Ejercicios Propuestos – Unidad IV

1. Universidad “Fermín Toro”
Vice-Rectorado Académico
Escuela de Ingeniería Mecánica
Cabudare
Nombre
Omar Mendoza
C.I: 20.670.309
Saia Seccion E

2. 1. 𝐷𝑎𝑑𝑜𝑠 𝑙𝑜𝑠 𝑐𝑜𝑛𝑗𝑢𝑛𝑡𝑜𝑠: 𝑋
= {2, 3, 0, −1,−9, −6, −2, 7,1, 5, 8, 9} ,
𝑌
= {3, −2, −6,5, 1, 0, 2, 5,−4, −3 4,6, 7, 9}
𝑦 𝑙𝑎 𝑟𝑒𝑙𝑎𝑐𝑖ó𝑛
𝑅1 = {(𝑥, 𝑦)/ (𝑥, 𝑦) 𝑋𝑥𝑌, 2𝑦
= −3𝑥 − 6}
𝑆𝑜𝑙𝑢𝑐𝑖ó𝑛
𝑋 = { −9,−6, −2, −1,0,1,2,3,5,7,8,9}
𝑌 = {−6, −4, −3,−2,0,1,2,3,4,5,6,7,9}
𝑃𝑎𝑟𝑎 𝑥 = −9
2𝑦 = −3(−9)− 6 => 27 − 6 => 21
2𝑦 = 21
𝑌 =
21
2
𝑃𝑎𝑟𝑎 𝑥 = −6
2𝑦 = −3(−6)− 6 => 18 − 6 => 12
2𝑦 = 12
𝑌 =
12
2
= 6
=> (−6,6) ∈ ℝ1
𝑃𝑎𝑟𝑎 𝑥 = −2
2𝑦 = −3(−2)− 6 => 6 − 6 => 0
2𝑦 = 0
𝑌 =
0
2
= 0 => (−2,0) ∈ ℝ1
𝑃𝑎𝑟𝑎 𝑥 = −1
2𝑦 = −3(−1)− 6 => 3 − 6 => −3
2𝑦 = 3
𝑌 = −
3
2
𝑃𝑎𝑟𝑎 𝑥 = 0
2𝑦 = −3(0)− 6 => 0 − 6 => −6
2𝑦 = −6
𝑌 = −
6
2
= −3
=> (0, −3) ∈ ℝ1
𝑃𝑎𝑟𝑎 𝑥 = 1
2𝑦 = −3(1)− 6 => −3 − 6 => −9
2𝑦 = −9
𝑌 = −
9
2
𝑃𝑎𝑟𝑎 𝑥 = 2
2𝑦 = −3(2) − 6 => −6 − 6 => −12
2𝑦 = −12 => 𝑌 = −
12
2
= −6
=> (2, −6) ∈ ℝ1

3. 𝑃𝑎𝑟𝑎 𝑥 = 5
2𝑦 = −3(5)− 6 => −15 − 6 => 21
2𝑦 = −21
𝑌 = −
21
2
𝑃𝑎𝑟𝑎 𝑥 = 7
2𝑦 = −3(7)− 6 => −21 − 6 => −27
2𝑦 = −27
𝑌 = −
27
2
𝑃𝑎𝑟𝑎 𝑥 = 8
2𝑦 = −3(8)− 6 => −24 − 6 => −30
2𝑦 = −30
𝑌 = −
30
2
= −15
Asi que viene dada por
R1= {(−6,6),(−2,0),(0, −3),(2 − 6)}
b) representación matricial
R1 -6 -4 -3 -2 0 1 2 3 4 5 6 7 9
-9 0 0 0 0 0 0 0 0 0 0 0 0 0
-6 0 0 0 0 0 0 0 0 0 0 1 0 0
-2 0 0 0 0 1 0 0 0 0 0 0 0 0
-1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0 0 0 0 0
2 1 0 0 0 0 0 0 0 0 0 0 0 0
3 0 0 0 0 0 0 0 0 0 0 0 0 0
5 0 0 0 0 0 0 0 0 0 0 0 0 0
7 0 0 0 0 0 0 0 0 0 0 0 0 0
8 0 0 0 0 0 0 0 0 0 0 0 0 0
9 0 0 0 0 0 0 0 0 0 0 0 0 0

4. Sabemos que R1 = {(−6,6),(−2,0),(0, −3),(2 − 6)}
6
5
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1
-2
-3
-4
-5
-6

