Sistema de Equações Lineares

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Sistema de Equações Lineares 
1) {
2𝑥 − 3𝑦 = 4
6𝑥 − 9𝑦 = 15
 (
2 −3
6 −9
⋮
4
15
)L1/2 (
1 −
3
2
6 −9
⋮
2
15
)-6L1+L2(
1 −
3
2
0 0
⋮
2
3
) 
{
𝑥 −
3
2
𝑦 = 2
𝑂𝑦 = 3
 Substituindo y, encontramos x. 𝑦 = 3 𝑥 =
13
2
 
É um Sistema Possível e Determinado (SPD). 
 
2) {
3𝑥 +2𝑦 −5𝑧 = 8
2𝑥 −4𝑦 −2𝑧 = −4
𝑥 −2𝑦 −3𝑧 = −4
 L3↔ L1(
1 −2 −3 ⋮
2 −4 −2 ⋮
3 2 −5 ⋮
−4
−4
8
) -2L1+L2, 
 
-3L1+L3(
1 −2 −3 ⋮
0 0 4 ⋮
0 8 4 ⋮
−4
4
20
)L2↔L3(
1 −2 −3 ⋮
0 8 4 ⋮
0 0 4 ⋮
−4
20
4
) 
 {
𝑥 −2𝑦 −3𝑧 = −4
8𝑦 +4𝑧 = 20
4𝑧 = 4
 Substituindo Z, encontramos X e Y. 
 𝑧 = 1, 𝑥 = 3, 𝑦 = 2 SPD! 
 
3) {
2𝑥 +4𝑦 +6𝑧 = −6
3𝑥 −2𝑦 −4𝑧 = −38
𝑥 +2𝑦 +3𝑧 = −3
 L1/2 (
1 2 3 ⋮
3 −2 −4 ⋮
1 2 3 ⋮
−3
−38
−3
) -3L1+L2, L3-L1 
(
1 2 3 ⋮
0 −8 −13 ⋮
0 0 0 ⋮
−3
−29
0
) {
𝑥 2𝑦 3𝑧 = −3
−8𝑦 −13𝑧 = −29 
 
 𝑧 = 0, 𝑥 =
41
4
, 𝑦 = −
29
8
 Substituindo Z, encontramos X e Y. 
Sistema Possível e Indeterminado (SPI). 
 
4) {
𝑥 +𝑦 −𝑧 = 0
2𝑥 −3𝑦 +𝑧 = 0
4𝑥 −4𝑦 −2𝑧 = 0
(
1 1 −1 ⋮
2 −3 1 ⋮
4 −4 −2 ⋮
0
0
0
) -2L1+L2, -4L1+L3 
(
1 1 −1 ⋮
0 −5 3 ⋮
0 −8 2 ⋮
0
0
0
)L2/5 (
1 1 −1 ⋮
0 −1 3/5 ⋮
0 −8 2 ⋮
0
0
0
)8L2-L3 
(
1 1 −1 ⋮
0 −1
3
5
⋮
0 0 −
14
5
⋮
0
0
0
) {
𝑥 +𝑦 −𝑧 = 0
−𝑦 +
3
5
𝑧 = 0
−
14
5
𝑧 = 0
 x=0, y=0,z=0 SPD! 
 
5) {
𝑥 −𝑧 = 0
3𝑥 +𝑦 +2𝑧 = 0
4𝑥 +2𝑦 +2𝑧 = 0
(
1 0 −1 ⋮
3 1 2 ⋮
4 2 2 ⋮
0
0
0
) -3L1+L2, -4L1+L3(
1 0 −1 ⋮
0 1 5 ⋮
0 2 6 ⋮
0
0
0
) 
-2L2+L3 (
1 0 −1 ⋮
0 1 5 ⋮
0 0 −4 ⋮
0
0
0
) {
𝑥 −𝑧 = 0
𝑦 +5𝑧 = 0
−4𝑧 = 0
 x=0, y=0, z=0 SPD! 
 
 
 
6) {
6𝑥 +2𝑦 +4𝑧 = 0
−9𝑥 −3𝑦 −6𝑧 = 0
 L1/6(
1
1
3
2
3
⋮ 0
−9 −3 −6 ⋮ 0
) 9L1+L2(
1
1
3
2
3
⋮ 0
0 0 0 ⋮ 0
) 
{
𝑥 +
1
3
𝑦 +
2
3
𝑧 = 0
𝑧 = 0
 Sistema Impossível (SI). 
 
7) {
𝑥 −𝑦 = 0
2𝑦 +4𝑧 = 6
𝑥 +𝑦 +4𝑧 = 6
(
1 −1 0 ⋮
0 2 4 ⋮
1 1 4 ⋮
0
6
6
) L3-L1 (
1 −1 0 ⋮
0 2 4 ⋮
0 2 4 ⋮
0
6
6
) L3-L2 
(
1 −1 0 ⋮
0 2 4 ⋮
0 0 0 ⋮
0
6
0
) {
𝑥 −𝑦 = 0
2𝑦 +4𝑧 = 0
 y=-2z, x=-2z SPI! 
 
