ejercicio 1. 1%. 0 3 75esfm.egormaximenko.com/linalg/task_bases_es.pdfejercicio 10. 1%. est an dadas...

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peaq

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No

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Algebra II, licenciatura. Tarea 4. Variante α.

Bases y dimension.

Nombre: Calificacion ( %):

Esta tarea vale 20 % de la calificacion parcial.

Ejercicio 1. 1 %.Construya una base del espacio S =

{x ∈ R5 : Ax = 04

}, donde A es la siguiente matriz. Haga

la comprobacion. 2 0 1 1 4

2 −1 −1 2 3

1 −2 −2 2 −44 −3 −1 4 −3

.Ejercicio 2. 1 %.Construya una base del espacio S =

{x ∈ R5 : Ax = 04


la comprobacion. −2 1 2 1 −12 2 −7 −3 2

4 −2 −4 −2 2

0 −3 5 2 −1

.Ejercicio 3. 1 %.Construya una base del siguiente espacio de polinomios. Haga la comprobacion.

S ={f ∈ P2(R) : f(2) = 2f(−1)

}.

Ejercicio 4. 1 %.Denotemos por P3(R) al espacio vectorial de los polinomios de grado 6 3 (incluso al polinomiocero). Se considera el conjunto S ⊂ P3(R) definido de la siguiente manera:

S ={f ∈ P3(R) : f(−3) = 0 y f ′(−2) = 0

}.

1. Muestre que S es un subespacio vectorial de P3(R).

2. Construya una base B del espacio S.

3. Haga las comprobaciones: verifique que los elementos de la base construida B pertenecena S, esto es, cumplen con las condiciones f(−3) = 0 y f ′(−2) = 0.

Tarea 4, variante α, pagina 1 de 4


{X ∈M2(R) : AX = XA

}, donde A es la siguiente matriz.

Haga la comprobacion.

A =

[1 −4

−2 2

].


{X ∈ M2(R) : tr(XA) = 0

}, donde A es la matriz dada.


A =

[4 2

−3 −2

].

Ejercicio 7. 1 %.Sea E la base canonica del espacio R3 y sea B la base que consta de los vectores dados b1, b2, b3.Escriba la matriz de cambio PE ,B, calcule la matriz de cambio PB,E y haga la comprobacion:PE ,BPB,E = I3.

b1 =

121

, b2 =

132

, b3 =

342

.Ejercicio 8. 1 %.Sean E y B las bases de R3 definidas en el ejercicio anterior, y sean u,w vectores de R3 talesque

uE =

−20

1

, wB =

3

−10

.Calcule uB y wE , calcule (u + w)E como uE + wE , calcule (u + w)B como uB + wB. Haga lacomprobacion: (u+w)E = PE ,B(u+w)B.

Ejercicio 9. 1 %.Transforme la matriz A en una matriz escalonada o pseudoescalonada y calcule su rango. Luegotransforme la matriz A> en una matriz escalonada o pseudoescalonada y calcule su rango.Indicacion: no se recomienda transformar A o A> en matrices pseudoescalonadas reducidas.

A =

−2 2 3 2 4

6 −4 2 2 4

−1 0 −4 −3 −63 −2 1 1 2

.


Ejercicio 10. 1 %.Estan dadas tres matrices A, B, C:

A =

−1 −5 5

0 0 0

−5 −5 0

0 5 0

, B =

4 2

2 1

6 3

, C =

[1 3

5 1

].

Sea F la lista de las columnas de la matriz A y sea S = `(F) el subespacio generado por F enel espacio V = Rm, donde m es el numero de los renglones de la matriz A. Determine el rangoy otras caracterısticas de F . De manera similar considere las columnas de las matrices B y C.Llene la siguiente tabla:

A B C

dim(V) = m|F | = n 3 2 2rango(F) = rdim(S)¿es F lin. indep.?¿genera F el espacio V?¿es F una base de V?

Ejercicio 11. 1 %.Con operaciones elementales por renglones transforme la matriz dada en una matriz escalonadareducida o pseudoescalonada reducida, encuentre un subsistema basico de las columnas y escribalas demas columnas como combinaciones lineales de las basicas. Haga la comprobacion.

1 4 1 1

−1 −5 −5 −4−1 −3 3 2

−2 −6 6 4

.

Ejercicio 12. 2 %.Halle una sublista basica de la lista de matrices A1, A2, A3, A4, A5 ∈M2(R). Escriba las demasmatrices como combinaciones lineales de las matrices que forman esa sublista basica. Haga lascomprobaciones.

A1 =

[3 2

4 −1

], A2 =

[1 −13 −2

], A3 =

[3 5

1 2

],

A4 =

[2 −15 −3

], A5 =

[1 1

1 0

].


Ejercicio 13. 2 %.Construya una base B del espacio generado por los polinomios f1, f2, f3, f4, f5. Represente cadauno de los polinomios f1, f2, f3, f4, f5 como una combinacion lineal de los elementos de la baseconstruida B y haga las comprobaciones.

f1 = −4+ 3x− 3x2 − x3, f2 = 1+ x+ x2 − 2x3, f3 = 2+ x+ x

2 − x3,

f4 = −2− 4x+ 2x2 − 2x3, f5 = −1+ x− 2x3.

Ejercicio 14. 1 %.En este y los siguientes ejercicios se consideran los vectores a1, a2, a3, b1, b2 y los subespaciosS1 = `(a1, a2, a3), S2 = `(b1, b2).

a1 =

3

−20

1

, a2 =

5

−24

3

, a3 =

3

−13

2

, b1 =

2

−11

1

, b2 =

1

1

3

1

.Construya una base del subespacio S1 + S2 como un subsistema basico del sistema de vectoresa1, a2, a3, b1, b2. Escriba los demas vectores como combinaciones lineales de los vectores queforman ese subsistema basico. Haga las comprobaciones.

Ejercicio 15. 1 %.Encuentre un sistema de ecuaciones lineales homogeneas que describa el subespacio S1. Hagalas comprobaciones. Calcule dim(S1).


Ejercicio 17. 1 %.Construya una base del subespacio S1 ∩S2. Comprueba que los vectores de esta base satisfacenlas ecuaciones homogeneas obtenidas en dos ejercicios anteriores. Haga la comprobacion de lasdimensiones con la formula de Grassmann.

Ejercicio 18. 1 %.Extienda la base de S1 ∩S2 obtenida en el ejercicio anterior a una base de S1; a una base de S2;a una base de S1 + S2.


Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante β.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. −1 1 1 1 2

−2 −1 −2 −3 −43 0 1 2 2

−3 1 1 2 3


{x ∈ R5 : Ax = 04


la comprobacion. 2 3 3 1 −13 1 4 1 3

−1 2 −1 0 −41 5 2 1 −5


S ={f ∈ P2(R) : f ′(−3) − 2f(−3) = 0

}.


S ={f ∈ P3(R) : f(3) = 0 y f ′′(−3) = 0

}.



3. Haga las comprobaciones: verifique que los elementos de la base construida B pertenecena S, esto es, cumplen con las condiciones f(3) = 0 y f ′′(−3) = 0.

Tarea 4, variante β, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[3 −11 2

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−2 4

−4 1

].


b1 =

322

, b2 =

212

, b3 =

101


uE =

322

, wB =

−1−14



A =

−3 −1 −2 −2 −32 2 1 −1 1

−3 1 1 −1 1

1 −1 1 3 2

.



A =

3 0 4

−2 −1 0

3 0 0

0 0 0

, B =

[3 −4

−5 2

], C =

2 −1−2 1

−2 1

.Sea F la lista de las columnas de la matriz A y sea S = `(F) el subespacio generado por F enel espacio V = Rm, donde m es el numero de los renglones de la matriz A. Determine el rangoy otras caracterısticas de F . De manera similar considere las columnas de las matrices B y C.Llene la siguiente tabla:

A B C



1 2 0 −42 3 −3 −34 6 −6 −6

−2 −2 6 −2

.


A1 =

[2 5

4 3

], A2 =

[−3 −5−1 −2

], A3 =

[−1 −2−1 −1

],

A4 =

[2 2

−2 0

], A5 =

[1 3

3 2

].



f1 = 1+ x+ 2x2 + 4x3, f2 = 1+ x+ 2x

2 + 3x3, f3 = 3− 2x+ x2 + 2x3,

f4 = −3− 3x2 + 2x3, f5 = 1+ 2x+ 3x2 + 3x3.


a1 =

4

1

3

5

, a2 =

−30

−3−6

, a3 =

3

1

2

3

, b1 =

1

0

1

2

, b2 =

3

3

1

−2







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 1 AJAS.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. −4 −2 −2 −1 −22 1 3 1 2

−3 −1 4 1 −1−1 0 3 1 −1


{x ∈ R5 : Ax = 04


la comprobacion. −4 −4 1 2 −32 4 −3 −5 1

−6 −4 −1 −1 −52 0 2 3 2


S ={f ∈ P2(R) : f ′(3) − 5f(3) = 0

}.


S ={f ∈ P3(R) : f(1) = 0 y f ′(−1) = 0

}.



3. Haga las comprobaciones: verifique que los elementos de la base construida B pertenecena S, esto es, cumplen con las condiciones f(1) = 0 y f ′(−1) = 0.

Tarea 4, variante 1 AJAS, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[−1 1

−2 4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−2 2

−4 3

].


b1 =

−12

1

, b2 =

211

, b3 =

232


uE =

211

, wB =

0

−21

.Calcule uB y wE , calcule (u − w)E como uE − wE , calcule (u − w)B como uB − wB. Haga lacomprobacion: (u−w)E = PE ,B(u−w)B.


A =

−1 3 −1 1 2

−2 4 1 1 −23 −5 −3 −1 6

−2 6 −2 2 4

.



A =

[−5 4

5 3

], B =

0 2 −2 −24 −1 0 0

0 −5 −2 0

, C =

[2 −1 1 −3

−4 2 −2 6

].


A B C



−2 −2 3 2

−2 −4 2 2

−2 0 4 2

−1 −2 1 1

.


A1 =

[−3 −25 −4

], A2 =

[3 2

−5 4

], A3 =

[1 1

−1 1

],

A4 =

[−1 −11 −1

], A5 =

[−3 −33 −3

].



f1 = 3+ 3x+ 2x2 + 2x3, f2 = 1+ 4x+ 4x

2 − 3x3, f3 = 3+ 4x+ 3x2 + x3,

f4 = −1− 4x− x2 − 3x3, f5 = 2+ 2x+ x2 + 2x3.


a1 =

−3−15

−2

, a2 =

3

−1−71

, a3 =

2

0

−41

, b1 =

1

−1−30

, b2 =

3

4

−13







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 2 BTCF.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. 1 −4 2 1 3

1 2 −1 −1 −5−2 0 2 1 2

−4 2 1 1 4


{x ∈ R5 : Ax = 04


la comprobacion. 0 1 −4 1 3

−8 −7 −8 −1 −18 8 4 2 4

4 4 2 1 2


S ={f ∈ P2(R) : f ′(−3) + 4f(−3) = 0

}.


S ={f ∈ P3(R) : f(3) = 0 y f ′(−2) = 0

}.




Tarea 4, variante 2 BTCF, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[−2 1

2 −4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[4 2

−1 1

].


b1 =

212

, b2 =

323

, b3 =

102


uE =

1

2

−2

, wB =

2

−1−2



A =

−1 1 2 −2 1

1 −1 −3 3 −11 1 1 1 −3

−1 2 4 −3 0

.



A =

2 1 1 −30 −4 3 −30 0 3 −3

, B =

[−6 2 4 −4−3 1 2 −2

], C =

1 2

1 3

0 −1


A B C



1 2 3 −32 3 5 1

−2 −4 −6 6

1 1 2 4

.


A1 =

[1 1

−3 1

], A2 =

[−4 1

2 6

], A3 =

[1 −11 −3

],

A4 =

[4 0

−4 −4

], A5 =

[1 2

−5 3

].



f1 = −4+ 3x− 3x2 + 3x3, f2 = 4+ 3x− 5x2 + x3, f3 = −1+ 3x− 2x2 + 4x3,

f4 = 1+ x− 2x2, f5 = 3+ 3x− 5x

2 + x3.


a1 =

−1−1−20

, a2 =

5

2

3

2

, a3 =

6

3

5

2

, b1 =

1

1

2

0

, b2 =

2

1

1

1







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 3 CSA.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. −3 −1 1 −3 −21 1 2 2 1

2 1 4 −1 1

2 0 −3 1 1


{x ∈ R5 : Ax = 04


la comprobacion. 1 1 2 3 −21 −2 3 −4 0

2 2 4 6 −4−2 1 −5 1 2


S ={f ∈ P2(R) : f ′(2) + 3f(2) = 0

}.


S ={f ∈ P3(R) : f(−2) = 0 y f ′′(1) = 0

}.



3. Haga las comprobaciones: verifique que los elementos de la base construida B pertenecena S, esto es, cumplen con las condiciones f(−2) = 0 y f ′′(1) = 0.

Tarea 4, variante 3 CSA, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[1 −4

−1 4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[1 3

−4 4

].


b1 =

2

−11

, b2 =

111

, b3 =

021


uE =

1

−20

, wB =

−13

−2



A =

1 2 1 4 2

2 2 −2 −6 6

1 1 −1 −3 3

2 3 0 1 5

.



A =

[1 −2 4 −41 4 3 −4

], B =

4 0 2

0 0 0

0 −5 −10 −5 0

, C =

3 3 −43 1 0

0 −3 0


A B C



5 1 1 −4−3 3 5 2

−4 1 2 3

1 2 3 −1

.


A1 =

[−5 3

1 2

], A2 =

[−6 2

−2 4

], A3 =

[−5 3

1 2

],

A4 =

[−3 2

1 1

], A5 =

[−6 3

0 3

].



f1 = 1+ x+ 2x2 + x3, f2 = −3+ 3x− x2 + 2x3, f3 = 1+ 2x+ 4x

2 + 3x3,

f4 = 2+ x− 2x3, f5 = 1+ x+ 3x

2 + 2x3.


a1 =

−12

1

2

, a2 =

1

3

3

5

, a3 =

0

5

4

7

, b1 =

−31

−1−1

, b2 =

0

2

2

3







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 4 CNKM.

Bases y dimension.




{x ∈ R5 : Ax = 04



1 0 −3 3 1

4 3 −2 1 3

−1 1 1 4 −2


{x ∈ R5 : Ax = 04


la comprobacion. 4 −2 1 −1 3

−2 4 −2 4 −54 4 −2 6 −4

−2 −2 1 −3 2


S ={f ∈ P2(R) :

1∫−2

f(x)dx = 0}.


S ={f ∈ P3(R) : f(−1) = 0 y f ′′(−1) = 0

}.



3. Haga las comprobaciones: verifique que los elementos de la base construida B pertenecena S, esto es, cumplen con las condiciones f(−1) = 0 y f ′′(−1) = 0.

Tarea 4, variante 4 CNKM, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[1 −13 4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−4 1

3 −2

].


b1 =

111

, b2 =

110

, b3 =

121


uE =

1

1

−2

, wB =

2

−1−2



A =

1 2 1 1 1

−2 −3 −2 −3 −30 1 1 2 4

1 1 1 2 2

.



A =

[2 −4

−4 2

], B =

3 −3 4 4

−4 2 −4 0

−5 −4 0 0

, C =

1 0

4 1

5 1


A B C



6 6 4 6

4 1 3 −47 4 5 −13 3 2 3

.