5. R1
X Y
-9 -6
-6 -4 c) dom(r1)={-6,-2, 0, 2}
-2 -3 rang(r1)={-6,-3, 0, 6}
-1 -2
0 0
1 1
2 2
3 3
4 4
5 5
7 6
8 7
9 9

6. Para 𝑼 = ℤ , 𝑿 = {𝟐, 𝟑, 𝟒, −𝟐, −𝟓, −𝟑, 𝟎, −𝟒, −𝟖 𝟓, 𝟔, 𝟕},
𝒀 = {𝟏𝟎, −𝟐𝟐, −𝟓𝟎, −𝟑𝟎, −𝟏𝟐, −𝟓, 𝟏𝟓, 𝟐𝟎, 𝟒 𝟏𝟏, 𝟏𝟐, 𝟏𝟑, 𝟏𝟒},
Escribir los elementos de la relación 𝑹2 ⊂ 𝑿 × 𝒀, donde
𝒙𝑹2 𝒚 𝒔𝒊 𝒚 𝒔𝒐𝒍𝒐 𝒔𝒊 𝒙 𝒅𝒊𝒗𝒊𝒅𝒆 (𝒆𝒙𝒂𝒄𝒕𝒂𝒎𝒆𝒏𝒕𝒆) 𝒂 𝒚.
a) Hallar Representación matricial, cartesiana y sagital de𝑹2
b) Dominio y rango de 𝑹2
para u= Z
𝒙 = {−𝟖, −𝟓,−𝟒, −𝟑,−𝟐, 𝟎, 𝟐, 𝟑, 𝟒, 𝟓, 𝟔, 𝟕}
𝒚 = {−𝟓𝟎,−𝟑𝟎,−𝟐𝟐,−𝟏𝟐,−𝟓, 𝟒, 𝟏𝟎, 𝟏𝟏, 𝟏𝟐, 𝟏𝟑, 𝟏𝟒, 𝟏𝟓, 𝟐𝟎}
R2= {(𝒙, 𝒚) ∈ 𝑋𝑥𝑌/𝑥𝑅2𝑦 𝑥 𝑑𝑖𝑣𝑖𝑑𝑒 𝑒𝑥𝑎𝑐𝑡𝑎𝑚𝑒𝑛𝑡𝑒 𝑎 𝑦} Los elementos de R2 son
R2={(-5,50),(-5,30),(-5,-5),(-5,10),(-5,15),(-5,20),(-4,-12),(-4,4),(-4,12),(-4,20),
(-3,-30),(-3,-12),(-3,12),(-3,15),(-2,50),(-2,-30),(-2,-22),
(-2,12),(2,4),(2,10),(2,12),(2,14),(2,20),(4,-22),(4,-12),(4,4),(4,12),(4,20),(5,-50),
(5,-30),(5,-5),(5,10),(5,15),(5,20),(6,-30),(6,-12),(6,12),(7,14)

7. R2
X Y
-8 50 Dom(R2) {-5,-4,-3,-2,3,4,5,6,7}
-5 30 rang(R2){-50,-30,-22,-12,-5,4,10,12,13,14,15,20}
-4 -22
-3 -12
-2 -5
0 4
3 10
4 11
5 12
6 13
7 14
15
20
M2 -50 -30 -22 -12 -5 4 10 11 12 13 14 15 20
-8 0 0 0 0 0 0 0 0 0 0 0 0 0
-5 1 1 0 0 1 0 1 0 0 0 0 1 1
-4 0 0 0 1 0 1 0 0 1 0 0 0 1
-3 0 1 0 1 0 0 0 0 1 0 0 1 0
-2 1 1 1 1 0 1 1 0 1 0 1 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0
3 0 1 0 1 0 0 0 0 0 1 0 1 0
4 0 0 0 1 0 1 0 0 1 0 0 0 1
5 1 1 0 0 1 0 1 0 0 0 0 1 1
6 0 1 0 1 0 0 0 0 1 0 0 0 0
7 0 0 0 0 0 0 0 0 0 0 1 0 0