 
8) {
𝑎1 +2𝑎2 = −4
−3𝑎1 +4𝑎2 = −18
2𝑎1 −𝑎2 = 7
(
1 2 ⋮ −4
−3 4 ⋮ −18
2 −1 ⋮ 7
) 3L1+L2, -2L1+L3 (
1 2 ⋮ −4
0 10 ⋮ −26
0 −5 ⋮ 15
) 
 
L2↔L3 (
1 2 ⋮ −4
0 −5 ⋮ 15
0 10 ⋮ −26
) -2L2-L3 (
1 2 ⋮ −4
0 −5 ⋮ 15
0 0 ⋮ −4
) {
𝑎1 +2𝑎2 = −4
−5𝑎2 = 15
 
 
a1=2, a2=-3 SPD! 
 
9) {
𝑎1 +2𝑎2 = 4
−3𝑎1 +4𝑎2 = 3
2𝑎1 −𝑎2 = −6
(
1 2 ⋮ 4
−3 4 ⋮ 3
2 −1 ⋮ −6
) 3L1+L2, -2L1+L3 (
1 2 ⋮ 4
0 10 ⋮ 15
0 −5 ⋮ −14
) 
 
2L3+L2 (
1 2 ⋮ 4
0 10 ⋮ 15
0 0 ⋮ −13
) {
𝑎1 +2𝑎2 = 4
10𝑎2 = 15
 a1=1, a2= 15/10 
 
 Sistema Impossível! 
 
10) {
𝑎1 +2𝑎2 = 𝑥
−3𝑎1 +4𝑎2 = 𝑦
2𝑎1 −𝑎2 = 𝑧
(
1 2 ⋮ 𝑥
−3 4 ⋮ 𝑦
2 −1 ⋮ 𝑧
) 3L1+L2, -2L1+L3 (
1 2 ⋮ 𝑥
0 10 ⋮ 3𝑥 + 𝑦
0 −5 ⋮ −2𝑥 + 𝑧
) 
L2/10 (
1 2 ⋮ 𝑥
0 1 ⋮
3𝑥+𝑦
10
0 −5 ⋮ −2𝑥 + 𝑧
) 5L2+L3 (
1 2 ⋮ 𝑥
0 1 ⋮
3𝑥+𝑦
10
0 0 ⋮ 2𝑥 + 𝑧 +
3𝑥+𝑦
10
) 
 
Para ter solução 2𝑥 + 𝑧 +
3𝑥+𝑦
10
=0 
 
 
11) {
𝑎1 +2𝑎2 = −1
−3𝑎1 +4𝑎2 = 𝑘
2𝑎1 −𝑎2 = −7
 (
1 2 ⋮ −1
−3 4 ⋮ 𝑘
2 −1 ⋮ −7
) 3L1+L2, -2L1+L3 (
1 2 ⋮ −1
0 10 ⋮ −3 + 𝑘
0 −5 ⋮ −5
) 
 L2/10 (
1 2 ⋮ −1
0 1 ⋮
−3+𝑘
10
0 −5 ⋮ −5
) 5L2+L3 (
1 2 ⋮ −1
0 10 ⋮
−3+𝐾
10
0 −5 ⋮
5(−3+𝐾)−5
10
) 
 {
𝑎1 +2𝑎2 = −1
𝑎2 =
−3+𝑘
10
0 =
5(−3+𝐾)−5
10
 k=13 
 
12) {
𝑥 −2𝑦 − 𝑧 = 𝑎
2𝑥 +𝑦 + 3𝑧 = 𝑏
4𝑥 −3𝑦 + 𝑧 = 𝑐
 (
1 −2 − 1 ⋮ 𝑎
2 1 3 ⋮ 𝑏
4 −3 1 ⋮ 𝑐
) -2L1+L2, -4L1+L3 
 
(
1 −2 −1 ⋮
0 5 5 ⋮
0 5 5 ⋮
𝑎
−2𝑎 + 𝑏
−4𝑎 + 𝑐
) L3+L2 (
1 −2 −1 ⋮
0 5 5 ⋮
0 0 0 ⋮
𝑎
−2𝑎 + 𝑏
−2𝑎 + 𝑏 − 𝑐
) 
 
Para ser compatível -2a+b=0 e -2a+b-c=0 
 
 
13) {
𝑥 −𝑦 − 𝑧 = 0
𝑥 −2𝑦 − 2𝑧 = 0
2𝑥 +𝑘𝑦 + 𝑧 = 0
 (
1 −1 − 1 ⋮ 0
1 −2 − 2 ⋮ 0
2 𝑘 1 ⋮ 0
) L2-L1, -2L1+L3 
 
(
1 −1 −1 ⋮
0 −1 −1 ⋮
0 2 + 𝑘 3 ⋮
0
0
0
) -2L2-L3 (
1 −1 −1 ⋮
0 −1 −1 ⋮
0 0 1 − 𝑘 ⋮
0
0
0
) {
𝑥 −𝑦 − 𝑧 = 0
−𝑦 − 𝑧 = 0
1 − 𝑘 = 0
 
 
Seria compatível e determinado se x=y=z=0