Taller de Algebra Lineal
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Transcript of 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
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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
12