Assignment 2 - Solution

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Môn: Applied Linear Algebra

Trường: Trường Đại học Quốc tế, Đại học Quốc gia Thành phố Hồ Chí Minh

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Profile Như Hoa
lOMoARcPSD|359747 69
APPLIED LINEAR ALGEBRA
ASSIGNMENT 2
EXERCISE 1:
4 8 −1
𝐴 = [ ], 𝐵 = [
0 1 0
are linearly independent or
not
0 −3
], 𝐶 = [
2 2
5 0
], 𝐷 = [
2 7
−1
],
1
Determine if the four
vectors:
SOLUTION
Assume that α
1
, α
2
, α
3
, α
4
∈ ℝ such that Aα
1
+ Bα
2
+ Cα
3
+ Dα
4
= 0
α
1
[
4
0
8
1] + α
2
[
−1
0 2
0
] + α
3
[
−3
2 2
5
] + α
4
[
0
7
−1
1 ] = [
0
0
0
0]
[4α
1
− α
2
− 3α
3
8α
1
+ 5α
3
− α
4
] = [0 0]
2α
3
+ 7α
4
α
1
+ 2α
2
+ 2α
3
+ α
4
0 0
4α
1
− α
2
− 3α
3
= 0
{ 8α1 + 5α3 − α4 = 0
2α
3
+ 7α
4
= 0 α
1
+
2α
2
+ 2α
3
+ α
4
= 0
4 −1 −3 0 0 1 2 2 1 0
⟨8 0 5 −1|00⟩ ⇒ ⟨80 00 52 −17 |00⟩ 𝑅1 ↔ 𝑅4
0 0 2 7
1 2 2 1 0 4 −1 −3 0 0
1 2 2 1 0
⟨0 −16 −11 −9|0⟩ 𝑅
2
→ −8𝑅
1
+ 𝑅
2
0 0 2 7 0 𝑅
4
→ −4𝑅
1
+ 𝑅
4
0 −9 −11 −4 0
1 2 2 1 0
⟨00 −16−9 −11−11 −9−4|00⟩ 𝑅
3
↔ 𝑅
4
0 0 2 7 0
lOMoARcPSD|359747 69
1 2 2 1 0
0 −16 −11 −9
⟨ |0⟩ 𝑅3 → −9 16 𝑅2 + 𝑅3
0 00
0 0 2 7 0
1
0
2
−11
1
−9
0
⟨0 0 |00⟩ 𝑅4 → 3277 𝑅3 + 𝑅4
0 0 0 0
α
1
= 0
{α
2
= 0
α
3
= 0 α
4
= 0
The equation has a unique solution (0,0,0,0) Linear independent
EXERCISE 2:
a = [5 1 2], b = [7 2 6] and c = [9 4 − 8] are linearly independent
Determine if the three vectors
or linearly dependent.
SOLUTION
Assume that α
1
, α
2
, α
3
, α
4
∈ ℝ such that aα
1
+ bα
2
+ cα
3
= 0
α
1
[5 1 2] + α
2
[7 2 6] + α
3
[9 4 −8] = [0 0 0]
[5α1 +7α2 +9α3 α1+2α2 +4α3 2α1+6α2 -8α3] = [0 0 0]
5α
1
+ 7α
2
+ 9α
3
= 0
{ α
1
+ 2α
2
+ 4α
3
= 0
2α
1
+ 6α
2
− 8α
3
= 0
5 7 9 0 1 2 4 0
⟨1 2 4 |0⟩ ⇒ ⟨5 7 9|0⟩ 𝑅
1
↔ 𝑅
2
2 6 −8 0 2 6 8 0
⟨10 −32 −114 |00⟩ 𝑅2 → −5𝑅1 + 𝑅2
0 2 −16 0 𝑅3 → −2𝑅1 + 𝑅2
lOMoARcPSD|359747 69
1 2 4 0
⟨0 −3 −11|0⟩
0 0 0
α
1
= 0
{α
2
= 0 The equation has a unique solution (0,0,0,0) Linear independent
α
3
= 0
EXERCISE 3:
v
1
= (1 , 1 , 2) , v
2
= (1 , 1 , 0) , v
3
= (0 , 2 , −3) .
{ v
1
, v
2
, v
3
} forms a basis of ℝ
3
.
b) If the answer of the above question is yes, find the coordinates of u = (4 , 8 , −9) with
respect
Given
(a) Determine
whether
( to { v
1
, v
2
,
v
3
}
SOLUTION
a)
Dim(ℝ
3
) = 3, { v
1
, v
2
, v
3
} contains 3 vectors (1)
Assume that α
1
, α
2
, α
3
, α
4
∈ ℝ such that v
1
α
1
+ v
2
α
2
+ v
3
α
3
= 0
1 1 0 0
α
1
[1] + α
2
[1] + α
3
[ 2 ] = [0]
2 0 −3 0
α
1
+ α
2
= 0
{α
1
+ α
2
+ 2α
3
= 0
2α
1
− 3α
3
= 0
1 1 0 0
⟨1 1 2 |0⟩
2 0 −3 0
⟨10 10 02 |00⟩ 𝑅𝑅32 → −2𝑅→ −𝑅11++𝑅𝑅23
0 −2 −3 0
1 1 0 0
⟨0 −2 −3|0⟩ 𝑅
2
↔ 𝑅
3
lOMoARcPSD|359747 69
0 0 2 0
α
1
= 0
{α
2
= 0 The equation has a unique solution (0,0,0,0) α
3
= 0
{ v
1
, v
2
, v
3
} are linearly independent (2)
From (1),(2) { v
1
, v
2
, v
3
} forms a basis of ℝ
3
b)
Assume that α
1
, α
2
, α
3
, α
4
∈ ℝ such that v
1
α
1
+ v
2
α
2
+ v
3
α
3
= u
1 1 0 4
α
1
[1] + α
2
[1] + α
3
[ 2 ] = [ 8 ]
2 0 −3 −9
α
1
+ α
2
= 4
{α
1
+ α
2
+ 2α
3
= 8
2α
1
− 3α
3
= −9
1 1 0 4
⟨1 1 2 | 8 ⟩
2 0 −3 −9
⟨10 10 02 | 44 ⟩ 𝑅𝑅32→ −2𝑅→ −𝑅11++𝑅𝑅23
0 −2 −3 −17
1 1 0 4
⟨0 −2 −3|−17⟩ 𝑅
2
↔ 𝑅
3
0 0 2 4
−3
α1 =
2
{α
2
= 11
2
α
3
= 2
−3 11
Hence, 𝑢 = 2 𝑣
1
+
2
𝑣
2
+ 2𝑣
3
Therefore,
𝑢
⁄{𝒗
𝟏
, 𝒗
2
, 𝒗
𝟑
} = ( , , 2)
| 1/4
| | Tải xuống