8. 3. Sean 𝑿 = {𝟎, 𝟏, −𝟔, −𝟒, 𝟏𝟎, −𝟕, 𝟓, 𝟒, −𝟏, 𝟐} , 𝒀 =
{−𝟑, −𝟔, 𝟎, 𝟏, −𝟓, −𝟐, −𝟏, 𝟑, 𝟐, 𝟒, 𝟓} y 𝒁 =
{−𝟒, 𝟎, 𝟏𝟎, 𝟏𝟏, 𝟐, 𝟏, 𝟓, 𝟔, −𝟓, −𝟐, −𝟔, −𝟑, 𝟑, 𝟕} y sean R y S las relaciones
𝑹 ⊂ 𝑿𝐱𝐘: 𝐱𝐑𝐲 ⇔ (−𝟑𝐱 + 𝟓𝐲)𝐞𝐬 𝐩𝐚𝐫
𝑺 ⊂ 𝒀𝐱𝐙: 𝐲𝐒𝐳 ⇔ (𝟓𝐲 − 𝐳)𝐞𝐬 𝐩𝐚𝐫
a) Hallar los elementos de R, S
b) 𝑺 ∘ 𝑹 y 𝑹−𝟏 ∘ 𝑺−𝟏
Solución:
𝑋 = {−7, −6,−4, −1,0,1,2,4,5,10}
𝑌 = {−6, −5, −3,−2, −1,0,1,2,3,4,5}
𝑍 = {−6, −5,−4, −3, −2,0,1,2,3,4,5,6,7,10,11}
a) Se busca los elementos de R
𝑃𝑎𝑟𝑎 𝑥 = −7 => −3(−7) + 5𝑦 => 21 + 5𝑦
𝑌 = −6 => 21 + 5(−6) => 21 − 30 = −9
𝑌 = −5 => 21 + 5(−5) => 21 − 25 = −4 => (−7,−5) ∈ 𝑅
𝑌 = −3 => 21 + 5(−3) => 21 − 15 = 6=>(−7,−3) ∈ 𝑅
𝑌 = −2 => 21 + 5(−2) => 21 − 10 = 11
𝑌 = −1 => 21 + 5(−1) => 21 − 5 = 16=>(−7,−1) ∈ 𝑅
𝑌 = 0 => 21 + 5(0) => 21 + 0 = 21
𝑌 = 1 => 21 + 5(1) => 21 + 5 = 26 => (−7,1) ∈ 𝑅
𝑌 = 2 => 21 + 5(2) => 21 + 10 = 31
𝑌 = 3 => 21 + 5(3) => 21 + 15 = 36 => (−7,3) ∈ 𝑅
𝑌 = 4 => 21 + 5(4) => 21 + 20 = 41
𝑌 = 5 => 21 + 5(5) => 21 + 25 = 46 => (−7,5) ∈ 𝑅

9. 𝑃𝑎𝑟𝑎 𝑥 = −6 => −3(−6) + 5𝑦 => 18 + 5𝑦
Como 18 es un número par para que 18+5y sea necesariamente 5y tiene que ser par
=> 𝑦 = −6, 𝑦 = −2, 𝑦 = 0, 𝑦 = 2, 𝑦 = 4 Caso 1
𝐴𝑠𝑖 (−6,−6),(−6, −2),(−6, 0),(−6,2)(−6,4) ∈ 𝑅
𝑃𝑎𝑟𝑎 𝑥 = −4 => −3(−4) + 5𝑦 => 12 + 5𝑦
Como 12 es un número par para que 12+5y sea necesariamente 5y tiene que ser par
=> 𝑦 = −6, 𝑦 = −2, 𝑦 = 0, 𝑦 = 2, 𝑦 = 4
𝐴𝑠𝑖 (−4,−6),(−4, −2),(−4, 0),(−4,2)(−4,4) ∈ 𝑅
𝑃𝑎𝑟𝑎 𝑥 = −1 => −3(−1) + 5𝑦 => 3 + 5𝑦
Como 3 es un número impar para que 3+5y sea necesariamente 5y tiene que ser par
=> 𝑦 = −5, 𝑦 = −3, 𝑦 = −1, 𝑦 = 1, 𝑦 = 3, 𝑦 = 5 Caso 2
𝐴𝑠𝑖 (−1,−5),(−1, −3),(−1, −1),(−1,1)(−1,3), (−1,5) ∈ 𝑅

