∫4x + 20

x2 − 10x + 41dx =

15. Dada A =

(1 50 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(5 −55 5

p(x) = x4 + 54 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Observaciones

Mβ5c β5

c(f) =

1 5 4 10 1 7 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 5, 1, 0), (4, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 5, 4, 1, 0), (7, 5, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 4 6 − 12− 3 6 6 − 12− 3 4 8 − 12− 2 1 9 − 11

; P−1 =

10 14 19 1810 6 10 65 1 1 15 5 9 12

∣∣∣∣∣∣∣∣ =

∫4x + 20

x2 − 10x + 41dx =

15. Dada A =

(1 50 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(5 −55 5

p(x) = x4 + 54 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Observaciones

Mβ5c β5

c(f) =

1 6 4 10 1 8 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 6, 1, 0), (4, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 6, 4, 1, 0), (8, 6, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 3 9 − 15− 3 5 9 − 15− 3 3 11 − 15− 2 0 12 − 14

; P−1 =

12 14 21 2512 6 11 96 1 1 16 5 10 16

∣∣∣∣∣∣∣∣ =

∫4x + 24

x2 − 12x + 52dx =

15. Dada A =

(1 60 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(6 −66 6

p(x) = x4 + 64 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Observaciones

Mβ5c β5

c(f) =

1 6 4 10 1 9 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 6, 1, 0), (4, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 6, 4, 1, 0), (9, 6, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 2 12 − 18− 3 4 12 − 18− 3 2 14 − 18− 2 − 1 15 − 17

; P−1 =

12 14 23 2412 6 12 86 1 1 16 5 11 16

∣∣∣∣∣∣∣∣ =

∫4x + 24

x2 − 12x + 52dx =

15. Dada A =

(1 60 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(6 −66 6

p(x) = x4 + 64 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Observaciones

Mβ5c β5

c(f) =

1 6 5 10 1 6 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 6, 1, 0), (5, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 6, 5, 1, 0), (6, 6, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 6 0 − 6− 3 8 0 − 6− 3 6 2 − 6− 2 3 3 − 5

; P−1 =

12 17 17 1412 7 9 76 1 1 16 6 8 7

∣∣∣∣∣∣∣∣ =

∫4x + 24

x2 − 12x + 61dx =

15. Dada A =

(1 60 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(6 −66 6

p(x) = x4 + 64 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Observaciones

Mβ5c β5

c(f) =

1 6 5 10 1 7 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 6, 1, 0), (5, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 6, 5, 1, 0), (7, 6, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 5 3 − 9− 3 7 3 − 9− 3 5 5 − 9− 2 2 6 − 8

; P−1 =

12 17 19 1312 7 10 66 1 1 16 6 9 7

∣∣∣∣∣∣∣∣ =

∫4x + 24

x2 − 12x + 61dx =

15. Dada A =

(1 60 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(6 −66 6

p(x) = x4 + 64 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Observaciones

Mβ5c β5

c(f) =

1 2 5 10 1 5 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 2, 1, 0), (5, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 2, 5, 1, 0), (5, 2, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 7 − 3 − 3− 3 9 − 3 − 3− 3 7 − 1 − 3− 2 4 0 − 2

; P−1 =

4 17 15 224 7 8 92 1 1 12 6 7 13

∣∣∣∣∣∣∣∣ =

∫4x + 8

x2 − 4x + 29dx =

15. Dada A =

(1 20 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(2 −22 2

p(x) = x4 + 24 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Observaciones

Mβ5c β5

c(f) =

1 2 5 10 1 1 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 2, 1, 0), (5, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 2, 5, 1, 0), (1, 2, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 11 − 15 9− 3 13 − 15 9− 3 11 − 13 9− 2 8 − 12 10

; P−1 =

4 17 7 194 7 4 82 1 1 12 6 3 11

∣∣∣∣∣∣∣∣ =

∫4x + 8

x2 − 4x + 29dx =

15. Dada A =

(1 20 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(2 −22 2

p(x) = x4 + 24 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Observaciones

Mβ5c β5

c(f) =

1 2 6 10 1 6 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 2, 1, 0), (6, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 2, 6, 1, 0), (6, 2, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 7 − 3 − 3− 3 9 − 3 − 3− 3 7 − 1 − 3− 2 4 0 − 2

