Taller de Algebra Lineal

Universidad de CartagenaFacultad de Ciencias Exactas

TALLER III DE ALGEBRA LINEAL(Aplicacion de la regla de cramer)

Alberto RodrıguezEiver Rodrıguez

15 de marzo de 2015

ALGEBRA LINEAL I 4 Sem. 2015

1. Desarrollo

1. Resolver los siguientes sistemas utilizando la regla de Cramer:

a.

2x1 + x2 + x3 = 63x1 − 2x2 − 3x3 = 58x1 + 2x2 + 5x3 = 11

Respuesta:

Sea A =

2 1 13 −2 −38 2 5

|A| =

∣∣∣∣∣∣2 1 13 −2 −38 2 5

∣∣∣∣∣∣ = 2(−10 + 6)− 1(15 + 24) + 1(6 + 16) = −25

Luego:

|B1| =

∣∣∣∣∣∣6 1 15 −2 −311 2 5

∣∣∣∣∣∣ = 6(−10 + 6)− 1(25 + 33) + 1(10 + 22) = −50

|B2| =

∣∣∣∣∣∣2 6 13 5 −38 11 5

∣∣∣∣∣∣ = 2(25 + 33)− 6(15 + 24) + 1(33− 40) = −125

|B3| =

∣∣∣∣∣∣2 1 63 −2 58 2 11

∣∣∣∣∣∣ = 2(−22− 10)− 1(33− 40) + 6(6 + 16) = 75

Como xi = |Bi||A|

x1 = |B1||A| = −50

−25 = 2

x2 = |B2||A| = −125

−25 = 5

x3 = |B3||A| = 75

−25 = −3

b.

x1 + x2 + x3 + x4 = 6

2x1 − x3 − x4 = 43x3 + 6x4 = 3x1 − x4 = 5

Respuesta:

Sea A =

1 1 1 12 0 −1 −10 0 3 61 0 0 −1

2


|A| =

∣∣∣∣∣∣∣∣1 1 1 12 0 −1 −10 0 3 61 0 0 −1

∣∣∣∣∣∣∣∣ = (−1)

∣∣∣∣∣∣2 −1 −10 3 61 0 −1

∣∣∣∣∣∣+ (0)

∣∣∣∣∣∣1 1 10 3 61 0 −1

∣∣∣∣∣∣− (0)

∣∣∣∣∣∣1 1 12 −1 −11 0 −1

∣∣∣∣∣∣+ (0)

∣∣∣∣∣∣1 1 12 −1 −10 3 6

∣∣∣∣∣∣ =

(−1)(2(−3) + 1(−6)− 1(−3)) = 9

Luego:

|B1| =

∣∣∣∣∣∣∣∣6 1 1 14 0 −1 −13 0 3 65 0 0 −1

∣∣∣∣∣∣∣∣ = (−1)

∣∣∣∣∣∣4 −1 −13 3 65 0 −1

∣∣∣∣∣∣+ (0)

∣∣∣∣∣∣6 1 13 3 65 0 −1

∣∣∣∣∣∣− (0)

∣∣∣∣∣∣6 1 14 −1 −15 0 −1

∣∣∣∣∣∣+ (0)

∣∣∣∣∣∣6 1 14 −1 −13 3 6

∣∣∣∣∣∣ =

(−1)(4(−3) + 1(−3− 30)− 1(−15)) = 30

|B2| =

∣∣∣∣∣∣∣∣1 6 1 12 4 −1 −10 3 3 61 5 0 −1

∣∣∣∣∣∣∣∣ = (1)

∣∣∣∣∣∣4 −1 −13 3 65 0 −1

∣∣∣∣∣∣ − (6)

∣∣∣∣∣∣2 −1 −10 3 61 0 −1

∣∣∣∣∣∣ + (1)

∣∣∣∣∣∣2 4 −10 3 61 5 −1

∣∣∣∣∣∣ − (1)

∣∣∣∣∣∣2 4 −10 3 31 5 0

∣∣∣∣∣∣ =

1[4(−3)+1(−3−30)−1(−15)]−6[2(−3)+1(−6)−1(−3)]+1[2(−3−30)−4(−6)−1(−3)]−1[2(−15)−4(−3)− 1(−3)] = 0

|B3| =

∣∣∣∣∣∣∣∣1 1 6 12 0 4 −10 0 3 61 0 5 −1

∣∣∣∣∣∣∣∣ = (−1)