10. 𝑃𝑎𝑟𝑎 𝑥 = 0 => −3(0) + 5𝑦 => 0 + 5𝑦
Como 0 es un número par para que 12+5y sea necesariamente 5y tiene que ser par y los valores de y serían los del caso
1
=> 𝑦 = −6, 𝑦 = −2, 𝑦 = 0, 𝑦 = 2, 𝑦 = 4
𝐴𝑠𝑖 (0,−6),(0, −2),(0, 0),(0, 2)(0,4) ∈ 𝑅
𝑃𝑎𝑟𝑎 𝑥 = 1, 𝑥 = 5 𝑠𝑒 𝑝𝑢𝑒𝑑𝑒 𝑑𝑒𝑐𝑖𝑟 𝑎𝑛𝑎𝑙𝑜𝑔𝑎𝑚𝑒𝑛𝑡𝑒 𝑜𝑏𝑡𝑒𝑛𝑖𝑑𝑜 𝑙𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝑦 𝑠𝑒𝑔𝑢𝑛 𝑙𝑜𝑠 𝑐𝑎𝑠𝑜 2 que son los impares
2 que son los impares
𝐴𝑠𝑖 (1,−5),(1, −3),(1, −1),(1, 1)(1,3),(1,5), (5,−5), (5,−3), (5,−1),(5, 1)(5,3),(5, 5) ∈ 𝑅
𝑝𝑎𝑟𝑎 𝑥 = 2, 𝑥 = 4, 𝑥 = 10 𝑠𝑒 𝑝𝑢𝑒𝑑𝑒 𝑑𝑒𝑐𝑖𝑟 𝑎𝑛𝑎𝑙𝑜𝑔𝑎𝑚𝑒𝑛𝑡𝑒 𝑜𝑏𝑡𝑒𝑛𝑖𝑑𝑜 𝑙𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝑦 𝑠𝑒𝑔𝑢𝑛 𝑙𝑜𝑠 𝑐𝑎𝑠𝑜 1 que son los pares
𝐴𝑠𝑖 (2,−6),(2, −2),(2, 0),(2, 2)(2,4),(4, −6),(4,−2), (4,0), (4,2)(4,4), (10,−6),(10, −2),(10,0), (10,2)(10,4) ∈ 𝑅
∈ 𝑅
Luego R viene dada por
R={(−7, −5),(−7,−3),(−7, −1),(−7,1),(−7,3),(−7,5),(−6, −6),(−6, −2),(−6,0), (−6,2)(−6,4),
(−4,−6), (−4,−2), (−4,0),(−4, 2)(−4,4),(−1, −5),(−1,−3), (−1,−1), (−1,1)(−1,3),(−1, 5),

11. (0,−6), (0,−2), (0,0), (0,2)(0,4),(1, −5),(1, −3),(1, −1),(1,1)(1,3), (1,5), (5,−5), (5,−3),(5, −1),(5, 1)(5,3),(5, 5),
(2,−6), (2,−2), (2,0), (2,2)(2,4),(4, −6),(4, −2),(4, 0),(4,2)(4,4), (10,−6), (10,−2),(10, 0),(10,2)(10,4)}
Elemento de S
𝑃𝑎𝑟𝑎 𝑦 = −6 => 5(−6) = −30 − 𝑍
Como 30 es par entonces Z tiene que ser par
Así que 𝑍 = −6 , 𝑍 = −4, 𝑍 = −2, 𝑍 = 0 , 𝑍 = 2, 𝑍 = 6, 𝑍 = 10
{(−6, −6),(−6,−4), (−6,−2), (−6,0),(−6,2),(−6,6),(−6,10)} ∈ 𝑆
Análogamente para 𝑌 = −2, 𝑌 = 0, 𝑌 = 4
{(−2, −6),(−2,−4), (−2,−2), (−2,0),(−2, 2),(−2,6),(−2,10), (0,−6),(0, −4),(0, −2),(0,0),(0,2), (0,6),(0,10),(4,−6), (4,−4),
(4,−2), (4,0),(4,2),(4,6), (4,10)} ∈ 𝑆

