Matriz Inversa e LI, LD

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Método da Matriz Inversa 
1) {
𝑥 +2𝑦 −2𝑧 = 0
2𝑥 +5𝑦 −4𝑧 = 3
3𝑥 +7𝑦 −5𝑧 = 7
 (
1 2 −2
2 5 −4
3 7 −5
|
1 0 0
0 1 0
0 0 1
) -2L1+L2, -3L1+L3 
 
(
1 2 −2
0 1 0
0 1 1
|
1 0 0
−2 1 0
−3 0 1
) -2L2+L1, L3-L2 (
1 0 −2
0 1 0
0 0 1
|
5 −2 0
−2 1 0
−1 −1 1
) 2L3+L1 
 
(
1 0 0
0 1 0
0 0 1
|
3 −4 2
−2 1 0
−1 −1 1
) (
3 −4 2
−2 1 0
−1 −1 1
) X(
0
3
7
)=(
0 −12 14
0 3 0
0 −3 7
)= (
2
3
4
) 
 
2) (
1 2 −2
2 5 −4
3 7 −5
|
1 0 0
0 1 0
0 0 1
) (
3 −4 2
−2 1 0
−1 −1 1
) X (
2
5
7
) = (
6 −20 14
−4 5 0
−2 −5 7
) = (
0
1
0
) 
 
 
 
Linearmente Independe e Linearmente Dependente 
 
1) Quais dos conjuntos de vetores são LD? 
a) (2,1,0,0), (1,0,2,1), (-1,2,0,-1) 
a1(2,1,0,0) + a2(1,0,2,1) + a3(-1,2,0,-1) = (0,0,0,0) 
{
2𝑎1 +𝑎2 −𝑎3 = 0
𝑎1 +2𝑎3 = 0
2𝑎2 = 0
𝑎2 −𝑎3 = 0
 a1=0, a2=0, a3=0 
 É Linearmente Independente (LI). 
 
b) (0,1,0,-1), (1,1,1,1), (-1,2,0,1), (1,2,1,0) 
a1(0,1,0,-1) + a2(1,1,1,1) + a3(-1,2,0,1) + a4(1,2,1,0) = (0,0,0,0) 
{
𝑎2 −𝑎3 +𝑎4 = 0
𝑎1 + 𝑎2 +2𝑎3 +2𝑎4 = 0
𝑎2 +𝑎4 = 0
−𝑎1 + 𝑎2 +𝑎3 = 0
 (
0 1 −1 1
1 1 2 2
0
−1
1
1
0
1
1
0
|
0
0
0
0
) L2↔L1 
 
(
1 1 2 2
0 1 −1 1
0
−1
1
1
0
1
1
0
|
0
0
0
0
) L4+L1 (
1 1 2 2
0 1 −1 1
0
0
1
0
0
−1
1
−2
|
0
0
0
0
) L3-L2 
 
(
1 1 2 2
0 1 −1 1
0
0
0
0
1
−1
0
−2
|
0
0
0
0
) L4+L3 (
1 1 2 2
0 1 −1 1
0
0
0
0
1
0
0
−2
|
0
0
0
0
) 
{
𝑎1 + 𝑎2 +2𝑎3 +2𝑎4 = 0
𝑎2 −𝑎3 +𝑎4 = 0
𝑎3 = 0
−2𝑎4 = 0
 a1=0, a2=0, a3=0, a4=0 
 Linearmente Independente (LI) 
 
C) (1,-1,0,0) (0,1,0,0) (0,0,1,-1) (1,2,1,-2) 
a1(1,-1,0,0) + a2(0,1,0,0) + a3(0,0,1,-1) + a4(1,2,1,-2) = (0,0,0,0) 
 
{
𝑎1 +𝑎4 = 0
−𝑎1 +𝑎2 +2𝑎4 = 0
𝑎3 +𝑎4 = 0
−𝑎3 −2𝑎4 = 0
 (
1 0 0 1
−1 1 0 2
0
0
0
0
1
−1
1
−2
|
0
0
0
0
) L2+L1 
 
(
1 0 0 1
0 1 0 1
0
0
0
0
1
−1
1
−2
|
0
0
0
0
) L4+L3 (
1 0 0 1
0 1 0 1
0
0
0
0
1
0
1
−1
|
0
0
0
0
) 
 
{
𝑎1 +𝑎4 = 0
+𝑎2 +𝑎4 = 0
𝑎3 +𝑎4 = 0
−𝑎4 = 0
 a1=0, a2=0, a3=0, a4=0 
Linearmente Independente! 
 
2) Determinar K para que seja LI: 
{(-1,0,2), (1,1,1), (k,-2,0)} 
a1(-1,0,2) + a2(1,1,1) + a3(k,-2,0) = (0,0,0,0) 
{
−𝑎1 + 𝑎2 +𝑘𝑎3 = 0
𝑎2 −2𝑎3 = 0
2𝑎1 +𝑎2 = 0
 (
−1 1 𝑘
0 1 −2
2 1 0
|
0
0
0
) 2L1+L3 
 (
−1 1 𝑘
0 1 −2
0 3 2𝑘
|
0
0
0
) -3L2+L3 (
−1 1 𝑘
0 1 −2
0 0 6 + 2𝑘
|
0
0
0
) {
−𝑎1 + 𝑎2 +𝑘 = 0
𝑎2 −2𝑎3 = 0
6 + 2𝑘 = 0
 
2K= -6 
K= -3