Taller de matematicas

1. Taller 3 Matemáticas II
Heidy Julitza Mosquera Capera
20202192265
Universidad Surcolombiana
Facultad de Economía y Administración
Neiva-Huila
2021-1

2. 1. Evalué los siguientes determinantes:
a) [
1 2 0 9
2 3 4 6
1 6 0 −1
0 −5 0 8
]
|𝐴| = −9 [
2 3 4
1 6 0
0 −5 0
] + 6[
1 2 0
1 6 0
0 −5 0
] − (−1)[
1 2 0
2 3 4
0 −5 0
]
+ 8 [
1 2 0
2 3 4
1 6 0
]
Regla de Sarrus para hallar los determinantes de 3x3
|𝐴| = −9 [
2 3 4 2 3
1 6 0 1 6
0 −5 0 0 −5
] + 6[
1 2 0 1 2
1 6 0 1 6
0 −5 0 0 −5
]
− (−1)[
1 2 0 1 2
2 3 4 2 3
0 −5 0 0 −5
]
+ 8 [
1 2 0 1 2
2 3 4 2 3
1 6 0 1 6
]
|𝐴| = −9 ((2 ∗ 6 ∗ 0) + (3 ∗ 0 ∗ 0) + (4 ∗ 1 ∗ (−5))
− (4 ∗ 6 ∗ 0) − (2 ∗ 0 ∗ (−5))− (3 ∗ 1 ∗ 0))
+ 6 ((1 ∗ 6 ∗ 0) + (2 ∗ 0 ∗ 0) + (0 ∗ 1 ∗ (−5))
− (0 ∗ 6 ∗ 0) − (1 ∗ 0 ∗ (−5))− (2 ∗ 1 ∗ 0))
+ 1 ((1 ∗ 3 ∗ 0) + (2 ∗ 4 ∗ 0) + (0 ∗ 2 ∗ (−5))
− (0 ∗ 3 ∗ 0) − (1 ∗ 4 ∗ (−5))− (2 ∗ 2 ∗ 0))
+ 8((1∗ 3 ∗ 0) + (2 ∗ 4 ∗ 1) + (0 ∗ 2 ∗ 6)
− (0 ∗ 3 ∗ 1) − (1 ∗ 4 ∗ 6) − (2 ∗ 2 ∗ 0))
|𝐴| = −9(−20) + 6(0) + 1(20) + 8(8− 24)
|𝐴| = 72

3. b) [
2 7 0 1
5 6 4 8
0 0 9 0
1 −3 1 4
]
|𝐴| = +0 [
5 6 8
0 0 0
1 −3 4
] − 4[
2 7 1
0 0 0
1 −3 4
] + 9[
2 7 1
5 6 8
1 −3 4
]
− 1 [
2 7 1
5 6 8
0 0 0
]
Regla de Sarrus para hallar los determinantes de 3x3
|𝐴| = −4 [
2 7 1 2 7
0 0 0 0 0
1 −3 4 1 −3
] + 9[
2 7 1 2 7
5 6 8 5 6
1 −3 4 1 −3
]
− 1 [
2 7 1 2 7
5 6 8 5 6
0 0 0 0 0
]
|𝐴| = −4(0) + 9(48+ 56 + (−15) − 6 − (−48) − 140)
− 1(0)
|𝐴| = 9(−9)
|𝐴| = −81
c)
[
1
6
2
7 −3 9
0 4 −6 1 0
4 −10 1 0 6
−8 10 0 3 1
1 0 2 5 8]
= 26445
2. Utilice la regla de Cramer para resolver los siguientes sistemas de
ecuaciones:
a)
𝑤1𝑃1 + 𝑤2𝑃2 = −𝑤0
𝑚1𝑃1 + 𝑚2𝑃2 = −𝑚0
𝐴 = (
1 1
1 1
)
𝑋 = (
𝑃1
𝑃2
)
𝐵 = (
−1
−1
)

4. |𝐴| = |
1 1
1 1
| = 1 − 1 = 0
No Podemos aplicar la regla de Cramer
b) −𝑥1 + 3𝑥2 + 2𝑥3 = 24
𝑥1 + 𝑥3 = 6
5𝑥2 − 𝑥3 = 8
𝐴 = (
−1 3 2
1 0 1
0 5 −1
)
𝑋 = (
𝑥1
𝑥2
𝑥3
)
𝐵 = (
24
6
8
)
|𝐴| = |
−1 3 2
1 0 1
0 5 −1
| = −1(0 − 5) − 3(−1 − 0) + 2(5− 0)
= 18
𝐴1 = (
24 3 2
6 0 1
8 5 −1
)
𝑥1 =
|
24 3 2
6 0 1
8 5 −1
|
|𝐴|
=
24(−5) − 3(−6 − 8) + 2(30)
18
= −
18
18
= −1
𝐴2 = (
−1 24 2
1 6 1
0 8 −1
)
𝑥2 =
|
−1 24 2
1 6 1
0 8 −1
|
|𝐴|
=
−1(−6− 8) − 24(−1) + 2(8)
18
=
54
18
= 3
𝐴3 = (
−1 3 24
1 0 6
0 5 8
)