∣∣∣∣∣∣2 4 −10 3 61 5 −1

∣∣∣∣∣∣ = −1[2(−3− 30)− 4(−6)− 1(−3)] = 39

|B4| =

∣∣∣∣∣∣∣∣1 1 1 62 0 −1 40 0 3 31 0 0 5

∣∣∣∣∣∣∣∣ = (−1)

∣∣∣∣∣∣2 −1 40 3 31 0 5

∣∣∣∣∣∣ = −1[2(15) + 1(−3) + 4(−3)] = −15

Como xi = |Bi||A|

x1 = |B1||A| = 30

9 = 103

x2 = |B2||A| = 0

9 = 0

x3 = |B3||A| = 39

9 = 133

x4 = |B4||A| = −15

9 = −53

c.

x1 − x4 = 72x2 + x3 = 2

4x1 − x2 = −33x3 − 5x4 = 2

Respuesta:

3


Sea A =

1 0 0 −10 2 1 04 −1 0 00 0 3 −5

|A| =

∣∣∣∣∣∣∣∣1 0 0 −10 2 1 04 −1 0 00 0 3 −5

∣∣∣∣∣∣∣∣ = (1)

∣∣∣∣∣∣2 1 0−1 0 00 3 −5

∣∣∣∣∣∣− (−1)

∣∣∣∣∣∣0 2 14 −1 00 0 3

∣∣∣∣∣∣ = 1[2(0)− 1(5) + 0(−3)] + 1[0− 2(12) +

1(0)] = −29

Luego:

|B1| =

∣∣∣∣∣∣∣∣7 0 0 −12 2 1 0−3 −1 0 02 0 3 −5

∣∣∣∣∣∣∣∣ = (7)

∣∣∣∣∣∣2 1 0−1 0 00 3 −5

∣∣∣∣∣∣ + (1)

∣∣∣∣∣∣2 2 1−3 −1 02 0 3

∣∣∣∣∣∣ = 7[2(0) − 1(5) + 0] + 1[2(−3) −

2(−9) + 1(2)] = −21

|B2| =

∣∣∣∣∣∣∣∣1 7 0 −10 2 1 04 −3 0 00 2 3 −5

∣∣∣∣∣∣∣∣ = (1)

∣∣∣∣∣∣2 1 0−3 0 02 3 −5

∣∣∣∣∣∣ + (4)

∣∣∣∣∣∣7 0 −12 1 02 3 −5

∣∣∣∣∣∣ = 1[2(0) − 1(15) + 0] + 4[7(−5) − 0 −

1(6− 2)] = −171

|B3| =

∣∣∣∣∣∣∣∣1 0 7 −10 2 2 04 −1 −3 00 0 2 −5

∣∣∣∣∣∣∣∣ = −(−1)

∣∣∣∣∣∣0 2 24 −1 −30 0 2

∣∣∣∣∣∣+ (−5)

∣∣∣∣∣∣1 0 70 2 24 −1 −3

∣∣∣∣∣∣ = 1[0− 2(8) + (0)]− 5[1(−6 +

2)− 0 + 7(−8)] = 284

|B4| =

∣∣∣∣∣∣∣∣1 0 0 70 2 1 24 −1 0 −30 0 3 2

∣∣∣∣∣∣∣∣ = (1)

∣∣∣∣∣∣2 1 2−1 0 −30 3 2

∣∣∣∣∣∣− (7)

∣∣∣∣∣∣0 2 14 −1 00 0 3

∣∣∣∣∣∣ = 1[2(0− (−9))− (−1)(2− 6)]− 7[−4(6−

0)] = 182

Como xi = |Bi||A|

x1 = |B1||A| = −21

−29 = 2129

x2 = |B2||A| = −171

−29 = 17129

x3 = |B3||A| = 284

−29 = − 28429

x4 = |B4||A| = 182

−29 = − 18229

2. Calcular los siguientes determinantes:

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a.

