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
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ALGEBRA LINEAL I 4 Sem. 2015
|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:
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ALGEBRA LINEAL I 4 Sem. 2015
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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ALGEBRA LINEAL I 4 Sem. 2015
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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ALGEBRA LINEAL I 4 Sem. 2015
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
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ALGEBRA LINEAL I 4 Sem. 2015
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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ALGEBRA LINEAL I 4 Sem. 2015
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
Por otro lado det(A) =
∣∣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
Por otro lado det(A) =
∣∣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
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ALGEBRA LINEAL I 4 Sem. 2015
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
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ALGEBRA LINEAL I 4 Sem. 2015
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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ALGEBRA LINEAL I 4 Sem. 2015
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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ALGEBRA LINEAL I 4 Sem. 2015
|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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