A1 =

[−2 1

−5 −3

], A2 =

[6 4

−6 2

], A3 =

[5 3

−4 2

],

A4 =

[3 0

3 3

], A5 =

[2 1

−1 1

].



f1 = −1+ x+ 3x2 + 2x3, f2 = −3+ x+ 4x2 + x3, f3 = −2+ x+ 3x2 + x3,

f4 = −2− 2x+ x2 − x3, f5 = 1− 4x+ 3x2 + 4x3.


a1 =

4

−43

2

, a2 =

6

−44

4

, a3 =

7

−65

4

, b1 =

3

−22

2

, b2 =

2

2

1

3







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 5 CNLE.

Bases y dimension.




{x ∈ R5 : Ax = 04



1 −2 2 −4 −1−2 4 −1 3 −31 −3 0 −2 2


{x ∈ R5 : Ax = 04


la comprobacion. −5 −1 2 −1 3

−3 −1 1 1 2

−2 0 1 −2 1

−4 −2 1 4 3


S ={f ∈ P2(R) : f ′(2) − 5f(2) = 0

}.


S ={f ∈ P3(R) : f(1) = 0 y f ′′(2) = 0

}.



3. Haga las comprobaciones: verifique que los elementos de la base construida B pertenecena S, esto es, cumplen con las condiciones f(1) = 0 y f ′′(2) = 0.

Tarea 4, variante 5 CNLE, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[2 1

−2 −4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−4 3

−1 1

].


b1 =

−12

1

, b2 =

211

, b3 =

232


uE =

132

, wB =

−1−32



A =

3 2 3 4 1

5 3 5 2 0

4 2 4 −4 −22 1 2 −2 −1

.



A =

1 2

1 3

0 −1

, B =

[6 4 −2 −4

−3 −2 1 2

], C =

1 −1 0 0

0 4 1 4

0 2 0 −1


A B C



−3 −2 1 −41 1 3 3

5 3 −5 5

2 1 −4 1

.


A1 =

[2 −11 3

], A2 =

[3 −12 4

], A3 =

[−1 −3−4 2

],

A4 =

[2 −11 3

], A5 =

[2 −11 3

].



f1 = 1− 2x− x2 + x3, f2 = 1− 4x+ 2x

2 + 3x3, f3 = 3− 2x− 4x2 − x3,

f4 = 2− 5x+ 3x3, f5 = 4− 3x+ x

2 − x3.


a1 =

−1−65

3

, a2 =

2

4

−2−2

, a3 =

−1−43

2

, b1 =

−1−21

1

, b2 =

1

−12

1







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 6 DPE.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. −1 4 −2 −4 −1−1 −4 2 1 −22 4 −3 2 3

1 4 −4 2 1


{x ∈ R5 : Ax = 04


la comprobacion. −2 2 2 −2 2

2 4 −1 1 4

1 −1 −1 1 −14 2 −3 3 2


S ={f ∈ P2(R) : f ′(3) − 3f(3) = 0

}.


S ={f ∈ P3(R) : f(3) = 0 y f ′(1) = 0

}.



3. Haga las comprobaciones: verifique que los elementos de la base construida B pertenecena S, esto es, cumplen con las condiciones f(3) = 0 y f ′(1) = 0.

Tarea 4, variante 6 DPE, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[1 −42 3

].


{X ∈ M2(R) : tr(XA) = 0



A =

[2 −21 3

].


b1 =

131

, b2 =

141

, b3 =

132


uE =

141

, wB =

−31

1



A =

1 2 1 −1 3

−4 1 0 −3 1

3 −1 1 3 1

−4 −1 −2 −2 −4

.



A =

−2 3 5 −2−5 0 −2 −20 0 −5 −3

, B =

1 1

2 1

−1 0

, C =

3 1 −52 1 0

5 0 0


A B C



2 2 3 −33 2 4 −14 2 5 1

1 0 1 2

.


A1 =

[3 5

−1 2

], A2 =

[−2 −2−2 0

], A3 =

[1 2

−1 1

],

A4 =

[3 5

−1 2

], A5 =

[−2 −1−4 1

].



f1 = 2+ 4x+ 2x2 + 2x3, f2 = 2+ 4x− 2x

2 − 4x3, f3 = 1+ 2x− 2x2 + 3x3,

f4 = 2+ 4x+ x2 + 2x3, f5 = 1+ 2x− x

2 + 2x3.


a1 =

1

−32

−2

, a2 =

1

−53

−5

, a3 =

−1−31

−7

, b1 =

0

−21

−3

, b2 =

1

−11

2







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 7 DEER.

Bases y dimension.




{x ∈ R5 : Ax = 04



0 3 2 −1 1

1 1 2 −2 2

2 −1 2 −3 3


{x ∈ R5 : Ax = 04


la comprobacion. −2 2 −4 2 −23 2 3 3 3

−2 −3 −1 −4 −2−1 1 −2 1 −1


S ={f ∈ P2(R) : f(3) = 2f(−2)

}.


S ={f ∈ P3(R) : f(3) = 0 y f ′(3) = 0

}.




Tarea 4, variante 7 DEER, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[2 3

−3 −2

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−2 1

−1 3

].


b1 =

212

, b2 =

112

, b3 =

101


uE =

−21

−1

, wB =

2

0

−3



A =

0 1 −1 5 1

6 1 3 3 1

3 1 1 4 1

3 −1 3 −6 −1

.



A =

[0 1 −3 2

1 1 −3 3

], B =

3 −61 −2

−2 4

, C =

0 0 0

2 1 2

1 0 −40 0 −2


A B C



1 −1 −3 1

4 −2 4 2

2 −1 2 1

−4 3 4 −3

.


A1 =

[1 −12 1

], A2 =

[2 −33 1

], A3 =

[4 −66 2

],

A4 =

[0 4

4 4

], A5 =

[−4 7

−5 −1

].



f1 = 3+ x+ 2x2 + 4x3, f2 = 4+ 2x− 3x

2 − x3, f3 = 2+ x2 + 3x3,

f4 = 3− x− 3x2 + x3, f5 = 2+ x+ 2x

2 + 3x3.


a1 =

6

5

2

1

, a2 =

2

−33

5

, a3 =

2

−12

3

, b1 =

2

1

1

1

, b2 =

1

−22

3







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 8 DLRTH.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. 3 1 −1 −3 −2

−1 3 3 3 2

−2 1 3 −1 3

−2 1 2 3 2


{x ∈ R5 : Ax = 04


la comprobacion. −1 1 −4 −3 4

4 2 3 2 −3−2 −4 5 4 −5−3 −3 1 1 −1


S ={f ∈ P2(R) : f(2) = 2f(−2)

}.


S ={f ∈ P3(R) : f(3) = 0 y f ′′(3) = 0

}.




Tarea 4, variante 8 DLRTH, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[−2 3

2 1

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−3 4

−4 3

].


b1 =

131

, b2 =

141

, b3 =

132


uE =

140

, wB =

−13

−2



A =

−4 2 3 1 2

−3 2 2 0 1

4 −3 −2 −2 −3−3 3 1 1 2

.



A =

3 −1 1 −2−4 −2 −5 0

4 0 2 0

, B =

4 3 4

1 −4 0

−3 0 0

, C =

0 −11 3

1 4


A B C



3 2 2 3

−2 −2 −4 4

1 1 2 −2−4 −3 −4 −1

.


A1 =

[−4 4

2 2

], A2 =

[−2 4

3 1

], A3 =

[−2 3

2 1

],

A4 =

[−4 5

3 2

], A5 =

[−2 2

1 1

].



f1 = 1+ 2x+ x2 + x3, f2 = −4+ 2x− 4x2 − 4x3, f3 = 1+ 4x+ 2x

3,

f4 = 1+ 3x− x2 + 3x3, f5 = 1− x+ 3x

2 − x3.


a1 =

5

1

−24

, a2 =

4

1

−13

, a3 =

2

−1−53

, b1 =

1

0

−11

, b2 =

4

2

3

1







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 9 DGGI.

Bases y dimension.




{x ∈ R5 : Ax = 04



1 1 1 4 0

2 −2 −2 −4 −12 2 4 −4 1


{x ∈ R5 : Ax = 04


la comprobacion. −3 4 −2 −1 −1−5 5 −3 −5 −42 −1 1 4 3

−1 3 −1 3 2


S ={f ∈ P2(R) : f ′(−1) − 3f(−1) = 0

}.


S ={f ∈ P3(R) : f(−3) = 0 y f ′′(3) = 0

}.




Tarea 4, variante 9 DGGI, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[2 −23 4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−2 3

−3 1

].


b1 =

214

, b2 =

112

, b3 =

021


uE =

−14

0

, wB =

−13

0



A =

1 −4 −1 1 3

4 1 −3 3 3

−5 3 4 −4 −6−7 −6 5 −5 −3

.



A =

−6 −32 1

6 3

, B =

[1 3 2 4

0 4 3 2

], C =

3 5 1

0 0 0

0 0 −4−2 0 0


A B C



−1 1 −1 2

−2 2 −3 3

−2 2 −2 4

−3 3 −4 5

.


A1 =

[3 −27 −1

], A2 =

[3 2

−1 −5

], A3 =

[4 3

−2 −7

],

A4 =

[4 −16 −3

], A5 =

[−3 0

−3 3

].



f1 = 3+ 3x+ x2 + 2x3, f2 = 2+ x+ 2x

2, f3 = 2+ 2x+ x2 + x3,

f4 = 3− 3x+ 4x2 − x3, f5 = −4+ x− x2 − 3x3.


a1 =

−7−5−21

, a2 =

4

3

2

0

, a3 =

7

5

2

−1

, b1 =

1

1

2

1

, b2 =

2

1

0

−1







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 10 DJRA.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. −1 3 2 3 −11 2 1 2 −2

−3 4 3 4 0

−1 3 3 4 4


{x ∈ R5 : Ax = 04



3 −2 5 4 2

2 −2 3 2 −13 −4 4 2 −5


S ={f ∈ P2(R) : f(−2) = 2f(2)

}.


S ={f ∈ P3(R) : f(−1) = 0 y f ′(−2) = 0

}.




Tarea 4, variante 10 DJRA, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[2 4

−3 −4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−2 1

−1 4

].


b1 =

−22

−1

, b2 =

111

, b3 =

121


uE =

2

−31

, wB =

−1−31



A =

3 −4 1 0 1

−3 −3 2 1 1

3 2 2 1 2

−3 3 3 2 2

.



A =

1 1

2 1

1 0

, B =

3 3 −3 2

4 0 −2 −3−5 0 −4 0

, C =

[2 −2 1 2

4 −4 2 4

].


A B C



4 3 3 2

−2 2 4 2

−1 1 2 1

3 4 5 3

.


A1 =

[−2 1

−3 4

], A2 =

[1 −12 −2

], A3 =

[−1 2

−3 2

],

A4 =

[−3 −63 6

], A5 =

[2 −13 −4

].



f1 = 1+ x+ 2x2 + 2x3, f2 = 1+ 2x+ 2x

2 + 4x3, f3 = 1− x+ 4x2 − 2x3,

f4 = −4+ 2x− x2 + 4x3, f5 = x+ x2 + 2x3.


a1 =

2

3

−41

, a2 =

3

4

−62

, a3 =

3

6

−60

, b1 =

1

2

−20

, b2 =

2

−1−34







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 11 FMFD.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. 2 −2 1 −2 −21 −1 2 1 1

−1 1 1 3 3

1 0 3 2 −3


{x ∈ R5 : Ax = 04



1 −2 3 −2 1

−4 0 −2 3 −53 2 −1 −1 4


S ={f ∈ P2(R) : f(2) = 2f(3)

}.


S ={f ∈ P3(R) : f(3) = 0 y f ′(−1) = 0

}.




Tarea 4, variante 11 FMFD, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[−2 2

−1 4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[1 −2

−3 4

].


b1 =

430

, b2 =

331

, b3 =

221


uE =

012

, wB =

1

−21



A =

2 −2 1 −4 −7

−2 −1 −4 1 1

0 1 1 1 2

−2 1 −2 3 5

.



A =

2 3 1

0 0 0

3 −4 0

−5 0 0

, B =

[2 −21 −2

], C =

1 −2−2 4

3 −6


A B C



4 −3 3 6

3 1 2 2

1 −4 1 4

5 6 3 0

.


A1 =

[−1 0

1 −1

], A2 =

[3 −1

−4 2

], A3 =

[2 1

−1 3

],

A4 =

[3 2

−1 5

], A5 =

[1 −2

−3 −1

].



f1 = 4+ 3x+ 2x2 + 3x3, f2 = 4+ 2x+ 2x

2 + 3x3, f3 = −4+ 3x− 4x2 − 4x3,

f4 = 1+ x+ x2 + x3, f5 = 4− 4x− 4x

2.


a1 =

5

3

−16

, a2 =

3

2

0

4

, a3 =

−1−2−4−4

, b1 =

1

1

1

2

, b2 =

2

1

−22







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 12 GDLD.

Bases y dimension.




{x ∈ R5 : Ax = 04



4 4 1 4 2

0 2 2 1 1

1 2 1 2 1


{x ∈ R5 : Ax = 04



−4 2 6 −6 −4−2 1 3 −3 −2−1 2 5 1 1


S ={f ∈ P2(R) : f ′(3) − 2f(3) = 0

}.


S ={f ∈ P3(R) : f(−2) = 0 y f ′(−2) = 0

}.




Tarea 4, variante 12 GDLD, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[3 1

4 −4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[2 4

−3 1

].


b1 =

121

, b2 =

140

, b3 =

131


uE =

−11

−4

, wB =

1

2

−3



A =

2 −3 1 1 −21 1 1 3 2

1 0 1 2 4

3 −4 2 2 4

.



A =

5 4 −53 0 −1

−1 0 0

, B =

[−1 2 1 2

−2 4 2 4

], C =

1 1

2 3

3 4


A B C



−1 −5 1 1

−6 −6 −6 −42 −2 4 3

3 3 3 2

.


A1 =

[4 −13 7

], A2 =

[1 1

2 3

], A3 =

[−4 7

3 −1

],

A4 =

[4 −5

−1 3

], A5 =

[1 2

3 4

].



f1 = 2− 2x− 3x2 + 4x3, f2 = 3− 4x− 3x

2 + 2x3, f3 = 1+ 4x+ x2 + 2x3,

f4 = 3+ 3x+ 2x2 − x3, f5 = 1+ 3x+ x

2 + x3.


a1 =

2

2

1

3

, a2 =

3

6

1

7

, a3 =

2

−42

−2

, b1 =

−12

−11

, b2 =

3

−12

1







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 13 GLMA.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. −1 2 1 1 −3−2 1 1 1 2

−1 3 3 4 −1−1 −2 −2 −3 3


{x ∈ R5 : Ax = 04


la comprobacion. 4 −8 −8 7 −3

−2 5 3 −4 2

2 −3 −5 3 −10 2 −2 −1 1


S ={f ∈ P2(R) : f ′(−2) − 2f(−2) = 0

}.


S ={f ∈ P3(R) : f(2) = 0 y f ′(−3) = 0

}.




Tarea 4, variante 13 GLMA, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[3 −2

−1 4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−3 1

−4 2

].


b1 =

212

, b2 =

323

, b3 =

102


uE =

212

, wB =

−21

1



A =

−3 7 −6 2 3

1 4 −2 1 2

−6 −5 0 −1 −3−4 3 −4 1 1

.



A =

[1 3 1 1

2 6 2 2

], B =

4 1

−2 0

5 1

, C =

5 0 −3 2

−5 1 3 0

−2 −3 0 0


A B C



2 −3 1 2

4 −6 2 4

−2 6 1 3

0 3 2 5

.