5. 𝑥2 =
|
−1 3 24
1 0 6
0 5 8
|
|𝐴|
=
−1(−30)− 3(8) + 24(5)
18
=
126
18
= 7
c) −𝑥 + 𝑦 + 𝑧 = 𝑎
𝑥 − 𝑦 + 𝑧 = 𝑏
𝑥 + 𝑦 − 𝑧 = 𝑐
𝐴 = (
−1 1 1
1 −1 1
1 1 −1
)
𝑋 = (
𝑥
𝑦
𝑧
)
𝐵 = (
𝑎
𝑏
𝑐
)
|𝐴| = |
−1 1 1
1 −1 1
1 1 −1
| = −1(0) − 1(−2) + 1(2) = 4
𝐴1 = (
𝑎 1 1
𝑏 −1 1
𝑐 1 −1
)
𝑥 =
|
𝑎 1 1
𝑏 −1 1
𝑐 1 −1
|
|𝐴|
=
𝑎(0) − 1(−1𝑏− 1𝑐) + 1(1𝑏 + 1𝑐)
4
=
1𝑏 + 1𝑐 + 1𝑏 + 1𝑐
4
=
2𝑏 + 2𝑐
4
=
𝑏 + 𝑐
2
𝐴2 = (
−1 𝑎 1
1 𝑏 1
1 𝑐 −1
)
𝑦 =
|
−1 𝑎 1
1 𝑏 1
1 𝑐 −1
|
|𝐴|
=
−1(−1𝑏− 1𝑐) − 𝑎(−2) + 1(1𝑐 − 1𝑏)
4
=
1𝑏 + 1𝑐 + 2𝑎 + 1𝑐 − 1𝑏
4
=
2𝑐 + 2𝑎
4
=
𝑐 + 𝑎
2
𝐴3 = (
−1 1 𝑎
1 −1 𝑏
1 1 𝑐
)

6. 𝑧 =
|
−1 1 𝑎
1 −1 𝑏
1 1 𝑐
|
|𝐴|
=
−1(−1𝑐 − 1𝑏) − 1(1𝑐 − 1𝑏) + 𝑎(2)
4
=
1𝑐 + 1𝑏 + 2𝑎 − 1𝑐 + 1𝑏
4
=
2𝑏 + 2𝑎
4
=
𝑏 + 𝑎
2
d) 4𝑥 + 3𝑦 − 2𝑧 = 1
𝑥 + 2𝑦 = 6
3𝑥 + 𝑧 = 4
𝐴 = (
4 3 −2
1 2 0
3 0 1
)
𝑋 = (
𝑥
𝑦
𝑧
)
𝐵 = (
1
6
4
)
|𝐴| = |
4 3 −2
1 2 0
3 0 1
| = 4(2)− 3(1) − 2(−6) = 17
𝐴1 = (
1 3 −2
6 2 0
4 0 1
)
𝑥 =
|
1 3 −2
6 2 0
4 0 1
|
|𝐴|
=
1(2) − 3(6) − 2(−8)
17
= 0
𝐴2 = (
4 1 −2
1 6 0
3 4 1
)
𝑦 =
|
4 1 −2
1 6 0
3 4 1
|
|𝐴|
=
4(6) − 1(1) − 2(4− 18)
17
=
51
17
= 3
𝐴3 = (
4 3 1
1 2 6
3 0 4
)

7. 𝑧 =
|
4 3 1
1 2 6
3 0 4
|
|𝐴|
=
4(8) − 3(4 − 18) + 1(−6)
17
=
68
17
= 4
3. Halle el Volumen de las siguientes estructuras geométricas por
medio de determinantes (Realice su procedimiento Paso a Paso y
argumentado):
a) La figura geométrica está formada por los puntos
𝑃1 (2,0,1),𝑃2 (1,3, 1),𝑃3 (1,4,2 ).
Establezca su configuración matricial y halle su volumen.
𝑉 = |𝑃1 ∗ (𝑃2 × 𝑃3)|
|𝑃1 ∗ (𝑃2 × 𝑃3)| = |
2 0 1
1 3 1
1 4 2
| = 2(6 − 4) − 0 + 1(4− 3) = 5
𝑉 = 5𝑢3
b) La figura geométrica está formada por los puntos
𝑃1 (2,2,2),𝑃2 (4,1,3),𝑃3 (5,3,4 ),𝑃4 (3,5,1 )

8. Tenga en cuenta que el área de la figura es 𝐴 =
1
6
.
Establezca su configuración matricial y halle su volumen.
𝑉 =
1
6
|𝑃1 𝑃2
⃗⃗⃗⃗⃗⃗⃗⃗ , 𝑃1𝑃3
⃗⃗⃗⃗⃗⃗⃗⃗ , 𝑃1 𝑃4
⃗⃗⃗⃗⃗⃗⃗⃗ |
𝑃1 𝑃2
⃗⃗⃗⃗⃗⃗⃗⃗ = 𝑃2 − 𝑃1 = (4,1,3) − (2,2,2) = (2,−1,1)
𝑃1 𝑃3
⃗⃗⃗⃗⃗⃗⃗⃗ = 𝑃3 − 𝑃1 = (5,3,4) − (2,2,2) = (3,1,2)
𝑃1 𝑃4
⃗⃗⃗⃗⃗⃗⃗⃗ = 𝑃4 − 𝑃1 = (3,5,1) − (2,2,2) = (1,3,−1)
|𝑃1 𝑃2
⃗⃗⃗⃗⃗⃗⃗⃗ , 𝑃1𝑃3
⃗⃗⃗⃗⃗⃗⃗⃗ , 𝑃1 𝑃4
⃗⃗⃗⃗⃗⃗⃗⃗ | = |
2 −1 1
3 1 2
1 2 −1
|
= 2(−1 − 4) − (−1)(−3− 2) + 1(6 − 1) = |−10|
𝑉 =
1
6
|−10| =
5
3
𝑢3