∣∣∣∣∣∣∣∣1 −1 2 34 0 2 5−1 2 3 75 1 0 4

∣∣∣∣∣∣∣∣Respuesta:∣∣∣∣∣∣∣∣

1 −1 2 34 0 2 5−1 2 3 75 1 0 4

∣∣∣∣∣∣∣∣ = −(−1)

∣∣∣∣∣∣4 2 5−1 3 75 0 4

∣∣∣∣∣∣− (2)

∣∣∣∣∣∣1 2 34 2 55 0 4

∣∣∣∣∣∣+ (1)

∣∣∣∣∣∣1 2 34 2 5−1 3 7

∣∣∣∣∣∣= 1[−2(−4− 35) + 3(16− 25)]− 2[−2(16− 25) + 2(4− 15)] + 1[1(14− 15)− 2(28 + 5) + 3(12 + 2)] = 34

b.

∣∣∣∣∣∣1 −1 23 4 2−2 3 4

∣∣∣∣∣∣Respuesta:∣∣∣∣∣∣

1 −1 23 4 2−2 3 4

∣∣∣∣∣∣ = 1(16− 6)− (−1)(12 + 4) + 2(9 + 8) = 60

3. Hallar la inversa de los siguientes matrices utilizando determinantes:

a.

2 1 0 00 −1 3 01 0 0 −23 0 −1 0

Respuesta:

Sea A =

2 1 0 00 −1 3 01 0 0 −23 0 −1 0

Por otro lado det(A) =

∣∣A∣∣ =

∣∣∣∣∣∣∣∣2 1 0 00 −1 3 01 0 0 −23 0 −1 0

∣∣∣∣∣∣∣∣ = −(−2)

∣∣∣∣∣∣2 1 00 −1 33 0 −1

∣∣∣∣∣∣ = 2[2(1)− 1(−9)] = 22

Sea B la matriz de cofactores de A esto es:

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B =

A11 A12 A13 A14

A21 A22 A23 A24

A31 A32 A33 A34

A41 A42 A43 A44

Como:

A11 = (−1)1+1

∣∣∣∣∣∣−1 3 00 0 −20 −1 0

∣∣∣∣∣∣ = −1(−2) = 2

A12 = (−1)1+2

∣∣∣∣∣∣0 3 01 0 −23 −1 0

∣∣∣∣∣∣ = −[−3(0− (−6))] = 18

A13 = (−1)1+3

∣∣∣∣∣∣0 −1 01 0 −23 0 0

∣∣∣∣∣∣ = −(−1)(0− (−6)) = 6

A14 = (−1)1+4

∣∣∣∣∣∣0 −1 31 0 03 0 −1

∣∣∣∣∣∣ = −[−(−1)(−1− 0)] = 1

A21 = (−1)2+1

∣∣∣∣∣∣1 0 00 0 −20 −1 0

∣∣∣∣∣∣ = −[1(0− 2)] = 2

A22 = (−1)2+2

∣∣∣∣∣∣2 0 01 0 −23 −1 0

∣∣∣∣∣∣ = 2(0− 2) = −4

A23 = (−1)2+3

∣∣∣∣∣∣2 1 01 0 −23 0 0

∣∣∣∣∣∣ = −[−1(0− (−6))] = 6

A24 = (−1)2+4

∣∣∣∣∣∣2 1 01 0 03 0 −1

∣∣∣∣∣∣ = −1(−1− 0) = 1

A31 = (−1)3+1

∣∣∣∣∣∣1 0 0−1 3 00 −1 0

∣∣∣∣∣∣ = 1(0− 0) = 0

A32 = (−1)3+2

∣∣∣∣∣∣2 0 00 3 03 −1 0

∣∣∣∣∣∣ = 0

6


A33 = (−1)3+3

∣∣∣∣∣∣2 1 00 −1 03 0 0

∣∣∣∣∣∣ = 0

A34 = (−1)3+4

∣∣∣∣∣∣2 1 00 −1 33 0 −1

∣∣∣∣∣∣ = −[2(1− 0)− 1(0− 9)] = −11

A41 = (−1)4+1

∣∣∣∣∣∣1 0 0−1 3 00 0 −2

∣∣∣∣∣∣ = −[1(−6− 0)] = 6

A42 = (−1)4+2

∣∣∣∣∣∣2 0 00 3 01 0 −2

∣∣∣∣∣∣ = 2(−6− 0) = −12

A43 = (−1)4+3

∣∣∣∣∣∣2 1 00 −1 01 0 −2

∣∣∣∣∣∣ = −[−2(−2− 0)] = −4

A44 = (−1)4+4

∣∣∣∣∣∣2 1 00 −1 31 0 0

∣∣∣∣∣∣ = 1(3− 0) = 3

Ası

B =

2 18 6 12 −4 6 10 0 0 −116 −12 −4 3

Luego Adj = Bt =

2 2 0 618 −4 0 −126 6 0 −41 1 −11 3

Ası, por teorema 5

A−1 = ( 1detA )adjA= ( 1

22 )