12. 𝑃𝑎𝑟𝑎 𝑦 = −5 => 5(−5) = −25 − 𝑍
Como -25 es par entonces Z tiene que ser impar
𝑍 = −5 , 𝑍 = −3, 𝑍 = 1, 𝑍 = 3, 𝑍 = 5, 𝑍 = 7, 𝑍 = 11
{(−5, −5),(−5,−3), (−5,1),(−5,3),(−5,5),(−5,7), (−5,11)}
Entonces análogamente para 𝑦 = −1, 𝑦 = 1, 𝑦 = 3, 𝑦 = 5
Quedara
Así {(−1,−5), (−1,−3),(−1,1), (−1,3),(−1,5),(−1,7),(−1,11),(1, −5),(1, −3),(1,1),(1,3),(1,5), (1,7),(1,11),
(3,−5), (3,−3), (3,1),(3,3),(3,5),(3,7), (3,11),(5,−5), (5,−3),(5,1), (5,3),(5,5),(5,7),(5,11)} ∈ 𝑆
Luego S Viene dada por
S={(−6, −6),(−6, −4),(−6,−2), (−6,0),(−6,2),(−6,6),(−6,10)(−2,−6),(−2, −4),(−2,−2), (−2,0),
(−2,2), (−2,6),(−2,10),(0, −6),(0,−4), (0,−2), (0,0),(0,2),(0,6),(0,10),(4, −6),(4, −4),(4, −2),(4,0),(4,2),(4,6), (4,10),
(−5,−5), (−5,−3), (−5,1),(−5,3),(−5,5),(−5,7),(−5,11),(−1,−5), (−1,−3), (−1,1),(−1,3),(−1,5),(−1,7),(−1,11),
(1,−5), (1,−3), (1,1),(1,3),(1,5),(1,7), (1,11),(3,−5), (3,−3),(3,1), (3,3),(3,5),(3,7),(3,11),
(5,−5), (5,−3), (5,1),(5,3),(5,5),(5,7), (5,11)}

13. 𝑏) 𝑆𝑜𝑅{
= {(−7,−5),(−7, −3),(−7,−1),(−7,1),(−7,3),(−7,7),(−7,11),(−6, −6),(−6,−4), (−6,−2), (−6,0),(−6,2),(−6,6),(−6,10),(−4, −6),
(−4,−4), (−4,−2), (−4,0),(−4,0),(−4,2),(−4,6),(−4,10),(1,5), (1,7),(1,11),(2,−6), (2,−4), (2,−2),(2,6), (2,10),(4,−6), (4,−2),
(4,6),(4,10),(5 − 5),(5 − 3),(5,1), (5,3),(5,7),(5,11),(10 − 6),(10,−4), (10,−2),(10,0),(10,2), (10,6),(10,10)}
( 𝑆𝑜𝑅)−1
Por teorema se obtiene que 𝑅−1
𝑜𝑆−1
= ( 𝑆𝑜𝑅)−1
quedando de la siguiente manera
( 𝑆𝑜𝑅)−1
{(−5,−7), (−3,−7),(−7, −1),(1, −7),(3,−7), (5,−7)(−6,−6), (−2,−6),(0, −6),(2, −6),(4,−6), (−6,−4), (−2,−4),
(0,−4), (2,−4), (4,−4), (−5,−1)(−3,−1),(1, −1),(1, −1),(3,−1), (5,−1), (−6,0),(−2,0),(0,0),(2,0),(4,0), (−5,1),(−3,1),(−1,1),
(1,1),(3,1),(5,1), (−6,2),(−2,2),(0,2),(2,2), (4,2)(−6,4),(−2,4),(0,4),(2,4), (4,4),(−5,5),(−3,5),(−1,5),(1,5), (3,5),(5,5),(−6,10)
(−2,10),(0,10),(2,10),(6,10),(10,10)
4. Sean 𝑿 = {𝟎, 𝟏, −𝟏, 𝟐} , 𝒀 = {−𝟐, −𝟑, 𝟐, 𝟒, 𝟓} y 𝒁 = {−𝟑, −𝟒, 𝟑, 𝟕} y sean R y S
las relaciones
𝑹 ⊂ 𝑿𝐱𝐘: 𝐱𝐑𝐲 ⇔ (𝐱 + 𝐲)𝐞𝐬 𝐧𝐮𝐦𝐞𝐫𝐨 𝐩𝐫𝐢𝐦𝐨
𝑺 ⊂ 𝒀𝐱𝐙: 𝐲𝐒𝐳 ⇔ (𝐲 + 𝐳)𝐞𝐬 𝐩𝐚𝐫
a) Hallar los elementos de R, S y 𝑺 ∘ 𝑹
b) Representación matricial de R y S
Solución
𝑋 = {−1,0,1,2} , 𝑦 = {−3, −2,2,4,5, }, 𝑧 = {−4,3,3,7}