; P−1 =

4 20 17 164 8 9 72 1 1 12 7 8 9

∣∣∣∣∣∣∣∣ =

∫4x + 8

x2 − 4x + 40dx =

15. Dada A =

(1 20 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(2 −22 2

p(x) = x4 + 24 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Observaciones

Mβ5c β5

c(f) =

1 2 5 10 1 2 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 2, 1, 0), (5, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 2, 5, 1, 0), (2, 2, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 10 − 12 6− 3 12 − 12 6− 3 10 − 10 6− 2 7 − 9 7

; P−1 =

4 17 9 174 7 5 62 1 1 12 6 4 11

∣∣∣∣∣∣∣∣ =

∫4x + 8

x2 − 4x + 29dx =

15. Dada A =

(1 20 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(2 −22 2

p(x) = x4 + 24 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Observaciones

Mβ5c β5

c(f) =

1 3 6 10 1 3 20 0 1 30 0 0 00 0 0 0

1 0 0 00 1 0 00 0 1 00 0 0 00 0 0 0

β = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β′ = {( , , , ), ( , , , ), ( , , , ), ( , , , )}

β4 = {(1, 1, 1, 1), (1, 3, 1, 0), (6, 1, 0, 0), (1, 0, 0, 0)},

β5 = {(1, 1, 1, 1, 1), (1, 3, 6, 1, 0), (3, 3, 1, 0, 0), (0, 1, 0, 0, 0), (1, 0, 0, 0, 0)}

Mβ4β5(f) =

cy Mβ5β5

Mβ4β4c

Mβ5β5

Ker f = {(x, y, z, t)β4 : }

βKer f = { }

dim Ker f =

Im f = {(x, y, z, t)β5 : }

βIm f = { }

dim Im f =

Mβ4β4(g) =

Ker g = {(x, y, z, t)β4 : }

βKer g = { }

dim Ker g =

Im g = {(x, y, z, t)β4 : }

βIm g = { }

dim Im g =

βKer g+Im g =

Mβ4β5(f ◦ g) =

f ◦ g(x, y, z, t)β4 = ( , , , , )β5

c(f) y calcula

− 1 10 − 12 6− 3 12 − 12 6− 3 10 − 10 6− 2 7 − 9 7

; P−1 =

6 20 11 226 8 6 93 1 1 13 7 5 13

∣∣∣∣∣∣∣∣ =

∫4x + 12

x2 − 6x + 45dx =

15. Dada A =

(1 30 1

1 00 0

(0 10 0

(0 01 0

(0 00 1

)}. Dada A =

(3 −33 3

p(x) = x4 + 34 =

a) βU = { }

b) V = {(x, y, z, t)β : }

c) βU∩V =

d) βU+V = { }

e) U + V = {(x, y, z, t) : }

Regalo de Navidad para Antonio José Fernández Soriano...

Documents

Transcript of Regalo de Navidad para Antonio José Fernández Soriano...

Presentación Congreso Soriano 2011

EGUREN FERNÁNDEZ y SORIANO - El Órgano Mental Del Lenguaje (en Introducción a Una Sintaxis Minimista)

Codigo de Etica Soriano

Informe de Laboratorio Soriano

CVN - Ignacio Martinez Soriano · 2d2b75a229b6b7caeb71f282a0e041b8 3 Ignacio Martinez Soriano Apellidos: Martinez Soriano Nombre: Ignacio DNI: 24248381X Fecha de nacimiento: 06/09/1967

ROJAS Soriano Raúl

Abel Soriano Bar Sawer

Presentación Congreso Soriano

MANUEL RICARDO PALMA SORIANO

Juan Soriano 2019

LIBRO Investigacion Accion SORIANO

ÓRGANO OFICIAL DE LA SOCIEDAD PERUANA DE … · oxicodona vs tramadol en pacientes de cirugía traumatológica ... Juan Urquizo Soriano Pedro Pardo Fernández Freddy Espinoza Mendoza

Diploma de MANUEL FERNÁNDEZ SORIANO

Sin Escala - Federico Soriano

PFC Irene Soriano

XXII Otoño Musical Soriano

Brochure Bufete Soriano

Carmen López Soriano

Pdi Soriano

Directores: D. Andrés Porras Piedra. Dña. Mª Luisa Soriano Martín. Alumno: Tomás Fernández Farao.