A1 =

[−6 −13 −4

], A2 =

[−2 0

1 −1

], A3 =

[−4 7

2 5

],

A4 =

[−2 3

1 2

], A5 =

[6 −2

−3 1

].



f1 = 2+ x+ 2x2 − 4x3, f2 = −4− 4x− x2 + x3, f3 = 1+ x

2 − 3x3,

f4 = 3+ 2x+ 2x2 − 4x3, f5 = −2− 2x− x2 + x3.


a1 =

1

−15

7

, a2 =

1

0

4

5

, a3 =

1

1

3

3

, b1 =

0

−11

2

, b2 =

1

1

4

4







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 14 GSLE.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. 1 3 −1 −4 1

−3 −2 3 1 −10 3 −1 −3 1

1 2 −3 1 1


{x ∈ R5 : Ax = 04



0 −2 −3 −3 −5−4 1 1 4 1

4 1 2 −1 4


S ={f ∈ P2(R) : f ′(−1) − 5f(−1) = 0

}.


S ={f ∈ P3(R) : f(−3) = 0 y f ′′(−3) = 0

}.




Tarea 4, variante 14 GSLE, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[−3 −14 3

].


{X ∈ M2(R) : tr(XA) = 0



A =

[3 −4

−2 2

].


b1 =

322

, b2 =

212

, b3 =

101


uE =

−11

−2

, wB =

−12

−1



A =

1 1 −1 0 1

−4 −3 4 3 −32 1 −2 −3 1

−2 1 2 4 2

.



A =

−1 −3 −3−3 0 −4−5 0 0

, B =

[4 5 3 2

3 3 −2 1

], C =

−2 2 3

0 3 4

0 0 0

0 2 0


A B C



−4 −2 0 −12 4 −2 3

−1 1 −1 1

3 3 −1 2

.


A1 =

[3 −52 7

], A2 =

[4 0

−4 −4

], A3 =

[1 −21 3

],

A4 =

[3 1

−4 −5

], A5 =

[3 −1

−2 −1

].



f1 = 3+ 2x+ 3x2 + 3x3, f2 = 1− x+ 3x

2 − x3, f3 = −1+ 4x+ x2 − 3x3,

f4 = 2+ 2x+ x2 + 3x3, f5 = −1− 3x− 2x3.


a1 =

0

1

−25

, a2 =

−43

2

3

, a3 =

−54

2

5

, b1 =

−21

2

−1

, b2 =

−21

1

1







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 15 KSJG.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. 1 1 2 −3 −21 2 −3 −1 2

0 1 3 2 2

1 1 −2 −3 −1


{x ∈ R5 : Ax = 04


la comprobacion. 4 1 1 −2 4

−8 −2 −2 4 −8−8 1 −7 −4 6

−2 1 −3 −3 5


S ={f ∈ P2(R) : f(−2) = 2f(1)

}.


S ={f ∈ P3(R) : f(3) = 0 y f ′′(−1) = 0

}.




Tarea 4, variante 15 KSJG, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[−3 −13 2

].


{X ∈ M2(R) : tr(XA) = 0



A =

[3 −24 2

].


b1 =

3

−10

, b2 =

211

, b3 =

111


uE =

201

, wB =

1

0

−2



A =

0 3 2 1 −22 1 −4 1 −42 4 −2 2 −61 2 −1 1 −3

.



A =

5 3 4

0 −2 0

0 0 5

, B =

[−4 −6 4 −22 3 −2 1

], C =

−1 1 −3 2

2 4 −2 0

−5 0 −1 0


A B C



−3 1 3 2

0 1 −1 1

3 −3 −1 −4−3 2 2 3

.


A1 =

[5 7

−3 2

], A2 =

[2 3

−1 1

], A3 =

[4 1

−7 −3

],

A4 =

[1 5

3 4

], A5 =

[−2 −6−2 −4

].



f1 = −3+ 2x+ x2 + 3x3, f2 = 1− x+ x3, f3 = 1− 4x+ 3x

2 + 2x3,

f4 = 1+ 3x− 4x2 + x3, f5 = 2− 4x+ 2x

2 + 3x3.


a1 =

−72

4

5

, a2 =

−4−13

0

, a3 =

−21

1

2

, b1 =

−30

2

1

, b2 =

2

1

−21







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 16 LSS.

Bases y dimension.




{x ∈ R5 : Ax = 04



2 4 1 −1 2

4 3 1 2 3

−4 2 −1 3 −4


{x ∈ R5 : Ax = 04


la comprobacion. −4 −2 5 5 3

−4 −4 1 1 1

0 1 2 2 1

4 3 −3 −3 −2


S ={f ∈ P2(R) : f ′(−1) + 3f(−1) = 0

}.


S ={f ∈ P3(R) : f(−1) = 0 y f ′′(−3) = 0

}.




Tarea 4, variante 16 LSS, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[3 −2

−3 1

].


{X ∈ M2(R) : tr(XA) = 0



A =

[2 −3

−2 3

].


b1 =

214

, b2 =

112

, b3 =

021


uE =

1

−4−1

, wB =

−23

−1



A =

2 −3 −1 1 3

0 −2 4 2 1

−1 −1 −2 −1 −21 3 −2 −1 1

.



A =

1 −22 −41 −2

, B =

[−1 2 3 −40 3 4 −2

], C =

5 4 −30 0 0

0 0 −5−1 0 0


A B C



3 −3 4 2

−4 −2 −6 4

2 1 3 −2−1 −5 −2 6

.


A1 =

[4 1

−5 3

], A2 =

[−1 −34 2

], A3 =

[−4 0

4 −4

],

A4 =

[1 5

−6 −4

], A5 =

[3 1

−4 2

].



f1 = 1− 4x+ 3x2 + x3, f2 = 2− 3x+ 3x

2 + x3, f3 = −3x+ x2 + x3,

f4 = 4− 3x− x2 + 4x3, f5 = 1+ x+ 2x

2 − x3.


a1 =

4

4

0

1

, a2 =

6

7

−11

, a3 =

−42

−6−4

, b1 =

2

3

−10

, b2 =

3

2

1

1







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 17 LPS.

Bases y dimension.




{x ∈ R5 : Ax = 04



2 1 3 3 2

−2 1 1 1 2

4 1 4 3 4


{x ∈ R5 : Ax = 04



4 −2 −2 −2 −6−2 1 1 1 3

2 −1 1 0 2


S ={f ∈ P2(R) : f(−2) = 2f(0)

}.


S ={f ∈ P3(R) : f(−2) = 0 y f ′′(−1) = 0

}.




Tarea 4, variante 17 LPS, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[−2 −44 3

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−1 4

3 −3

].


b1 =

101

, b2 =

212

, b3 =

213


uE =

021

, wB =

2

1

−2



A =

1 −1 0 3 1

6 2 1 3 5

4 4 1 −3 3

3 5 1 −6 2

.



A =

0 0 0

0 0 −5−4 2 0

−2 0 0

, B =

1 2

−1 −22 4

, C =

[4 1 −2 −33 1 3 1

].


A B C



−2 −1 3 −2−2 2 4 2

−1 1 2 1

1 2 −1 3

.


A1 =

[1 3

2 5

], A2 =

[1 2

1 3

], A3 =

[−2 −3−1 −4

],

A4 =

[3 −1

−4 −5

], A5 =

[0 2

2 4

].



f1 = −1+ 3x2 + 2x3, f2 = −1− 2x− 4x2 − 3x3, f3 = 3+ 3x+ 2x2 + 2x3,

f4 = 1+ x+ x2 + x3, f5 = 4+ 2x− 4x

2 − 2x3.


a1 =

4

3

1

2

, a2 =

−34

3

1

, a3 =

−13

2

1

, b1 =

−21

1

0

, b2 =

2

2

1

1







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 18 MPDD.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. 2 2 3 −3 −13 2 3 −2 1

3 3 3 −2 −1−3 −1 −3 2 −3


{x ∈ R5 : Ax = 04



−2 −2 3 −2 2

4 4 −6 4 −4−4 −1 4 −2 3


S ={f ∈ P2(R) : f ′(−1) + 5f(−1) = 0

}.


S ={f ∈ P3(R) : f(1) = 0 y f ′(1) = 0

}.




Tarea 4, variante 18 MPDD, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[4 2

−2 −1

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−4 1

−2 3

].


b1 =

121

, b2 =

021

, b3 =

−13

2


uE =

−30

1

, wB =

1

−31



A =

−4 1 0 3 3

−3 −1 2 −4 −2−4 1 1 3 −33 1 −1 4 −4

.



A =

[1 3 1 −12 6 2 −2

], B =

1 1

−2 −23 2

, C =

[4 −11 4

].


A B C



−3 −3 −4 6

1 0 1 −2−3 −6 −5 6

1 3 2 −2

.


A1 =

[2 1

2 3

], A2 =

[−1 2

4 1

], A3 =

[−3 2

4 −1

],

A4 =

[1 −2

−4 −1

], A5 =

[3 2

4 5

].



f1 = 1− x+ 2x2 + 2x3, f2 = −2+ 3x− 4x2 − 4x3, f3 = 2+ x+ 2x

2 + 4x3,

f4 = −2− 4x+ 2x2 − 4x3, f5 = 1− 2x+ 3x2 + 2x3.


a1 =

3

1

−44

, a2 =

7

5

4

4

, a3 =

4

3

3

2

, b1 =

3

2

1

2

, b2 =

3

2

3

1







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 19 MHF.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. 4 4 −1 2 −3

−2 1 1 1 0

4 2 −1 1 −22 3 −1 1 −2


{x ∈ R5 : Ax = 04



1 −5 −4 3 −1−2 4 1 −3 1

−4 2 −5 −3 1


S ={f ∈ P2(R) : f ′(−2) − 4f(−2) = 0

}.


S ={f ∈ P3(R) : f(−3) = 0 y f ′′(2) = 0

}.




Tarea 4, variante 19 MHF, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[3 4

−3 −4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[3 −24 −3

].


b1 =

142

, b2 =

131

, b3 =

032


uE =

132

, wB =

−13

−2



A =

4 1 2 2 4

−2 1 −1 1 0

2 −4 1 −5 −42 2 1 3 4

.



A =

[1 4 −3 1

4 1 −3 3

], B =

−1 2

1 −21 −2

, C =

−5 1 1

0 2 0

−4 0 0

0 0 0


A B C



−6 2 −4 −24 4 4 1

3 −1 2 1

2 −6 0 1

.


A1 =

[−1 −12 0

], A2 =

[−1 2

−4 −3

], A3 =

[2 1

−2 1

],

A4 =

[3 1

−2 2

], A5 =

[5 1

−2 4

].



f1 = 3+ 4x+ 2x2 − 4x3, f2 = 3+ 3x+ 2x

2 − 2x3, f3 = 1+ x+ x2,

f4 = 3+ 2x+ 4x2 + 4x3, f5 = −3+ 2x+ x2 − 2x3.


a1 =

4

5

2

3

, a2 =

2

3

0

1

, a3 =

3

4

1

2

, b1 =

1

2

−10

, b2 =

1

2

−2−1







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 20 MME.

Bases y dimension.




{x ∈ R5 : Ax = 04



1 1 1 −1 1

4 1 1 −2 2

−2 1 2 1 4


{x ∈ R5 : Ax = 04



−3 −2 1 1 −22 2 1 −2 1

4 4 2 −4 2


S ={f ∈ P2(R) : f ′(1) − 2f(1) = 0

}.


S ={f ∈ P3(R) : f(1) = 0 y f ′′(1) = 0

}.




Tarea 4, variante 20 MME, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[1 −3

−1 2

].


{X ∈ M2(R) : tr(XA) = 0



A =

[4 −1

−4 2

].


b1 =

324

, b2 =

112

, b3 =

201


uE =

213

, wB =

−23

2



A =

3 3 4 1 3

−3 1 4 −3 3

2 2 1 4 1

−1 1 1 2 1

.



A =

−2 4

1 −23 −6

, B =

[−3 2

−5 2

], C =

[−4 1 1 −12 4 3 −3

].


A B C



4 −4 1 3

−1 0 1 2

3 −4 2 5

2 −4 3 7

.


A1 =

[1 −2

−4 3

], A2 =

[2 −3

−7 5

], A3 =

[−1 −20 1

],

A4 =

[−4 −35 −1

], A5 =

[4 6

−2 −2

].



f1 = 4+ 4x− 2x2 + 2x3, f2 = −2+ x+ x2 + 2x3, f3 = 1− x+ 2x

2 + x3,

f4 = x+ x2 + 2x3, f5 = 2− 3x+ 4x

2 + x3.


a1 =

3

1

6

4

, a2 =

1

−52

4

, a3 =

2

0

4

3

, b1 =

1

−12

2

, b2 =

1

2

2

1







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 21 MRCK.

Bases y dimension.




{x ∈ R5 : Ax = 04



1 −4 1 1 −23 2 2 2 3

2 0 2 1 4


{x ∈ R5 : Ax = 04



1 2 4 4 3

−1 −4 −7 −3 −4−1 2 2 −6 −1


S ={f ∈ P2(R) : f ′(−3) − 4f(−3) = 0

}.


S ={f ∈ P3(R) : f(−3) = 0 y f ′′(−1) = 0

}.




Tarea 4, variante 21 MRCK, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[−2 2

3 −3

].


{X ∈ M2(R) : tr(XA) = 0



A =

[3 −1

−4 1

].


b1 =

111

, b2 =

211

, b3 =

2

−1−2


uE =

2

−1−2

, wB =

2

−1−1



A =

1 −1 0 1 1

2 1 −1 1 1

5 −2 −1 4 4

4 −1 −1 3 3

.



A =

−2 −21 2

1 1

, B =

1 −5 5

−2 0 −1−2 0 0

, C =

0 2 5 −14 0 −4 2

−1 0 0 −3


A B C



2 2 3 4

−3 −3 2 3

−6 −6 4 6

−5 −5 −1 −1

.


A1 =

[−2 −13 4

], A2 =

[−6 −3−4 −1

], A3 =

[6 3

−1 −4

],

A4 =

[4 2

3 1

], A5 =

[2 1

2 1

].



f1 = 3+ 4x+ x2 + 4x3, f2 = 3+ 3x+ x

2 + 2x3, f3 = 1+ x+ x3,

f4 = −2+ 2x+ 2x2 + 4x3, f5 = 3+ 3x+ 4x2 − x3.


a1 =

1

−1−2−1

, a2 =

3

7

2

3

, a3 =

2

3

0

1

, b1 =

3

2

−20

, b2 =

1

3

1

1







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 22 PHJL.

Bases y dimension.




{x ∈ R5 : Ax = 04



−3 1 0 −1 1

3 4 2 1 2

3 −1 2 1 3


{x ∈ R5 : Ax = 04



−4 6 −4 −2 −20 2 −6 −3 −22 −3 2 1 1


S ={f ∈ P2(R) : f(3) = 2f(0)

}.


S ={f ∈ P3(R) : f(2) = 0 y f ′′(3) = 0

}.




Tarea 4, variante 22 PHJL, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[−4 4

−2 3

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−4 4

−1 3

].


b1 =

301

, b2 =

311

, b3 =

2

−11


uE =

2

−21

, wB =

−20

3



A =

1 −1 1 0 3

2 −2 3 1 −22 −2 2 1 3

−3 3 −3 −2 −3

.



A =

1 2

0 1

1 1

, B =

[2 −1 −2 1

−4 2 4 −2

], C =

3 2 1 5

−5 4 0 1

3 5 0 0


A B C



5 −4 3 1

3 0 2 −22 −4 1 3

1 4 1 −5

.