14. A ) elementos de R
𝑃𝑎𝑟𝑎 𝑥 = −1
𝑌 = −3 => −1 − 3 = −4
𝑌 = −2 => −1 − 2 = −3
Y 𝑌 = 2 => −1 + 2 = 1
𝑌 = 4 => −1 + 4 = 3 => (−1,4) ∈ 𝑅
𝑌 = 5 => −1 + 5 = 4
𝑃𝑎𝑟𝑎 𝑥 = 0
𝑌 = −3 => 0 − 3 = −3
𝑌 = −2 => 0− 2 = −2
Y 𝑌 = 2 => 0 + 2 = 2 => (0,2) ∈ 𝑅
𝑌 = 4 => 0 + 4 = 4
𝑌 = 5 => 0 + 5 = 5 => (0,5) ∈ 𝑅
𝑃𝑎𝑟𝑎 𝑥 = 1
𝑌 = −3 => 1 − 3 = −2
𝑌 = −2 => 1− 2 = −1
Y 𝑌 = 2 => 1 + 2 = 3 => (1,2) ∈ 𝑅
𝑌 = 4 => 1 + 4 = 5 => (0,5) ∈ 𝑅
𝑌 = 5 => 1 + 5 = 6

15. 𝑃𝑎𝑟𝑎 𝑥 = 2
𝑌 = −3 => 2 − 3 = −1
𝑌 = −2 => 2− 2 = 0
Y 𝑌 = 2 => 2 + 2 = 4
𝑌 = 4 => 2 + 4 = 6
𝑌 = 5 => 2 + 5 = 7 => (2,5) ∈ 𝑅
Entonces R
R{(-1,4),(-1,5),(0,2),(0,5),(1,2),(1,4),(2,5)}
Entonces los elementos de S Sera
𝑃𝑎𝑟𝑎 𝑦 = −3
𝑍 = −4 => −3 − 4 = −7
𝑍 = −3 => −3 − 3 = −6 => (−3,−3) ∈ 𝑆
Y 𝑍 = 3 => −3 + 3 = 0 => (−3,3) ∈ 𝑆
𝑍 = 7 => −3 + 7 = 4 => (−3, −7) ∈ 𝑆

16. 𝑃𝑎𝑟𝑎 𝑦 = −2
𝑍 = −4 => −2 − 4 = −6 => (−2,−4) ∈ 𝑆
𝑍 = −3 => −2 − 3 = −5
Y 𝑍 = 3 => −2 + 3 = 1
𝑍 = 7 => −2 + 7 = 5
𝑃𝑎𝑟𝑎 𝑦 = 2
𝑍 = −4 => 2 − 4 = −2 => (2,−4) ∈ 𝑆
𝑍 = −3 => 2 − 3 = −1
Y 𝑍 = 3 => 2 + 3 = 5
𝑍 = 7 => 2 + 7 = 9
𝑃𝑎𝑟𝑎 𝑦 = 4
𝑍 = −4 => 4 − 4 = 0 => (4,−4) ∈ 𝑆
𝑍 = −3 => 4 − 3 = 1
Y 𝑍 = 3 => 4 + 3 = 7
𝑍 = 7 => 4 + 7 = 11

17. 𝑃𝑎𝑟𝑎 𝑦 = 5
𝑍 = −4 => 5 − 4 = 9
𝑍 = −3 => 5 − 3 = 2 => (5, −3) ∈ 𝑆
Y 𝑍 = 3 => 5 + 3 = 8 => (5,3) ∈ 𝑆
𝑍 = 7 => 5 + 7 = 12 => (5,7) ∈ 𝑆
Asi que S viene dada por
S={(-3,-3),(-3,3),(-3,7),(-2,-4),(2,-4),(4,-4),(5,-3),(5,3),(5,7)}
Entonces será que
R{(-1,4),(-1,5),(0,2),(0,5),(1,2),(1,4),(2,5)}
S={(-3,-3),(-3,3),(-3,7),(-2,-4),(2,-4),(4,-4),(5,-3),(5,3),(5,7)}
SoR={(-1,-4),(-1,-3),(-1,3),(-1,7),(0,-4),(0,-3),(0,3),(0,7),(1,-4),(2,-3),(2,3),(2,7)}
b) representación matricial de R
Mr=
-3 -2 2 4 5
−1 0 0 0 1 1
0 0
1 0
2 0
0 1 0 1
0 1 1 0
0 0 0 1

18. Representación matricial de S
Ms=
-4 -3 3 7
-3 0 1 1 1
-2 1 0 0 0
2 1 0 0 0
4 1 0 0 0
5 0 1 1 1