2 2 0 618 −4 0 −126 6 0 −41 1 −11 3

=

1/11 1/11 0 3/11

9/11 −2/11 0 −6/11

3/11 3/11 0 −2/11

1/22 1/22 −1/2 3/22

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b.

3 −1 2 41 1 0 3−2 4 1 56 −4 1 2

Respuesta:

Sea A =

3 −1 2 41 1 0 3−2 4 1 56 −4 1 2


∣∣A∣∣=

∣∣∣∣∣∣∣∣3 −1 2 41 1 0 3−2 4 1 56 −4 1 2

∣∣∣∣∣∣∣∣ = (2)

∣∣∣∣∣∣1 1 3−2 4 56 −4 2

∣∣∣∣∣∣+(1)

∣∣∣∣∣∣3 −1 41 1 36 −4 2

∣∣∣∣∣∣−(1)

∣∣∣∣∣∣3 −1 41 1 3−2 4 5

∣∣∣∣∣∣ =

2[1(8 + 20)− 1(−4− 30) + 3(8− 24)] + [3(2 + 12)− (−1)(2− 18) + 4(−4− 6)]− [3(5− 12)− (−1)(5−(−6)) + 4(4− (−2))] = 0

Como det(A) = 0 Por Teorema 5. A no es invertible.

c.

1 0 −1 10 2 2 −34 −1 −1 0−2 1 4 0

Respuesta:

Sea A =

1 0 −1 10 2 2 −34 −1 −1 0−2 1 4 0


∣∣A∣∣ =

∣∣∣∣∣∣∣∣1 0 −1 10 2 2 −34 −1 −1 0−2 1 4 0

∣∣∣∣∣∣∣∣ = (−1)

∣∣∣∣∣∣0 2 24 −1 −1−2 1 4

∣∣∣∣∣∣ + (−3)

∣∣∣∣∣∣1 0 −14 −1 −1−2 1 4

∣∣∣∣∣∣ =

−[0− 2(16− 2) + 2(4− 2)]− 3[1(−4 + 1)− 1(4− 2)] = 39

Sea B la matriz de cofactores de A esto es:

B =

A11 A12 A13 A14

A21 A22 A23 A24

A31 A32 A33 A34

A41 A42 A43 A44

Como:

A11 = (−1)1+1

∣∣∣∣∣∣2 2 −3−1 −1 01 4 0

∣∣∣∣∣∣ = −3(−4− (−1)) = 9

A12 = (−1)1+2

∣∣∣∣∣∣0 2 −34 −1 0−2 4 0

∣∣∣∣∣∣ = −[−3(16− 2)] = 42

8


A13 = (−1)1+3

∣∣∣∣∣∣0 2 −34 −1 0−2 1 0

∣∣∣∣∣∣ = −3(4− 2) = −6

A14 = (−1)1+4

∣∣∣∣∣∣0 2 24 −1 −1−2 1 4

∣∣∣∣∣∣ = −[−2(16− 2) + 2(4− 2)] = 24

A21 = (−1)2+1

∣∣∣∣∣∣0 −1 1−1 −1 01 4 0

∣∣∣∣∣∣ = −[1(−4 + 1)] = 3

A22 = (−1)2+2

∣∣∣∣∣∣1 −1 14 −1 0−2 4 0

∣∣∣∣∣∣ = 1(16− 2) = 14

A23 = (−1)2+3

∣∣∣∣∣∣1 0 14 −1 0−2 1 0

∣∣∣∣∣∣ = −[1(4− 2)] = −2

A24 = (−1)2+4

∣∣∣∣∣∣1 0 −14 −1 −1−2 1 4

∣∣∣∣∣∣ = 1(−4 + 1)− 1(4− 2) = −5

A31 = (−1)3+1

∣∣∣∣∣∣0 −1 12 2 −31 4 0

∣∣∣∣∣∣ = −(−1)(0 + 3) + 1(8− 2) = 9

A32 = (−1)3+2

∣∣∣∣∣∣1 −1 10 2 −3−2 4 0

∣∣∣∣∣∣ = −[1(0 + 12)− 2(3− 2)] = −10

A33 = (−1)3+3

∣∣∣∣∣∣1 0 10 2 −3−2 1 0

∣∣∣∣∣∣ = 1(0 + 3) + 1(0 + 4) = 7

A34 = (−1)3+4

∣∣∣∣∣∣1 0 −10 2 2−2 1 4

∣∣∣∣∣∣ = −[1(8− 2)− 1(0 + 4)] = −2

A41 = (−1)4+1

∣∣∣∣∣∣0 −1 12 2 −3−1 −1 0

∣∣∣∣∣∣ = −[−(−1)(0− 3) + 1(−2 + 2)] = 3

9


A42 = (−1)4+2

∣∣∣∣∣∣1 −1 10 2 −34 −1 0

∣∣∣∣∣∣ = 1(0− 3) + 4(3− 2) = 1

A43 = (−1)4+3

∣∣∣∣∣∣1 0 10 2 −34 −1 0

∣∣∣∣∣∣ = −[1(0− 3) + 1(0− 8)] = 11

A44 = (−1)4+4

∣∣∣∣∣∣1 0 −10 2 24 −1 −1

∣∣∣∣∣∣ = 1(−2 + 2)− 1(0− 8) = 8

Ası

B =

9 42 −6 243 14 −2 −59 −10 7 −23 1 11 8

Luego Adj = Bt =

9 3 9 342 14 −10 1−6 −2 7 1124 −5 −2 8

Ası, por teorema 5

A−1 = ( 1detA )adjA= ( 1

39 )

9 3 9 342 14 −10 1−6 −2 7 1124 −5 −2 8

=

3/13 1/13 3/13 1/13

14/13 14/39 −10/39 1/39

−2/13 −2/39 7/39 11/39

8/13 −5/39 −2/39 8/39

4.

Figura 1: Triangulo

i. Usando trigonometria elemental muestre que:

c cosA + a cosC = bb cosA + a cosB = cc cosB + b cosC = a

Respuesta:Trazemos una perpendicular desde el vertice del angulo B hasta un punto en la recta formada por losvertices de los angulos A y C; como se muestra en la figura 2:

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Figura 2: Triangulo para hallar b

De donde b = d + e = c cosA + a cosC

Ahora trazemos una perpendicular desde el vertice del angulo C hasta un punto en la recta forma-da por los vertices de los angulos A y B; como se muestra en la figura 3:

Figura 3: Triangulo para hallar c

De donde b = f + g = b cosA + a cosB

Por ultimo trazemos una perpendicular desde el vertice del angulo A hasta un punto en la rectaformada por los vertices de los angulos B y C; como se muestra en la figura 4:

Figura 4: Triangulo para hallar a

De donde a = r + s = c cosB + b cosC

ii. Use la regla de Cramer para resolver el sistema i.

Respuesta:

Ordenando las ecuaciones tenemos:

c cosA + 0 + a cosC = bb cosA + a cosB + 0 = c0 + c cosB + b cosC = a

Hallemos cosA, cosB, cosC

Sea A =

c 0 ab a 00 c b

Asi:

|A| =

c 0 ab a 00 c b

= c(ab− 0) + a(bc− 0) = 2abc

Luego:

|B1| =

∣∣∣∣∣∣b 0 ac a 0a c b

∣∣∣∣∣∣ = b(ab− 0) + a(c2 − a2) = ab2 + ac2 − a3

|B2| =

∣∣∣∣∣∣c b ab c 00 a b

∣∣∣∣∣∣ = c(cb− 0)− b(b2 − a2) = c2b− b3 + a2b

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|B3| =

∣∣∣∣∣∣c 0 bb a c0 c a

∣∣∣∣∣∣ = c(a2 − c2) + b(bc− 0) = ca2 − c3 + b2c

Por lo tanto:

cosA = ab2+ac2−a3

2abc = b2+c2−a2

2bc

cosB = c2b−b3+a2b2abc = c2−b2+a2

2ac

cosC = ca2−c3+b2c2abc = a2−c2+b2

2ab

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