A1 =

[−5 2

3 −7

], A2 =

[0 4

−4 −4

], A3 =

[−2 1

1 −3

],

A4 =

[−3 4

−1 −7

], A5 =

[−2 3

−1 −5

].



f1 = −2− x+ 2x2 + 2x3, f2 = −2− 4x− x2 + 2x3, f3 = 4+ 2x− 4x2 − 4x3,

f4 = 1+ x+ 2x2 + 4x3, f5 = 1+ x+ x

2 + 2x3.


a1 =

0

−13

2

, a2 =

1

−37

5

, a3 =

−53

1

−1

, b1 =

1

−11

1

, b2 =

−21

2

1







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 23 RAJA.

Bases y dimension.




{x ∈ R5 : Ax = 04



1 2 4 3 1

1 0 4 2 4

1 4 4 4 −2


{x ∈ R5 : Ax = 04


la comprobacion. 5 −4 4 −3 −24 −2 3 −3 −3

−2 −2 −1 3 5

1 −2 1 0 1


S ={f ∈ P2(R) : f ′(3) − 4f(3) = 0

}.


S ={f ∈ P3(R) : f(1) = 0 y f ′(3) = 0

}.




Tarea 4, variante 23 RAJA, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[1 2

−2 3

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−4 −31 2

].


b1 =

−11

2

, b2 =

101

, b3 =

113


uE =

3

−1−1

, wB =

−1−31



A =

3 1 −2 −3 3

−2 2 2 −3 5

−6 −2 4 6 −6−5 1 4 0 2

.



A =

−4 3 −42 0 2

0 0 −2

, B =

3 2

1 1

−2 −1

, C =

−4 −3 0 4

0 0 −1 2

−2 0 4 0


A B C



1 3 4 3

−2 2 3 4

3 1 1 −1−5 1 2 5

.


A1 =

[0 −2

−1 −1

], A2 =

[2 −6

−4 −2

], A3 =

[3 7

2 5

],

A4 =

[4 2

−1 3

], A5 =

[1 3

1 2

].



f1 = 1+ 2x+ x2 + x3, f2 = −1+ x+ 4x2 + 2x3, f3 = 2+ 3x+ x

3,

f4 = −1+ 2x− 2x2 + 3x3, f5 = −1+ x+ 3x2 + 2x3.


a1 =

−13

5

4

, a2 =

−24

7

5

, a3 =

4

0

−22

, b1 =

1

1

1

2

, b2 =

2

1

0

2







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 24 RCE.

Bases y dimension.




{x ∈ R5 : Ax = 04



0 1 1 1 1

1 −1 −2 −1 −2−1 3 4 3 4


{x ∈ R5 : Ax = 04



−1 3 −4 −2 2

−4 −4 2 2 −22 2 −1 −1 1


S ={f ∈ P2(R) : f ′(−1) − 4f(−1) = 0

}.


S ={f ∈ P3(R) : f(−1) = 0 y f ′(1) = 0

}.



3. Haga las comprobaciones: verifique que los elementos de la base construida B pertenecena S, esto es, cumplen con las condiciones f(−1) = 0 y f ′(1) = 0.

Tarea 4, variante 24 RCE, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[−1 −21 4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−4 −32 3

].


b1 =

111

, b2 =

210

, b3 =

−12

4


uE =

−30

2

, wB =

2

−10



A =

1 2 1 3 −42 2 1 3 1

0 −1 1 2 −21 −2 2 4 1

.



A =

[4 1

3 −2

], B =

−2 −41 2

−2 −4

, C =

3 4 2

0 0 4

−2 0 0

0 0 0


A B C



−4 1 −7 3

−1 2 −3 5

3 1 4 2

2 3 1 7

.


A1 =

[2 1

1 −3

], A2 =

[3 2

1 −5

], A3 =

[0 3

−3 −3

],

A4 =

[−1 −43 5

], A5 =

[−4 1

−5 3

].



f1 = 2− x+ x2 + 3x3, f2 = 2+ x+ 3x

2 + x3, f3 = 3+ x+ 2x2 + 2x3,

f4 = 1+ x+ x2, f5 = −2− 3x− 4x2 + x3.


a1 =

3

1

5

7

, a2 =

1

−4−7−2

, a3 =

2

1

4

5

, b1 =

−11

1

−1

, b2 =

2

−11

4







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 25 RDIDJ.

Bases y dimension.




{x ∈ R5 : Ax = 04



1 −1 4 −2 4

1 2 2 −3 1

1 1 1 −1 −1


{x ∈ R5 : Ax = 04



−4 6 5 −5 2

2 −4 3 3 −21 −1 −4 1 0


S ={f ∈ P2(R) : f(−3) = 2f(3)

}.


S ={f ∈ P3(R) : f(−3) = 0 y f ′′(1) = 0

}.




Tarea 4, variante 25 RDIDJ, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[2 1

−4 −2

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−4 2

−1 1

].


b1 =

232

, b2 =

122

, b3 =

331


uE =

−3−12

, wB =

−21

1



A =

2 −2 −1 5 1

0 2 −3 1 3

−2 4 −2 −4 2

1 −2 1 2 −1

.



A =

[6 −6 4 2

3 −3 2 1

], B =

−1 0

1 2

1 1

, C =

−4 0 −4 −54 1 0 0

0 2 0 5


A B C



−7 −5 −4 1

5 3 3 −42 2 1 3

3 1 2 −7

.


A1 =

[3 1

−1 4

], A2 =

[1 −1

−1 2

], A3 =

[−4 −40 −4

],

A4 =

[3 5

1 2

], A5 =

[2 4

1 1

].



f1 = 1+ x+ 3x2 − 3x3, f2 = 2+ x+ 3x

2 − 2x3, f3 = −3− 3x+ 2x2 − 2x3,

f4 = 4− 2x+ 2x2 + 4x3, f5 = 2+ x+ 4x

2 − 3x3.


a1 =

6

4

0

−2

, a2 =

−3−73

5

, a3 =

−1−42

3

, b1 =

−3−20

1

, b2 =

2

−11

1







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 26 SRGA.

Bases y dimension.




{x ∈ R5 : Ax = 04



3 −1 4 4 −1−2 1 −2 −4 1

2 2 4 4 1


{x ∈ R5 : Ax = 04


la comprobacion. 1 −4 −2 −2 2

2 −3 1 −3 5

0 −5 −5 −1 −1−1 −1 −3 1 −3


S ={f ∈ P2(R) : f ′(−2) + 3f(−2) = 0

}.


S ={f ∈ P3(R) : f(−3) = 0 y f ′(1) = 0

}.




Tarea 4, variante 26 SRGA, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[3 1

4 2

].


{X ∈ M2(R) : tr(XA) = 0



A =

[1 2

4 −1

].


b1 =

423

, b2 =

313

, b3 =

101


uE =

212

, wB =

1

−23



A =

−3 −2 0 1 2

2 −2 −1 1 2

−2 −1 2 2 3

3 −1 1 2 3

.



A =

−2 1 −2 5

4 5 0 −10 3 0 −4

, B =

[3 −1 1 2

−6 2 −2 −4

], C =

[4 −2

−1 4

].


A B C



2 1 1 1

1 2 1 3

3 3 2 4

1 −1 0 −2

.


A1 =

[−2 5

−3 4

], A2 =

[2 −31 −4

], A3 =

[−2 −24 4

],

A4 =

[1 −21 −2

], A5 =

[−3 3

0 6

].



f1 = 1+ 2x+ 3x2 − 2x3, f2 = 3+ x+ 2x

2 + x3, f3 = 2+ 2x− x2 + 3x3,

f4 = 2+ x+ x2 + x3, f5 = 3+ 2x+ 2x

2 + x3.


a1 =

1

2

1

−1

, a2 =

4

5

1

2

, a3 =

5

6

1

3

, b1 =

1

1

0

1

, b2 =

1

2

1

−2







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 27 TMMA.

Bases y dimension.




{x ∈ R5 : Ax = 04



−1 −2 2 −3 −1−3 2 −1 2 1

−2 −1 2 0 1


{x ∈ R5 : Ax = 04



2 −2 6 6 3

1 −2 4 2 2

2 −4 8 4 4


S ={f ∈ P2(R) : f ′(2) − 4f(2) = 0

}.


S ={f ∈ P3(R) : f(−2) = 0 y f ′′(3) = 0

}.




Tarea 4, variante 27 TMMA, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[4 1

2 3

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−1 −21 2

].


b1 =

114

, b2 =

0

−11

, b3 =

122


uE =

120

, wB =

−12

1



A =

1 3 5 3 −61 0 2 1 −4

−2 6 2 2 4

−1 3 1 1 2

.



A =

2 2 4

−2 0 1

2 0 0

, B =

−1 1 4 4

0 4 3 −10 0 1 1

, C =

[1 3 1 −1

−2 −6 −2 2

].


A B C



−4 6 −5 −44 −3 1 1

0 −3 4 3

−4 −3 7 5

.


A1 =

[4 0

−2 −2

], A2 =

[−6 2

5 3

], A3 =

[−2 1

2 1

],

A4 =

[4 3

1 −2

], A5 =

[−6 1

4 3

].



f1 = −2+ x+ 3x2 − x3, f2 = 3+ 4x− 2x2 − x3, f3 = 3− x− 4x

2 + x3,

f4 = −1+ x+ 2x2 − x3, f5 = 1+ 2x+ 2x2 − 3x3.


a1 =

−21

2

1

, a2 =

−75

1

−1

, a3 =

−32

1

0

, b1 =

−21

2

1

, b2 =

−21

4

2







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 28 TELD.

Bases y dimension.




{x ∈ R5 : Ax = 04



2 4 1 −2 3

−1 −2 3 2 −11 2 4 0 2


{x ∈ R5 : Ax = 04


la comprobacion. −6 5 −1 −1 −2−2 3 1 −1 1

4 −2 2 0 3

4 −6 −2 2 −2


S ={f ∈ P2(R) : f ′(−1) + 4f(−1) = 0

}.


S ={f ∈ P3(R) : f(1) = 0 y f ′′(−1) = 0

}.




Tarea 4, variante 28 TELD, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[1 −13 −4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−3 4

−1 1

].


b1 =

134

, b2 =

111

, b3 =

034


uE =

−22

3

, wB =

0

−31



A =

−1 2 −1 1 −1−3 −2 2 −2 2

2 1 1 1 −14 0 −1 1 −1

.



A =

[−4 1 −1 4

−4 1 −1 5

], B =

[−2 −34 1

], C =

0 0 0

−2 −5 −40 −1 −20 −2 0


A B C



−2 −2 −2 −4−4 2 2 6

1 1 1 2

−1 2 2 5

.


A1 =

[−5 −3−2 1

], A2 =

[−3 −3−3 0

], A3 =

[2 4

5 1

],

A4 =

[−1 1

2 1

], A5 =

[3 5

6 1

].



f1 = x− x2 + 2x3, f2 = 1+ 2x+ x

2 + x3, f3 = −2+ 3x− x2 + 4x3,

f4 = 1− 2x− 4x2 + 2x3, f5 = 2+ 4x+ 3x

2 + x3.


a1 =

4

−27

−5

, a2 =

3

−15

−4

, a3 =

−33

−63

, b1 =

1

1

1

−2

, b2 =

1

2

0

−2







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 29 UTAV.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. −1 1 4 −1 −13 1 2 1 −51 1 3 0 −31 1 1 1 −4


{x ∈ R5 : Ax = 04



1 4 1 3 3

−3 −8 −2 −1 −52 4 1 −2 2


S ={f ∈ P2(R) :

1∫−1

f(x)dx = 0}.


S ={f ∈ P3(R) : f(−2) = 0 y f ′(1) = 0

}.




Tarea 4, variante 29 UTAV, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[4 −2

−1 1

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−4 2

−1 1

].


b1 =

111

, b2 =

110

, b3 =

233


uE =

−10

4

, wB =

−32

0



A =

0 −3 5 −5 2

3 −3 −2 −4 4

−2 1 3 1 −2−1 2 −1 3 −2

.



A =

[−1 −2 1 −22 4 −2 4

], B =

1 0

3 2

1 1

, C =

3 1 0

2 0 −30 0 3


A B C



−5 1 3 1

4 1 2 1

−1 2 5 2

3 3 7 3

.


A1 =

[−1 −7−3 4

], A2 =

[7 −1

−4 −3

], A3 =

[−3 −11 2

],

A4 =

[1 −5

−3 2

], A5 =

[−2 0

1 1

].



f1 = 2+ 4x+ x2 + x3, f2 = 2+ 3x+ 2x

2 − x3, f3 = 1+ 3x+ x2 + x3,

f4 = −1+ 4x+ 2x2 + 3x3, f5 = −4− x+ 3x3.


a1 =

1

3

2

3

, a2 =

1

4

3

5

, a3 =

2

6

4

6

, b1 =

0

1

1

2

, b2 =

2

3

1

−1







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 30 VNDI.

Bases y dimension.




{x ∈ R5 : Ax = 04



2 −2 2 −3 1

1 −3 0 3 1

1 −1 −1 3 1


{x ∈ R5 : Ax = 04


la comprobacion. 2 1 1 1 −12 −3 3 2 −44 2 2 2 −20 −4 2 1 −3


S ={f ∈ P2(R) : f(−3) = 2f(1)

}.


S ={f ∈ P3(R) : f(3) = 0 y f ′′(1) = 0

}.




Tarea 4, variante 30 VNDI, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[−4 3

−3 4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−1 4

2 3

].


b1 =

2

−11

, b2 =

111

, b3 =

021


uE =

302

, wB =

1

−31



A =

−2 −3 −2 1 −13 3 3 3 1

3 4 −2 −3 1

1 1 −4 −2 0

.



A =

[−4 3 1 2

2 4 −2 3

], B =

0 −1 5

1 0 3

0 0 −1

, C =

3 −61 −22 −4


A B C



3 1 4 −35 2 4 −42 1 0 −17 3 4 −5

.


A1 =

[2 3

1 −3

], A2 =

[1 6

−4 3

], A3 =

[3 3

3 −6

],

A4 =

[3 5

1 −4

], A5 =

[−1 −42 −1

].



f1 = −3x− x2 + 2x3, f2 = −2+ 3x+ 4x2 + x3, f3 = 1+ 2x+ 4x2 + 2x3,

f4 = −4− 4x− x2 + 3x3, f5 = −1+ 2x+ 3x2 + x3.


a1 =

1

−10

1

, a2 =

3

3

4

1

, a3 =

5

4

6

2

, b1 =

1

2

2

0

, b2 =

2

−11

2







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 31 VHA.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. −2 0 1 2 −41 2 1 1 4

−1 1 2 4 −1−2 1 1 1 −3


{x ∈ R5 : Ax = 04


la comprobacion. −2 2 2 0 −24 −1 3 1 −35 −2 2 1 −26 −3 1 1 −1


S ={f ∈ P2(R) : f ′(−2) − 5f(−2) = 0

}.


S ={f ∈ P3(R) : f(2) = 0 y f ′′(−3) = 0

}.




Tarea 4, variante 31 VHA, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[4 −23 1

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−1 2

4 3

].


b1 =

301

, b2 =

311

, b3 =

2

−11


uE =

1

−31

, wB =

2

−30



A =

1 −3 1 1 3

3 −4 −3 4 0

2 −1 −4 3 −3−1 −2 5 −2 6

.



A =

[−2 1 2 3

−4 2 4 6

], B =

1 1

−1 −22 3

, C =

0 −2 −35 5 0

0 5 0


A B C



2 1 0 5

2 −6 2 6

1 4 −1 2

−1 3 −1 −3

.


A1 =

[−1 −10 −1

], A2 =

[1 3

1 2

], A3 =

[2 4

1 3

],

A4 =

[−6 2

4 −2

], A5 =

[−5 3

4 −1

].



f1 = −3− 4x− x2 − 4x3, f2 = 1+ 3x− 2x2 + x3, f3 = 1+ 4x− 3x

2 + x3,

f4 = x+ x2 + x3, f5 = 4− 4x− 2x

2 − x3.


a1 =

3

5

−43

, a2 =

1

2

−22

, a3 =

0

1

−23

, b1 =

2

3

−21

, b2 =

2

3

−32







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 32 VMJJ.

Bases y dimension.




{x ∈ R5 : Ax = 04



1 −1 −4 1 −42 2 2 −2 1

3 3 2 3 1


{x ∈ R5 : Ax = 04



4 1 2 2 5

4 3 1 4 2

−8 −6 −2 −8 −4


S ={f ∈ P2(R) : f(−1) = 2f(0)

}.


S ={f ∈ P3(R) : f(1) = 0 y f ′(2) = 0

}.




Tarea 4, variante 32 VMJJ, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[4 −41 −2

].


{X ∈ M2(R) : tr(XA) = 0



A =

[2 −43 −2

].


b1 =

124

, b2 =

112

, b3 =

111


uE =

−20

3

, wB =

2

−31



A =

4 3 −1 2 3

3 0 −5 −1 −1−3 4 −1 3 4

1 3 4 3 4

.



A =

5 −1 1 3

2 −1 1 0

0 1 0 0

, B =

[−2 3

−1 1

], C =

[4 4 −2 4

−2 −2 1 −2

].


A B C



−3 1 2 −3−3 3 4 3

−6 2 4 −60 1 1 3

.


A1 =

[5 3

−7 2

], A2 =

[3 −1

−7 4

], A3 =

[4 4

−4 0

],

A4 =

[2 1

−3 1

], A5 =

[−3 −24 −1

].



f1 = 2+ 4x− 3x2 − x3, f2 = 1− x− x

2, f3 = 1+ 4x+ 2x2 + 3x3,

f4 = 1+ 2x+ x2 + 2x3, f5 = 2− 4x− 3x

2 − x3.


a1 =

3

1

−42

, a2 =

4

2

−63

, a3 =

−60

6

−3

, b1 =

1

1

−21

, b2 =

−12

−21







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 33 GMOA.

Bases y dimension.




{x ∈ R5 : Ax = 04



−4 −1 1 1 1

−1 4 1 3 2

−3 −2 1 1 4


{x ∈ R5 : Ax = 04



1 −4 −2 2 3

1 4 −4 2 1

1 0 −3 2 2


S ={f ∈ P2(R) :

0∫−2

f(x)dx = 0}.


S ={f ∈ P3(R) : f(−1) = 0 y f ′(2) = 0

}.




Tarea 4, variante 33 GMOA, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[−1 −21 3

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−3 −12 4

].


b1 =

111

, b2 =

−12

1

, b3 =

−22

1


uE =

−21

1

, wB =

0

−21



A =

−4 2 −4 5 −3−2 1 −5 3 0

−2 1 1 2 −3−4 2 2 4 −6

.



A =

[−5 2

2 5

], B =

[3 1 0 2

1 −1 4 1

], C =

3 −4 4

−2 −2 0

0 4 0

0 0 0


A B C



2 4 2 3

2 −1 −3 −4−6 −2 4 5

−4 −3 1 1

.


A1 =

[3 4

−2 −2

], A2 =

[−1 −46 −2

], A3 =

[−3 −1−4 5

],

A4 =

[4 2

4 −6

], A5 =

[2 3

−2 −1

].



f1 = 3+ 3x+ 4x3, f2 = 2+ x+ x

2 + 2x3, f3 = −1− 2x+ x2 + 4x3,

f4 = 2+ x+ x2 + x3, f5 = 3+ 2x+ x

2 + 4x3.


a1 =

2

2

−1−1

, a2 =

1

4

4

−5

, a3 =

2

6

5

−7

, b1 =

−10

2

−1

, b2 =

2

3

1

−3







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 34 VALA.

Bases y dimension.




{x ∈ R5 : Ax = 04



−3 4 −1 −4 4

3 −4 2 4 −11 2 1 2 0


{x ∈ R5 : Ax = 04


la comprobacion. −2 0 −7 4 −11 −3 3 −2 2

2 −6 6 −4 4

1 3 4 −2 −1


S ={f ∈ P2(R) : f ′(1) + 4f(1) = 0

}.


S ={f ∈ P3(R) : f(−3) = 0 y f ′(−1) = 0

}.




Tarea 4, variante 34 VALA, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[4 1

3 2

].


{X ∈ M2(R) : tr(XA) = 0



A =

[3 −34 2

].


b1 =

120

, b2 =

121

, b3 =

1

1

−1


uE =

1

0

−4

, wB =

−21

−1



A =

1 1 1 −4 0

−2 3 −3 −2 −23 −2 4 −2 2

2 4 2 4 1

.



A =

6 5

−4 −31 1

, B =

−1 4 4 −4−2 5 −4 0

2 1 0 0

, C =

[4 4

−3 −2

].


A B C



1 2 1 2

4 0 3 −2−3 2 −2 4

2 4 2 4

.


A1 =

[−5 −12 3

], A2 =

[3 5

1 −4

], A3 =

[5 −3

−4 −1

],

A4 =

[−7 −13 4

], A5 =

[−2 −4−1 3

].



f1 = 1− 2x− 3x2 − 4x3, f2 = 4+ 4x+ x

2 − 2x3, f3 = 2+ 3x+ 2x2 + x3,

f4 = 3+ 3x+ x2 − x3, f5 = −4− 4x− 4x2 − 4x3.


a1 =

3

2

4

1

, a2 =

2

4

4

−6

, a3 =

5

4

7

0

, b1 =

1

0

1

2

, b2 =

1

2

2

−2







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 35 ZPJ.

Bases y dimension.




{x ∈ R5 : Ax = 04



−2 1 −2 −1 2

2 −1 3 2 −13 0 1 3 −1


{x ∈ R5 : Ax = 04


la comprobacion. 6 −6 −4 2 −42 −3 4 1 3

4 −3 −8 1 −73 −3 −2 1 −2


S ={f ∈ P2(R) : f(0) = 2f(−3)

}.


S ={f ∈ P3(R) : f(−2) = 0 y f ′(3) = 0

}.




Tarea 4, variante 35 ZPJ, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[2 4

−1 −2

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−4 2

−3 1

].


b1 =

111

, b2 =

233

, b3 =

223


uE =

1

−21

, wB =

2

1

−2



A =

5 1 3 4 −16 −6 4 6 −22 4 1 1 0

3 −3 2 3 −1

.



A =

[1 3 −2 0

2 5 3 −3

], B =

2 1

6 3

6 3

, C =

[3 −53 −1

].


A B C



1 5 0 −13 3 1 1

5 1 2 3

2 −2 1 2

.


A1 =

[−2 4

3 −1

], A2 =

[6 −4

−1 3

], A3 =

[6 2

5 3

],

A4 =

[2 1

2 1

], A5 =

[2 −2

−1 1

].



f1 = 1+ x+ x2 + x3, f2 = 1+ 2x+ 2x

2 + x3, f3 = −1+ 3x− x2 + 3x3,

f4 = 3+ x2 + 2x3, f5 = 3− 3x− 3x

2 + 3x3.


a1 =

3

1

−45

, a2 =

2

2

−46

, a3 =

2

1

−34

, b1 =

1

1

−23

, b2 =

−11

−12







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 36 BRMF.

Bases y dimension.




{x ∈ R5 : Ax = 04



1 −4 2 −1 −32 −2 2 1 1

1 2 0 2 4


{x ∈ R5 : Ax = 04


la comprobacion. −2 −4 −2 −2 −41 2 1 1 2

−2 −4 1 1 3

1 2 −2 −2 −5


S ={f ∈ P2(R) : f(2) = 2f(1)

}.


S ={f ∈ P3(R) : f(2) = 0 y f ′′(2) = 0

}.




Tarea 4, variante 36 BRMF, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[1 2

−3 −1

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−2 4

1 3

].


b1 =

122

, b2 =

−23

1

, b3 =

−12

1


uE =

1

−2−1

, wB =

0

1

−3



A =

−3 −2 1 −2 1

−1 2 3 1 −42 1 1 1 3

4 0 −4 1 3

.



A =

[2 4 −6 −61 2 −3 −3

], B =

[−5 4

5 −2

], C =

−5 −1 3 1

3 5 0 −4−5 0 0 −2


A B C



2 2 5 2

1 2 3 −22 4 6 −41 0 2 4

.


A1 =

[−6 4

−2 4

], A2 =

[3 −21 −2

], A3 =

[−1 2

1 −2

],

A4 =

[6 −33 −6

], A5 =

[−1 3

2 −4

].



f1 = −2− 2x− 4x2 + 2x3, f2 = −1− 3x− 4x2 − 2x3, f3 = 2+ x+ 3x2 + x3,

f4 = −1+ 2x+ x2 + x3, f5 = 1+ x+ 2x2 + x3.


a1 =

−4−26

2

, a2 =

5

4

−35

, a3 =

4

3

−33

, b1 =

2

1

−3−1

, b2 =

3

2

−22







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 37 RMIA.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. 1 −1 −2 2 2

0 −2 −1 1 1

−2 −3 −1 −4 −31 4 3 2 1


{x ∈ R5 : Ax = 04


la comprobacion. 4 −1 −3 0 −1

−1 −1 5 −1 2

3 −2 2 −1 1

−6 4 −4 2 −2


S ={f ∈ P2(R) : f ′(2) + 2f(2) = 0

}.


S ={f ∈ P3(R) : f(2) = 0 y f ′(1) = 0

}.




Tarea 4, variante 37 RMIA, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[−1 3

2 −3

].


{X ∈ M2(R) : tr(XA) = 0



A =

[2 4

3 −3

].


b1 =

134

, b2 =

111

, b3 =

034


uE =

310

, wB =

2

−3−1



A =

−2 4 −2 2 4

1 −4 4 1 1

−1 2 −1 1 2

0 −2 3 2 3

.



A =

−2 −13 2

1 1

, B =

[4 −45 −1

], C =

−1 −3 −5 −51 −2 5 0

0 −2 −1 0


A B C



6 2 −6 −43 1 −3 −24 1 −3 2

−5 −1 3 −6

.


A1 =

[4 −7

−3 −1

], A2 =

[2 −5

−3 1

], A3 =

[4 −13 −7

],

A4 =

[−1 5

4 −3

], A5 =

[1 0

1 −2

].



f1 = 3+ x+ 2x2 + 4x3, f2 = 2+ 3x+ x

2 + 3x3, f3 = 1+ 4x+ x2 + x3,

f4 = 2+ 4x+ x2 + 3x3, f5 = −1+ 2x− 3x2 + x3.


a1 =

4

−2−22

, a2 =

2

1

2

2

, a3 =

6

1

3

5

, b1 =

0

2

3

1

, b2 =

2

2

3

2







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 38 MRHA.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. −1 1 1 2 −34 2 4 1 1

−1 −3 −4 −4 2

3 1 2 0 4


{x ∈ R5 : Ax = 04



1 5 2 2 −5−2 −2 −2 −6 2

1 1 1 3 −1


S ={f ∈ P2(R) : f(−2) = 2f(−3)

}.


S ={f ∈ P3(R) : f(1) = 0 y f ′′(−3) = 0

}.




Tarea 4, variante 38 MRHA, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[2 −33 1

].


{X ∈ M2(R) : tr(XA) = 0



A =

[4 −1

−3 2

].


b1 =

113

, b2 =

0

1

−1

, b3 =

114


uE =

0

−12

, wB =

−12

2



A =

−1 4 2 −4 1

2 1 0 −2 1

3 1 1 −4 2

2 −3 −3 4 −1

.



A =

[3 −3 2 4

2 4 1 3

], B =

−1 −3 −41 −5 0

0 0 0

2 0 0

, C =

4 1 −22 0 −30 0 −4


A B C



1 1 −1 3

1 2 1 3

1 3 3 3

1 4 5 3

.


A1 =

[3 −7

−1 −4

], A2 =

[2 −13 1

], A3 =

[−1 1

−1 0

],

A4 =

[4 −53 −1

], A5 =

[5 −37 2

].



f1 = 1− 4x− 3x2 + 2x3, f2 = 3+ x+ 2x

2 − 3x3, f3 = 3x+ 2x2 − x3,

f4 = 3+ x+ x2 − x3, f5 = 4+ x+ x

2 − x3.


a1 =

6

4

3

3

, a2 =

7

5

3

4

, a3 =

−7−5−3−4

, b1 =

4

2

3

1

, b2 =

2

1

1

1







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 39 AVMA.

Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. −2 2 −1 1 −31 3 −1 1 −10 4 −1 2 −33 1 1 1 1


{x ∈ R5 : Ax = 04


la comprobacion. −2 4 −2 −2 −6−3 −6 −5 −3 3

1 4 2 1 −32 2 3 2 0


S ={f ∈ P2(R) : f(0) = 2f(3)

}.


S ={f ∈ P3(R) : f(−2) = 0 y f ′(−3) = 0

}.




Tarea 4, variante 39 AVMA, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[3 1

−4 −2

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−3 −11 2

].


b1 =

111

, b2 =

110

, b3 =

121


uE =

0

2

−1

, wB =

−20

1



A =

−2 3 −2 2 1

−4 6 −4 4 2

−6 3 −5 3 2

−4 0 −3 1 1

.



A =

6 3

2 1

4 2

, B =

0 1 −10 0 0

5 −3 0

4 0 0

, C =

3 2 −30 −5 5

0 0 3


A B C



4 −6 −6 6

2 −3 −3 3

3 2 −1 4

1 5 2 1

.


A1 =

[−3 4

7 1

], A2 =

[4 1

−3 5

], A3 =

[−1 3

4 2

],

A4 =

[1 2

1 3

], A5 =

[5 1

−4 6

].



f1 = 1− 4x− 4x2 − x3, f2 = 4− x+ 4x

2 + x3, f3 = 4+ 2x2 − 2x3,

f4 = 3− x+ 3x2 + x3, f5 = 2− x+ 3x

2 + 2x3.


a1 =

−15

−15

, a2 =

1

−51

−5

, a3 =

−27

−36

, b1 =

1

1

3

3

, b2 =

1

0

3

2







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 40 MCCO.

Bases y dimension.




{x ∈ R5 : Ax = 04



−3 3 1 1 1

−4 2 −3 −3 −24 −1 4 3 2


{x ∈ R5 : Ax = 04



3 3 1 1 3

−3 −2 −6 −1 −46 6 2 2 6


S ={f ∈ P2(R) : f ′(−2) + 2f(−2) = 0

}.


S ={f ∈ P3(R) : f(−1) = 0 y f ′(−3) = 0

}.




Tarea 4, variante 40 MCCO, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[1 2

−2 −1

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−4 −21 3

].


b1 =

344

, b2 =

343

, b3 =

233


uE =

231

, wB =

−1−13



A =

1 −3 −1 −2 0

1 1 2 2 −41 4 1 1 −2

−1 −3 3 4 −4

.



A =

−2 −25 4

4 3

, B =

−2 2 −30 −1 0

4 0 0

, C =

[6 4 −2 −2

−3 −2 1 1

].


A B C



−1 0 1 2

−4 −2 −2 −42 1 1 2

1 1 2 4

.


A1 =

[3 1

−2 5

], A2 =

[1 2

1 0

], A3 =

[2 5

3 −1

],

A4 =

[3 5

2 1

], A5 =

[−4 −31 −5

].



f1 = 1− x− 2x2 − 2x3, f2 = −2+ x+ x2 + 3x3, f3 = −1+ x2 + x3,

f4 = −3+ x+ x2 + 4x3, f5 = −3− x− 4x2 + 2x3.


a1 =

−3−5−1−3

, a2 =

−40

2

1

, a3 =

−7−13

1

, b1 =

−3−11

0

, b2 =

−14

3

3







Engra

peaq

uı

No

dobl

e

Algebra II, licenciatura. Tarea 4. Variante 41.

Bases y dimension.




{x ∈ R5 : Ax = 04



3 1 2 −1 3

2 1 1 2 2

2 3 1 −4 3


{x ∈ R5 : Ax = 04



−1 −1 5 −3 −1−2 4 7 −6 7

−1 1 4 −3 2


S ={f ∈ P2(R) : f(3) = 2f(2)

}.


S ={f ∈ P3(R) : f(1) = 0 y f ′′(3) = 0

}.




Tarea 4, variante 41, pagina 1 de 4


{X ∈M2(R) : AX = XA



A =

[2 −2

−1 4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−1 1

3 2

].


b1 =

120

, b2 =

121

, b3 =

1

1

−1


uE =

121

, wB =

−23

1



A =

1 −1 −4 0 −31 −3 2 4 1

−1 2 1 −2 1

−2 4 2 −4 2

.



A =

−2 −1 1

0 0 0

0 0 −10 4 0

, B =

−2 0 0

0 0 3

0 −5 0

, C =

[2 1 2 1

1 2 5 4

].


A B C



3 4 2 1

4 5 −1 −1−2 −3 −5 −31 1 −3 −2

.


A1 =

[−3 −2−2 1

], A2 =

[−2 4

−4 −2

], A3 =

[1 −22 1

],

A4 =

[5 6

2 −3

], A5 =

[2 −23 1

].



f1 = 1+ x+ x3, f2 = 3+ x+ 2x

2 + 2x3, f3 = −1− 3x+ 2x2 − 2x3,

f4 = 3+ x+ 4x2 + 2x3, f5 = 3+ x+ x

2 + 2x3.


a1 =

6

1

4

3

, a2 =

3

−13

1

, a3 =

3

2

1

2

, b1 =

0

−32

−1

, b2 =

2

1

1

1







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04



1 −3 1 1 1

−2 −4 −3 −4 −31 3 1 2 −2


{x ∈ R5 : Ax = 04



4 6 3 3 5

2 4 2 3 3

−2 −6 −3 −6 −4


S ={f ∈ P2(R) : f(−2) = 2f(3)

}.


S ={f ∈ P3(R) : f(−1) = 0 y f ′(3) = 0

}.






{X ∈M2(R) : AX = XA



A =

[2 −24 −1

].


{X ∈ M2(R) : tr(XA) = 0



A =

[2 −13 4

].


b1 =

344

, b2 =

343

, b3 =

233


uE =

013

, wB =

2

−31



A =

2 2 0 1 1

−4 4 4 1 3

−1 3 2 1 2

3 2 3 2 −1

.



A =

−3 6

1 −22 −4

, B =

[2 4 2 3

0 1 3 1

], C =

5 4 −30 0 3

0 0 0

2 0 0


A B C



−4 −2 −4 2

1 1 2 −12 2 4 −23 2 4 −2

.


A1 =

[2 5

−7 3

], A2 =

[−4 1

3 5

], A3 =

[−2 −13 1

],

A4 =

[−4 0

4 4

], A5 =

[1 2

−3 1

].



f1 = 1+ x− x3, f2 = 2+ 2x+ x

2 − x3, f3 = 2+ 3x+ 2x2 − x3,

f4 = 2+ x+ 4x2 + 3x3, f5 = −1+ 4x+ x2 − 3x3.


a1 =

0

−64

2

, a2 =

6

6

5

4

, a3 =

4

3

4

3

, b1 =

0

−32

1

, b2 =

1

−23

2







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04



−3 −1 0 1 2

2 −2 1 1 1

−1 −3 1 2 3


{x ∈ R5 : Ax = 04



−2 1 −2 −2 3

0 2 1 2 5

2 1 3 4 2


S ={f ∈ P2(R) : f ′(−3) + 5f(−3) = 0

}.


S ={f ∈ P3(R) : f(−1) = 0 y f ′′(2) = 0

}.






{X ∈M2(R) : AX = XA



A =

[3 −24 2

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−3 −24 1

].


b1 =

343

, b2 =

334

, b3 =

102


uE =

0

2

−1

, wB =

−10

2



A =

4 −3 3 −2 4

−3 1 −2 5 −31 −2 1 3 1

5 −5 4 1 5

.



A =

0 0 0

3 4 −10 5 1

0 0 3

, B =

[4 2

4 −2

], C =

−2 1

4 −2−2 1


A B C



−3 1 0 3

−2 −4 −2 −41 2 1 2

4 1 1 −1

.


A1 =

[5 3

7 2

], A2 =

[1 −35 4

], A3 =

[2 1

3 1

],

A4 =

[3 0

6 3

], A5 =

[3 2

4 1

].



f1 = 2− 3x+ x2 + x3, f2 = 1− 2x+ x

2 + x3, f3 = 3− 2x− x3,

f4 = 3+ x− 2x2 − 4x3, f5 = −1+ 4x+ 4x2 − 3x3.


a1 =

5

5

−2−7

, a2 =

−11

2

3

, a3 =

3

2

−2−5

, b1 =

1

−1−2−3

, b2 =

2

4

1

−1







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04



0 −4 2 −2 1

1 −1 3 −3 2

1 2 2 2 1


{x ∈ R5 : Ax = 04


la comprobacion. 4 4 −8 2 −21 −2 0 1 −12 2 −4 1 −1

−4 2 4 −3 3


S ={f ∈ P2(R) :

2∫−2

f(x)dx = 0}.


S ={f ∈ P3(R) : f(−3) = 0 y f ′(3) = 0

}.






{X ∈M2(R) : AX = XA



A =

[2 −1

−2 3

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−3 −12 4

].


b1 =

111

, b2 =

110

, b3 =

121


uE =

231

, wB =

2

3

−2



A =

2 −4 3 3 −33 −1 4 3 1

3 −4 4 4 0

−2 1 −3 −2 2

.



A =

[1 −3 2 −41 1 3 −1

], B =

−6 −32 1

2 1

, C =

0 0 0

5 5 5

4 −4 0

0 4 0


A B C



−1 −1 −2 −2−1 2 3 2

−2 1 1 0

−3 3 4 2

.


A1 =

[5 −1

−4 −2

], A2 =

[2 −1

−1 −2

], A3 =

[−5 3

2 6

],

A4 =

[−1 −34 −6

], A5 =

[−3 2

1 4

].



f1 = 1+ x+ x2 + x3, f2 = 3+ 2x+ 4x

2, f3 = 2− 2x+ 3x2 − 4x3,

f4 = 3+ 3x+ 4x2 + x3, f5 = −4+ 3x− 2x2 − x3.


a1 =

4

−23

1

, a2 =

2

0

1

−1

, a3 =

5

−34

2

, b1 =

2

0

1

−1

, b2 =

2

1

1

−2







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. 3 3 −1 4 −31 0 −1 1 −3

−2 −2 −2 −2 −21 2 −1 2 −1


{x ∈ R5 : Ax = 04


la comprobacion. −6 −3 −5 −2 −36 −3 4 1 3

4 0 3 1 2

2 3 2 1 1


S ={f ∈ P2(R) : f ′(3) + 5f(3) = 0

}.


S ={f ∈ P3(R) : f(−2) = 0 y f ′(−1) = 0

}.






{X ∈M2(R) : AX = XA



A =

[2 −4

−3 3

].


{X ∈ M2(R) : tr(XA) = 0



A =

[1 −42 −2

].


b1 =

1

0

−3

, b2 =

111

, b3 =

112


uE =

014

, wB =

1

−13



A =

−1 1 4 3 −41 1 3 2 1

2 2 6 4 2

3 1 2 1 6

.



A =

−5 −3 −2−2 4 0

5 0 0

, B =

2 3 1

−2 0 0

0 0 0

0 3 0

, C =

2 −1−2 1

2 −1


A B C



2 3 5 −21 −3 2 3

2 −6 4 6

−1 −6 −3 5

.


A1 =

[−2 −3−4 −5

], A2 =

[−2 −10 1

], A3 =

[−1 1

3 5

],

A4 =

[−2 1

4 7

], A5 =

[4 4

4 4

].



f1 = 3− x+ 2x2 + 2x3, f2 = 3+ 2x

2 + x3, f3 = 1− x+ x2 + x3,

f4 = −1+ x− 4x2 + 2x3, f5 = 3− 2x+ 3x2 + 2x3.


a1 =

3

1

−32

, a2 =

5

−33

0

, a3 =

2

3

−63

, b1 =

4

−10

1

, b2 =

1

1

−21







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04



3 2 3 −2 2

−4 −3 −4 3 −3−2 1 4 −3 0


{x ∈ R5 : Ax = 04



−2 1 −4 1 −4−6 3 −4 4 −66 −3 −4 −5 0


S ={f ∈ P2(R) : f ′(2) + 5f(2) = 0

}.


S ={f ∈ P3(R) : f(−1) = 0 y f ′(−1) = 0

}.






{X ∈M2(R) : AX = XA



A =

[1 4

2 −4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−4 −21 2

].


b1 =

2

−11

, b2 =

111

, b3 =

021


uE =

−13

0

, wB =

−13

−2



A =

0 1 3 −4 1

−3 3 4 −4 2

2 −3 −1 2 −1−1 2 2 −3 1

.



A =

1 5 2 −33 −4 4 0

2 −5 0 0

, B =

5 1 −5−1 1 0

0 2 0

, C =

−2 −23 2

4 3


A B C



1 1 5 −32 2 6 6

1 1 3 3

1 1 4 0

.


A1 =

[1 −34 −5

], A2 =

[3 4

−1 −2

], A3 =

[1 −23 −4

],

A4 =

[−1 0

−1 2

], A5 =

[−1 −54 −3

].



f1 = 1+ 4x+ 2x2 − 3x3, f2 = 4x+ 3x

2 − 4x3, f3 = 1+ 4x+ x2 − 3x3,

f4 = 1+ 3x+ 2x2 − 2x3, f5 = 1− x− 2x

2 + 2x3.


a1 =

−2−2−1−1

, a2 =

5

4

−15

, a3 =

7

6

0

6

, b1 =

3

2

−24

, b2 =

1

1

0

1







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. 1 4 1 −1 −1

−1 1 1 1 1

−1 3 1 −2 0

1 1 −1 −4 −2


{x ∈ R5 : Ax = 04



1 −1 −2 −2 2

2 −2 −4 −4 4

−1 −2 3 0 −3


S ={f ∈ P2(R) : f ′(2) − 3f(2) = 0

}.


S ={f ∈ P3(R) : f(2) = 0 y f ′(3) = 0

}.






{X ∈M2(R) : AX = XA



A =

[1 2

−2 −4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[4 3

−2 −4

].


b1 =

114

, b2 =

101

, b3 =

113


uE =

112

, wB =

0

2

−1



A =

2 −2 1 4 2

6 −2 2 −6 −43 −1 1 −3 −25 −3 2 1 0

.



A =

1 −3 4 5

0 −3 2 3

0 0 3 −1

, B =

[−3 5

−3 3

], C =

[−2 2 6 6

−1 1 3 3

].


A B C



1 −1 3 4

1 −4 1 1

2 −5 4 5

0 3 2 3

.


A1 =

[−3 7

1 5

], A2 =

[1 −11 1

], A3 =

[4 −35 6

],

A4 =

[0 −1

−1 −2

], A5 =

[−1 −1−3 −5

].



f1 = 1− 2x+ 3x3, f2 = −2+ 4x− 3x2 − 2x3, f3 = 1− 2x+ 2x

2 − 2x3,

f4 = 1− 2x+ x2 + x3, f5 = 1− 2x+ 2x

2 − x3.


a1 =

−7−51

−2

, a2 =

0

−32

3

, a3 =

−1−53

4

, b1 =

−1−21

1

, b2 =

0

−21

2







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. −4 −3 2 3 −23 2 −3 −2 1

2 1 4 3 −12 1 −4 −1 0


{x ∈ R5 : Ax = 04



1 4 1 −2 1

−1 2 −3 2 −1−2 −2 −4 4 −2


S ={f ∈ P2(R) : f ′(−2) − 3f(−2) = 0

}.


S ={f ∈ P3(R) : f(1) = 0 y f ′(−2) = 0

}.






{X ∈M2(R) : AX = XA



A =

[2 1

−4 −3

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−2 −13 2

].


b1 =

423

, b2 =

111

, b3 =

122


uE =

2

−10

, wB =

−12

0



A =

2 1 1 2 2

−1 −1 1 1 3

3 2 0 1 −1−4 1 2 3 −1

.



A =

−3 0 4 −4−4 2 0 4

0 −1 0 0

, B =

[−4 −45 4

], C =

[2 2 2 −21 1 1 −1

].


A B C



−3 −6 −5 −31 −4 1 3

2 1 3 3

1 5 2 0

.


A1 =

[1 1

2 0

], A2 =

[4 −22 6

], A3 =

[2 1

3 1

],

A4 =

[3 4

7 −1

], A5 =

[−4 3

−1 −7

].



f1 = 2+ 2x+ x2 + x3, f2 = 4+ x+ x

2 + 2x3, f3 = 2+ 3x+ x2 + x3,

f4 = −4+ 2x− 2x2 − 2x3, f5 = 2+ 4x+ 4x2 + x3.


a1 =

3

6

5

2

, a2 =

−2−4−3−1

, a3 =

2

4

3

1

, b1 =

1

2

2

1

, b2 =

1

1

2

2







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04



1 1 −2 1 3

2 1 −3 2 3

1 1 1 4 4


{x ∈ R5 : Ax = 04



0 4 −5 −7 1

1 −1 1 4 3

−2 −2 3 −1 −7


S ={f ∈ P2(R) : f(1) = 2f(−3)

}.


S ={f ∈ P3(R) : f(−2) = 0 y f ′′(−2) = 0

}.






{X ∈M2(R) : AX = XA



A =

[−1 −24 3

].


{X ∈ M2(R) : tr(XA) = 0



A =

[2 −3

−1 1

].


b1 =

214

, b2 =

112

, b3 =

011


uE =

102

, wB =

0

1

−4



A =

−5 1 2 4 2

−3 1 0 3 2

−1 1 −2 2 2

1 1 −4 1 2

.



A =

0 0 0

2 5 0

0 1 3

0 0 −4

, B =

−2 −4−2 −41 2

, C =

[3 0 3 2

1 −2 −4 1

].


A B C



−5 3 −1 −5−3 1 0 −12 −2 1 4

−1 −1 1 3

.


A1 =

[−4 −4−4 0

], A2 =

[7 −32 5

], A3 =

[4 2

3 1

],

A4 =

[3 1

2 1

], A5 =

[1 3

2 −1

].



f1 = 1+ x+ x2 + x3, f2 = −2+ 2x+ 4x2 − 4x3, f3 = 1− 3x− 2x

3,

f4 = 1+ 2x+ x2 + 2x3, f5 = −2− x− 4x2 + x3.


a1 =

−56

1

4

, a2 =

1

−2−10

, a3 =

−45

1

3

, b1 =

−34

1

2

, b2 =

−24

2

1







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04



−3 0 −1 2 1

−1 1 −4 3 3

−2 1 −3 2 3


{x ∈ R5 : Ax = 04


la comprobacion. −1 3 1 3 −3−3 1 1 −1 1

−2 2 1 1 −1−1 −1 0 −2 2


S ={f ∈ P2(R) : f ′(2) − 2f(2) = 0

}.


S ={f ∈ P3(R) : f(−2) = 0 y f ′(2) = 0

}.






{X ∈M2(R) : AX = XA



A =

[3 −32 −4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[1 2

−3 −2

].


b1 =

223

, b2 =

−10

1

, b3 =

112


uE =

311

, wB =

1

0

−2



A =

2 3 3 3 1

1 −4 4 −1 1

−2 4 −4 −2 −11 3 3 0 1

.



A =

1 −5 2

0 0 0

0 5 −20 0 −4

, B =

1 −2−3 6

3 −6

, C =

[−5 3

−2 3

].


A B C



−3 −2 3 4

−1 −4 1 2

−1 1 1 1

2 −2 −2 −2

.


A1 =

[1 3

1 2

], A2 =

[2 4

1 3

], A3 =

[−4 −21 −3

],

A4 =

[4 4

0 4

], A5 =

[1 −5

−3 −2

].



f1 = −4x+ 5x2 + x3, f2 = 4+ x+ 5x2 + 2x3, f3 = −1+ x− 3x2 − x3,

f4 = −1− x+ 4x2 + 4x3, f5 = 2+ x+ 2x2 + x3.


a1 =

−51

2

5

, a2 =

3

5

3

4

, a3 =

1

3

2

3

, b1 =

−3−10

1

, b2 =

1

1

1

1







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04



−1 1 0 −2 3

−4 1 −1 1 4

−4 1 1 4 1


{x ∈ R5 : Ax = 04



2 −2 −2 2 1

−2 4 −2 1 1

4 −4 −4 4 2


S ={f ∈ P2(R) :

0∫−1

f(x)dx = 0}.


S ={f ∈ P3(R) : f(2) = 0 y f ′(−1) = 0

}.






{X ∈M2(R) : AX = XA



A =

[1 3

−1 2

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−3 2

−2 3

].


b1 =

122

, b2 =

212

, b3 =

201


uE =

3

−11

, wB =

0

2

−1



A =

5 0 −1 −1 3

2 −2 −3 3 1

3 2 2 −4 2

4 −4 −6 6 2

.



A =

2 0 −3

−2 −3 0

0 −4 0

0 0 0

, B =

0 3 −55 0 −15 0 0

, C =

−3 −6−2 −41 2


A B C



−1 −2 −4 −21 −4 1 2

1 0 3 2

1 −2 2 2

.


A1 =

[4 −2

−2 0

], A2 =

[−2 −11 1

], A3 =

[6 5

−3 −4

],

A4 =

[−6 −13 2

], A5 =

[6 −5

−3 1

].



f1 = 3+ x+ x2 + 2x3, f2 = −2+ 4x− 3x2 + 2x3, f3 = 2− 2x+ 2x

2 + 3x3,

f4 = 2− 2x+ 2x2 − 4x3, f5 = −2x+ x2 + x3.


a1 =

4

5

1

1

, a2 =

5

6

0

2

, a3 =

4

4

−44

, b1 =

3

4

2

0

, b2 =

3

4

1

1







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. −1 −1 1 1 1

3 1 3 1 1

2 0 −4 2 1

4 1 −3 2 1


{x ∈ R5 : Ax = 04



1 3 1 4 5

3 3 1 4 3

−4 −3 −1 −4 −2


S ={f ∈ P2(R) :

2∫0

f(x)dx = 0}.


S ={f ∈ P3(R) : f(3) = 0 y f ′′(2) = 0

}.






{X ∈M2(R) : AX = XA



A =

[−3 −24 2

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−2 −34 3

].


b1 =

121

, b2 =

140

, b3 =

131


uE =

−12

−3

, wB =

−12

−1



A =

0 1 1 −2 2

4 3 2 1 −12 1 1 −1 3

4 2 1 3 −3

.



A =

2 3 2 2

0 5 5 −30 5 0 5

, B =

[2 −24 2

], C =

3 5

−2 −31 2


A B C



4 3 5 −13 0 4 2

−2 3 −3 −5−1 −3 −1 3

.


A1 =

[1 1

2 2

], A2 =

[−2 −1−4 −3

], A3 =

[−2 4

−4 2

],

A4 =

[1 −32 −2

], A5 =

[1 2

2 3

].



f1 = −1+ 3x− x2 − 4x3, f2 = 4+ 4x+ 3x2 − x3, f3 = 1+ 2x+ x

2 − x3,

f4 = −2− x− 2x2 − x3, f5 = 1+ x+ x2.


a1 =

2

3

0

−2

, a2 =

−6−44

2

, a3 =

5

5

−2−3

, b1 =

1

4

2

−3

, b2 =

2

4

1

−3







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04



2 1 0 3 −1−1 1 2 2 −33 1 1 3 −4


{x ∈ R5 : Ax = 04



−2 1 −2 −4 1

0 −5 −2 −4 −33 −4 2 4 −3


S ={f ∈ P2(R) : f(−2) = 2f(−1)

}.


S ={f ∈ P3(R) : f(2) = 0 y f ′′(−2) = 0

}.






{X ∈M2(R) : AX = XA



A =

[3 −4

−1 4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[3 −42 1

].


b1 =

1

0

−1

, b2 =

111

, b3 =

112


uE =

322

, wB =

−32

−2



A =

−1 2 −5 −3 2

−4 4 −6 −4 −4−1 0 2 1 −42 −2 3 2 2

.



A =

−6 −3−4 −22 1

, B =

2 2 3

−5 0 −1−2 0 0

0 0 0

, C =

[2 1

−1 −5

].


A B C



3 −4 1 2

−4 2 2 −31 2 −3 1

−5 0 5 −4

.


A1 =

[−2 −1−3 −6

], A2 =

[4 2

−2 −4

], A3 =

[6 3

2 4

],

A4 =

[2 1

1 2

], A5 =

[−2 −12 4

].



f1 = −2+ x+ x2 + x3, f2 = 4+ x− 3x3, f3 = −3− x− x2 + 4x3,

f4 = 3− 2x− x2 − 3x3, f5 = −3+ x+ x2 + 2x3.


a1 =

−1−35

3

, a2 =

−2−44

2

, a3 =

−2−57

4

, b1 =

−1−22

1

, b2 =

2

2

0

1







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04



1 −2 2 −1 −31 2 0 2 2

1 −2 1 1 −4


{x ∈ R5 : Ax = 04



−4 −8 −2 4 8

2 4 1 −2 −46 4 2 −2 −8


S ={f ∈ P2(R) : f ′(3) + 2f(3) = 0

}.


S ={f ∈ P3(R) : f(−1) = 0 y f ′′(−2) = 0

}.






{X ∈M2(R) : AX = XA



A =

[1 2

−3 4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−3 3

4 1

].


b1 =

111

, b2 =

110

, b3 =

233


uE =

013

, wB =

1

−20



A =

3 1 3 4 0

3 −4 −4 3 −12 4 −1 2 −11 −3 4 2 1

.



A =

3 −3 5 5

−5 5 0 −35 0 0 1

, B =

−1 0

4 1

3 1

, C =

[6 2 6 2

3 1 3 1

].


A B C



2 −4 4 6

1 6 3 −11 −2 2 3

2 4 5 2

.


A1 =

[3 −4

−6 1

], A2 =

[−1 2

2 0

], A3 =

[1 4

−2 3

],

A4 =

[2 6

−4 5

], A5 =

[1 2

−2 2

].



f1 = 1+ 2x+ x2 + x3, f2 = −3+ 2x2 − 2x3, f3 = 1+ x+ 3x

2 − 2x3,

f4 = 2+ 2x+ x2 + x3, f5 = −1+ 3x+ x2 + 2x3.


a1 =

7

5

−13

, a2 =

4

2

−22

, a3 =

5

4

0

2

, b1 =

1

2

2

0

, b2 =

4

3

−12







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. 3 1 1 3 −13 1 1 0 −21 1 2 −1 2

1 1 2 2 3


{x ∈ R5 : Ax = 04



−4 −1 −5 3 4

−4 −1 −2 0 1

4 1 4 −2 −3


S ={f ∈ P2(R) : f(−1) = 2f(3)

}.


S ={f ∈ P3(R) : f(2) = 0 y f ′(−2) = 0

}.






{X ∈M2(R) : AX = XA



A =

[−3 2

3 −2

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−1 1

2 −3

].


b1 =

111

, b2 =

110

, b3 =

121


uE =

031

, wB =

1

−20



A =

5 −4 1 4 −11 −2 4 1 −14 −2 −3 3 0

6 −6 5 5 −2

.



A =

−5 0 4

0 −2 −30 −1 0

, B =

1 1

2 1

−2 −2

, C =

[2 2 −2 −6

−1 −1 1 3

].


A B C



2 4 3 4

1 0 1 2

1 4 2 2

3 4 4 6

.


A1 =

[2 3

1 1

], A2 =

[3 4

2 1

], A3 =

[0 −22 −2

],

A4 =

[2 5

−1 3

], A5 =

[4 2

6 −2

].



f1 = 1− x− x2, f2 = −1+ 2x+ 4x2 + x3, f3 = 1+ 3x+ 3x

2 − 4x3,

f4 = 2+ 2x+ 3x2 − 3x3, f5 = 3+ 2x+ 3x

2 − 4x3.


a1 =

1

2

−31

, a2 =

2

3

−53

, a3 =

3

3

−66

, b1 =

0

1

−1−1

, b2 =

1

1

−21







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. −2 −2 −1 1 0

3 2 2 2 4

1 2 1 2 4

−1 2 −1 −4 −4


{x ∈ R5 : Ax = 04


la comprobacion. −2 1 1 −2 −2−2 4 5 −4 1

0 3 4 −2 3

4 −2 −2 4 4


S ={f ∈ P2(R) : f ′(−3) − 3f(−3) = 0

}.


S ={f ∈ P3(R) : f(−3) = 0 y f ′(2) = 0

}.






{X ∈M2(R) : AX = XA



A =

[3 −22 −4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−1 2

−3 1

].


b1 =

2

−11

, b2 =

113

, b3 =

012


uE =

−31

−2

, wB =

−11

−3



A =

−4 1 2 2 1

2 2 3 3 1

1 2 4 3 1

3 −1 −1 −2 −1

.



A =

0 0 0

−5 3 −1−4 0 4

0 0 4

, B =

[5 −2 3 4

2 4 1 1

], C =

−4 1 4

0 2 −10 0 5


A B C



1 0 2 4

2 1 3 1

−3 −2 −4 2

−1 −1 −1 3

.


A1 =

[2 −11 −2

], A2 =

[1 2

3 4

], A3 =

[5 −23 −4

],

A4 =

[2 −11 −2

], A5 =

[1 −3

−2 −6

].



f1 = 3+ 4x− 2x2 + 2x3, f2 = 3+ x+ 4x

2 + x3, f3 = −3− 4x+ 2x2 + 3x3,

f4 = 3+ x+ 4x2, f5 = 2+ x+ 2x

2 − x3.


a1 =

2

3

3

3

, a2 =

0

3

2

4

, a3 =

2

−3−1−5

, b1 =

4

3

4

2

, b2 =

1

2

2

2







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04



2 1 4 −2 1

3 2 4 −4 2

2 2 4 −1 1


{x ∈ R5 : Ax = 04



−2 −5 −4 −4 1

1 7 −4 5 7

1 1 4 1 −3


S ={f ∈ P2(R) : f ′(1) − 5f(1) = 0

}.


S ={f ∈ P3(R) : f(2) = 0 y f ′(2) = 0

}.






{X ∈M2(R) : AX = XA



A =

[1 −4

−3 4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−2 3

2 −1

].


b1 =

124

, b2 =

112

, b3 =

111


uE =

221

, wB =

−12

0



A =

1 3 2 2 1

2 0 −6 −5 1

3 3 −4 −3 2

−2 −6 −4 −4 −2

.



A =

0 −21 1

1 2

, B =

[−3 5

−1 3

], C =

3 −4 3 3

−3 0 −2 0

0 0 2 0


A B C



2 −3 5 4

2 −1 3 2

2 −2 4 3

0 −1 1 1

.


A1 =

[4 −2

−1 1

], A2 =

[6 2

6 4

], A3 =

[1 1

2 1

],

A4 =

[4 0

2 2

], A5 =

[1 3

5 2

].



f1 = 3+ 4x+ 4x3, f2 = −4+ 5x+ 4x2 + x3, f3 = 3+ 5x+ x

2 + 4x3,

f4 = −2− 5x− 4x2 − x3, f5 = 1+ 2x+ x2 + x3.


a1 =

5

3

6

−1

, a2 =

3

2

4

0

, a3 =

−3−1−23

, b1 =

1

1

2

1

, b2 =

−11

2

4







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. −1 3 −2 1 −32 −2 2 1 3

1 2 1 3 3

−1 0 1 −1 3


{x ∈ R5 : Ax = 04



3 −1 1 1 4

−1 3 −3 −1 −22 2 −2 0 2


S ={f ∈ P2(R) :

2∫−1

f(x)dx = 0}.


S ={f ∈ P3(R) : f(1) = 0 y f ′′(−2) = 0

}.






{X ∈M2(R) : AX = XA



A =

[4 1

−3 −1

].


{X ∈ M2(R) : tr(XA) = 0



A =

[1 4

−2 −3

].


b1 =

2

1

−1

, b2 =

012

, b3 =

111


uE =

−2−12

, wB =

1

2

−1



A =

1 1 4 4 1

1 0 3 −3 2

2 1 −1 −2 4

1 1 −4 1 2

.



A =

−5 −3 −1 −31 2 0 1

0 2 0 −4

, B =

[−2 2 2 −4−1 1 1 −2

], C =

4 −4 2

0 −3 2

0 3 0


A B C



5 3 3 −24 2 −2 −22 1 −1 −1

−1 −1 −5 0

.


A1 =

[−2 1

2 3

], A2 =

[2 −1

−3 −4

], A3 =

[−2 1

1 2

],

A4 =

[4 −20 −2

], A5 =

[6 −3

−2 −5

].



f1 = 1+ x− x2 − x3, f2 = 1+ 2x+ 3x

2 + 2x3, f3 = 4+ x− x2 + 2x3,

f4 = 2+ x− x2, f5 = 2+ 2x+ x

2 + x3.


a1 =

1

4

1

3

, a2 =

1

7

2

4

, a3 =

−45

3

−5

, b1 =

1

1

0

2

, b2 =

2

4

1

4







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. −4 0 1 2 −4−4 2 5 7 2

4 1 2 3 2

−4 1 2 2 4


{x ∈ R5 : Ax = 04


la comprobacion. −3 −3 2 1 −2−2 −1 3 1 2

−5 −4 5 2 0

4 2 −6 −2 −4


S ={f ∈ P2(R) : f(2) = 2f(−3)

}.


S ={f ∈ P3(R) : f(2) = 0 y f ′′(−1) = 0

}.






{X ∈M2(R) : AX = XA



A =

[−4 −31 2

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−2 4

3 2

].


b1 =

112

, b2 =

012

, b3 =

−21

3


uE =

2

0

−1

, wB =

−13

−1



A =

0 3 6 2 −56 −7 2 −4 3

−3 2 −4 1 1

−3 5 2 3 −4

.



A =

−1 0 5 −40 −3 −5 −40 −5 −5 0

, B =

[−1 −2 −3 1

−2 −4 −6 2

], C =

1 2

1 1

0 1


A B C



−6 1 −1 −14 2 6 4

−2 3 5 3

2 1 3 2

.


A1 =

[2 1

2 1

], A2 =

[2 2

3 1

], A3 =

[6 0

3 3

],

A4 =

[−4 −4−6 −2

], A5 =

[−4 4

2 −2

].



f1 = −4− 4x− 4x2 + 3x3, f2 = 1+ x+ x2 + 3x3, f3 = x− x

2 − 3x3,

f4 = 1+ x+ x2 + 4x3, f5 = 2+ x+ 3x

2 + 3x3.


a1 =

6

3

−36

, a2 =

4

3

−23

, a3 =

6

5

−34

, b1 =

2

2

−11

, b2 =

1

0

−12







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04



1 2 −4 −3 −11 1 0 2 3

1 1 −3 −2 2


{x ∈ R5 : Ax = 04



2 −3 5 3 2

−2 4 2 1 2

4 −7 3 2 0


S ={f ∈ P2(R) : f ′(−3) + 3f(−3) = 0

}.


S ={f ∈ P3(R) : f(3) = 0 y f ′(2) = 0

}.






{X ∈M2(R) : AX = XA



A =

[3 4

−4 −3

].


{X ∈ M2(R) : tr(XA) = 0



A =

[4 −1

−2 3

].


b1 =

101

, b2 =

223

, b3 =

213


uE =

132

, wB =

2

−32



A =

2 2 2 1 1

1 2 3 1 −3−1 1 −2 1 3

−2 1 −1 1 −1

.



A =

−2 0

2 1

3 1

, B =

2 −2 1 0

0 −3 2 −30 2 0 3

, C =

[1 2 −1 2

2 4 −2 4

].


A B C



0 2 2 2

−1 4 3 3

1 −2 −1 −1−1 3 2 2

.


A1 =

[2 −5

−4 3

], A2 =

[2 −5

−4 3

], A3 =

[2 −6

−4 4

],

A4 =

[3 1

−6 −4

], A5 =

[1 −2

−2 1

].



f1 = 2+ x− x2 + x3, f2 = 3+ x+ x

2 + 2x3, f3 = 3+ x+ 2x3,

f4 = 1− 2x− 3x2 + 3x3, f5 = 1+ 2x− x

2 − x3.


a1 =

3

4

1

0

, a2 =

4

5

1

1

, a3 =

2

1

−15

, b1 =

2

3

1

−1

, b2 =

2

3

2

−3







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. 2 1 −4 1 −41 1 −2 1 2

4 1 −1 2 −4−4 −2 1 −3 −4


{x ∈ R5 : Ax = 04



2 −2 −2 6 −46 −2 −4 −6 −4

−1 1 1 −3 2


S ={f ∈ P2(R) : f(1) = 2f(2)

}.


S ={f ∈ P3(R) : f(2) = 0 y f ′′(1) = 0

}.






{X ∈M2(R) : AX = XA



A =

[−2 −31 2

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−4 4

1 −2

].


b1 =

121

, b2 =

021

, b3 =

−13

2


uE =

410

, wB =

1

2

−2



A =

−2 −2 −5 −3 −21 1 −4 1 −1

−4 −4 3 −5 0

3 3 1 4 1

.



A =

[−5 3

3 −1

], B =

[−2 1 3 1

4 4 2 3

], C =

−5 −3 −10 0 0

0 −3 −30 1 0


A B C



4 −1 1 −43 −1 1 −41 −1 1 −42 −1 1 −4

.


A1 =

[1 −7

−4 −3

], A2 =

[2 4

1 3

], A3 =

[−2 4

3 1

],

A4 =

[6 −2

−4 2

], A5 =

[3 5

1 4

].



f1 = 3+ 4x+ x2 + 2x3, f2 = 3+ 5x+ 2x

2 + 2x3, f3 = 1+ x+ x3,

f4 = 2+ 4x+ 2x2 − 4x3, f5 = −3− x+ 2x2 − 3x3.


a1 =

0

4

1

−4

, a2 =

−52

3

3

, a3 =

−32

2

1

, b1 =

−12

1

−1

, b2 =

−21

1

1







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. 2 2 1 1 −24 2 1 2 −12 1 1 1 0

4 1 1 2 1


{x ∈ R5 : Ax = 04



−5 −5 −6 −1 −23 2 2 3 1

1 −1 −2 5 0


S ={f ∈ P2(R) : f ′(1) + 3f(1) = 0

}.


S ={f ∈ P3(R) : f(−1) = 0 y f ′′(1) = 0

}.






{X ∈M2(R) : AX = XA



A =

[−4 −21 3

].


{X ∈ M2(R) : tr(XA) = 0



A =

[3 2

−1 −3

].


b1 =

323

, b2 =

−1−11

, b3 =

322


uE =

213

, wB =

−21

2



A =

−1 3 4 −3 3

1 −1 −2 1 −3−1 2 3 −2 3

4 3 4 −4 4

.



A =

1 −2−2 4

1 −2

, B =

0 0 0

3 −4 4

0 −1 1

0 4 0

, C =

[−2 3

3 5

].


A B C



−2 −3 −2 0

1 1 −1 −3−1 −2 −3 −33 4 1 −3

.


A1 =

[2 −35 −7

], A2 =

[−3 −74 −1

], A3 =

[3 6

−3 0

],

A4 =

[1 −23 −4

], A5 =

[−1 1

−2 3

].



f1 = 3+ 2x+ 2x2 − x3, f2 = −3− 3x+ 2x2 − 2x3, f3 = 1+ x+ x

2 − x3,

f4 = 3+ 3x− x2 + x3, f5 = 2x+ x

2 − 3x3.


a1 =

−2−57

4

, a2 =

−42

2

0

, a3 =

−1−45

3

, b1 =

−1−12

1

, b2 =

2

−20

1







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04



2 1 2 1 −11 −2 −3 −2 −1

−3 1 1 1 2


{x ∈ R5 : Ax = 04


la comprobacion. −3 −2 1 3 −24 1 −1 −2 3

2 −2 0 2 2

−2 −3 1 4 −1


S ={f ∈ P2(R) : f(−1) = 2f(−3)

}.


S ={f ∈ P3(R) : f(−2) = 0 y f ′′(−3) = 0

}.






{X ∈M2(R) : AX = XA



A =

[3 4

2 −4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[1 3

2 −2

].


b1 =

111

, b2 =

110

, b3 =

121


uE =

141

, wB =

2

−2−1



A =

−4 5 2 2 −31 2 −3 1 −25 −3 −5 −1 1

2 4 −6 2 −4

.



A =

5 5 −5 −1−1 −2 2 0

0 1 4 0

, B =

−2 −1 5

0 2 −20 0 4

, C =

[−3 1 1 1

6 −2 −2 −2

].


A B C



1 2 4 −33 7 4 −2

−2 −5 0 −11 3 −4 4

.


A1 =

[4 5

1 6

], A2 =

[3 4

1 5

], A3 =

[1 −3

−4 −7

],

A4 =

[4 2

−2 0

], A5 =

[4 3

−1 2

].



f1 = 2− x+ 3x2 + x3, f2 = x− x

2 + x3, f3 = 1− 4x+ 4x2 − 4x3,

f4 = 3+ 2x− x2 + 3x3, f5 = 1+ x− x

2 + x3.


a1 =

−26

2

4

, a2 =

2

6

−55

, a3 =

1

5

−34

, b1 =

1

1

−21

, b2 =

1

4

−33







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04



3 −4 −4 1 0

1 2 3 1 3

3 −2 −2 2 4


{x ∈ R5 : Ax = 04



−4 −4 −2 −6 −22 2 1 3 1

3 2 3 5 2


S ={f ∈ P2(R) : f ′(−2) + 5f(−2) = 0

}.


S ={f ∈ P3(R) : f(−3) = 0 y f ′(−3) = 0

}.






{X ∈M2(R) : AX = XA



A =

[1 −3

−2 4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[1 4

−3 −2

].


b1 =

111

, b2 =

−12

1

, b3 =

−22

1


uE =

211

, wB =

−32

−1



A =

−1 −1 3 2 1

−4 −1 −3 −1 −13 −3 2 2 1

0 3 4 1 1

.



A =

−2 2 −34 0 2

0 0 −2

, B =

−2 −24 3

1 1

, C =

2 −3 0 1

−1 −3 1 0

−2 0 2 0


A B C



4 2 −3 5

1 1 0 2

3 1 −3 3

5 3 −3 7

.


A1 =

[0 3

3 −6

], A2 =

[−4 1

5 −2

], A3 =

[1 2

1 −4

],

A4 =

[1 3

2 −6

], A5 =

[3 −3

−6 6

].



f1 = 1+ 2x+ 2x2 + 4x3, f2 = −2+ 3x+ x2 + 4x3, f3 = −2− 3x+ 2x2 − x3,

f4 = −1+ 2x+ x2 + 3x3, f5 = 4+ 2x2 + 2x3.


a1 =

−4−12

−5

, a2 =

5

−1−47

, a3 =

2

−1−23

, b1 =

1

1

0

1

, b2 =

1

−1−12







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04



3 −4 4 2 −12 −4 2 1 1

1 −3 1 0 3


{x ∈ R5 : Ax = 04



−2 −2 4 6 −2−1 1 2 3 2

1 3 −2 −3 4


S ={f ∈ P2(R) : f(3) = 2f(−1)

}.


S ={f ∈ P3(R) : f(−3) = 0 y f ′′(−2) = 0

}.






{X ∈M2(R) : AX = XA



A =

[−1 4

−4 3

].


{X ∈ M2(R) : tr(XA) = 0



A =

[1 4

−1 3

].


b1 =

142

, b2 =

131

, b3 =

032


uE =

−3−12

, wB =

1

−31



A =

4 3 5 −2 6

3 1 1 0 1

−1 −2 −4 2 −5−2 1 3 −2 4

.



A =

−1 0

2 1

1 1

, B =

[3 1 −1 2

−6 −2 2 −4

], C =

3 −2 2

−5 0 −52 0 0


A B C



−2 0 1 −1−4 2 1 −35 −3 −1 4

−3 1 1 −2

.


A1 =

[4 2

6 4

], A2 =

[2 1

3 2

], A3 =

[−6 −31 −1

],

A4 =

[4 2

−2 0

], A5 =

[2 1

1 1

].



f1 = −1+ 4x+ 4x2 + 3x3, f2 = 2+ 4x+ x2 + 3x3, f3 = 1+ 3x+ x

2 + 2x3,

f4 = 2+ 3x+ x2 + 3x3, f5 = 1+ 2x

2 + 3x3.


a1 =

0

2

3

−4

, a2 =

1

3

4

−6

, a3 =

−51

4

−2

, b1 =

−11

2

−2

, b2 =

−21

2

−1







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04


la comprobacion. −1 4 −1 3 2

4 −4 −2 −4 −21 2 −2 1 1

2 1 −1 1 1


{x ∈ R5 : Ax = 04



−4 2 3 3 −34 0 1 1 −1

−4 1 1 1 −1


S ={f ∈ P2(R) : f ′(3) + 3f(3) = 0

}.


S ={f ∈ P3(R) : f(1) = 0 y f ′(−3) = 0

}.






{X ∈M2(R) : AX = XA



A =

[−2 2

1 −4

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−1 2

−3 1

].


b1 =

111

, b2 =

433

, b3 =

−11

0


uE =

021

, wB =

−31

−1



A =

−4 1 −3 1 1

−4 1 −1 1 2

−4 1 1 1 3

−3 3 4 4 3

.



A =

1 1 3

0 −3 −30 5 0

, B =

2 4

2 4

1 2

, C =

0 −5 −30 0 0

−3 −2 0

0 −3 0


A B C



−4 2 1 3

6 −5 −1 −24 −6 0 2

−2 −1 1 4

.


A1 =

[1 2

−3 1

], A2 =

[1 2

−3 1

], A3 =

[4 −2

−2 −6

],

A4 =

[2 5

−7 3

], A5 =

[−3 −36 0

].



f1 = 2+ 3x+ 2x2 + x3, f2 = 4− 4x+ 2x

2 − 4x3, f3 = −4− 2x− 4x2 + 2x3,

f4 = 3+ 4x+ 3x2 + x3, f5 = 1+ 3x+ 2x

2.


a1 =

2

−17

−4

, a2 =

1

−25

−5

, a3 =

1

−14

−3

, b1 =

0

−11

−2

, b2 =

1

1

3

1







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04



1 −4 0 1 −32 4 1 3 2

2 1 1 2 −4


{x ∈ R5 : Ax = 04



−6 2 −6 4 −6−3 1 −3 2 −3−6 3 −1 5 −2


S ={f ∈ P2(R) : f ′(2) + 4f(2) = 0

}.


S ={f ∈ P3(R) : f(−1) = 0 y f ′′(3) = 0

}.






{X ∈M2(R) : AX = XA



A =

[4 3

−1 −3

].


{X ∈ M2(R) : tr(XA) = 0



A =

[1 −2

−1 4

].


b1 =

111

, b2 =

121

, b3 =

1

−12


uE =

0

−31

, wB =

2

−1−2



A =

−1 5 −2 0 4

−3 4 2 −2 3

−1 −6 6 −2 −52 1 −4 2 1

.



A =

4 −4 −1 4

2 5 −5 0

4 0 2 0

, B =

−3 −4 5

5 4 0

−4 0 0

, C =

[−2 6 −2 −21 −3 1 1

].


A B C



−1 −1 −2 1

1 2 1 −20 1 −1 −12 3 3 −3

.


A1 =

[2 −21 1

], A2 =

[2 −31 2

], A3 =

[2 3

1 −4

],

A4 =

[4 −42 2

], A5 =

[4 −52 3

].



f1 = 3− x− 4x2 − 3x3, f2 = 3+ x+ 2x

2 + x3, f3 = 3+ 2x+ 2x2,

f4 = 3− 3x− x2 + 2x3, f5 = 4+ 2x+ 3x

2 + x3.


a1 =

4

3

−1−6

, a2 =

3

2

0

−4

, a3 =

−7−3−56

, b1 =

1

1

−1−2

, b2 =

2

1

0

−2







Engra

peaq

uı

No

dobl

e


Bases y dimension.




{x ∈ R5 : Ax = 04



2 0 2 1 −31 1 1 2 1

3 1 3 3 −2


{x ∈ R5 : Ax = 04



−2 1 −3 −1 1

−1 −3 5 4 −4−5 −1 −1 2 −2


S ={f ∈ P2(R) : f ′(−3) − 5f(−3) = 0

}.


S ={f ∈ P3(R) : f(3) = 0 y f ′(−3) = 0

}.






{X ∈M2(R) : AX = XA



A =

[3 −4

−2 2

].


{X ∈ M2(R) : tr(XA) = 0



A =

[−3 3

4 −1

].


b1 =

241

, b2 =

131

, b3 =

121


uE =

231

, wB =

2

−31



A =

3 4 2 1 3

1 1 1 2 −41 1 1 1 −2

−2 −3 −1 −1 −3

.



A =

4 1 −4−2 0 3

0 0 5

, B =

[6 2 −2 −63 1 −1 −3

], C =

2 5

1 2

0 1


A B C



3 −4 5 1

−1 4 −2 1

1 4 1 3

2 0 3 2

.


A1 =

[−2 3

1 1

], A2 =

[−3 5

1 2

], A3 =

[−4 5

3 1

],

A4 =

[−1 5

−3 4

], A5 =

[−3 4

2 1

].



f1 = 2− x+ 4x2 + x3, f2 = −4− x− 3x2 − 4x3, f3 = 1+ 4x+ 2x

2 − 4x3,

f4 = 1− x+ 2x2 + x3, f5 = 1− 3x+ 3x

2 + 2x3.


a1 =

1

3

4

2

, a2 =

−21

1

1

, a3 =

4

5

7

3

, b1 =

−21

1

1

, b2 =

4

1

2

0







ejercicio 1. 1%. 0 3 75esfm.egormaximenko.com/linalg/task_bases_es.pdfejercicio 10. 1%. est an dadas...

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