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lOMoARcPSD|359 747 69 APPLIED LINEAR ALGEBRA ASSIGNMENT 2 EXERCISE 1: Determine if the four 4 8 −1 vectors: 𝐴 = [ ], 𝐵 = [ 0 1 0 0 −3 5 0 −1 are linearly independent or ], 𝐶 = [ ], 𝐷 = [ ], not 2 2 2 7 1 SOLUTION
Assume that α1, α2, α3, α4 ∈ ℝ such that Aα1 + Bα2 + Cα3 + Dα4 = 0
 α1 [40 81] + α2 [−10 20] + α3 [−32 25] + α4 [07 −11 ] = [00 00]  [4α1 − α2 − 3α3 8α1 + 5α3 − α4 ] = [0 0] 2α3 + 7α4 α1 + 2α2 + 2α3 + α4 0 0 4α1 − α2 − 3α3 = 0  { 8α1 + 5α3 − α4 = 0 2α3 + 7α4 = 0 α1 + 2α2 + 2α3 + α4 = 0 4 −1 −3 0 0 1 2 2 1 0  ⟨8 0 5 −1|00⟩ ⇒ ⟨80 00 52 −17 |00⟩ 𝑅1 ↔ 𝑅4 0 0 2 7 1 2 2 1 0 4 −1 −3 0 0 1 2 2 1 0  ⟨0
−16 −11 −9|0⟩ 𝑅2 → −8𝑅1 + 𝑅2 0
0 2 7 0 𝑅4 → −4𝑅1 + 𝑅4 0 −9 −11 −4 0 1 2 2 1 0  ⟨00 −16−9 −11−11
−9−4|00⟩ 𝑅3 ↔ 𝑅4 0 0 2 7 0 lOMoARcPSD|359 747 69 1 2 2 1 0 0 −16 −11 −9  ⟨ |0⟩
𝑅3 → −9 16 𝑅2 + 𝑅3 0 00 0 0 2 7 0 1 2 2 1 0 −16 −11 −9 0  ⟨0 0
|00⟩ 𝑅4 → 3277 𝑅3 + 𝑅4 0 0 0 0 α1 = 0  {α2 = 0 α3 = 0 α4 = 0
 The equation has a unique solution (0,0,0,0)  Linear independent EXERCISE 2:
Determine if the three vectors or linearly dependent.
a = [5 1 2], b = [7 2 6] and c = [9 4 − 8] are linearly independent SOLUTION
Assume that α1, α2, α3, α4 ∈ ℝ such that aα1 + bα2 + cα3 = 0
 α1 [5 1 2] + α2 [7 2 6] + α3 [9 4 −8] = [0 0 0]
 [5α1 +7α2 +9α3 α1+2α2 +4α3 2α1+6α2 -8α3] = [0 0 0] 5α1 + 7α2 + 9α3 = 0  { α1 + 2α2 + 4α3 = 0 2α1 + 6α2 − 8α3 = 0 5 7 9 0 1 2 4 0  ⟨1 2 4 |0⟩ ⇒ ⟨5 7 9|0⟩ 𝑅1 ↔ 𝑅2 2 6 −8 0 2 6 8 0  ⟨10
−32 −114 |00⟩ 𝑅2 → −5𝑅1 + 𝑅2 0
2 −16 0 𝑅3 → −2𝑅1 + 𝑅2 lOMoARcPSD|359 747 69 1 2 4 0  ⟨0 −3 −11|0⟩ 0 0 0 α1 = 0
 {α2 = 0  The equation has a unique solution (0,0,0,0)  Linear independent α3 = 0 EXERCISE 3: Given
v1 = (1 , 1 , 2) , v2 = (1 , 1 , 0) , v3 = (0 , 2 , −3) . (a) Determine whether
{ v1 , v2, v3 } forms a basis of ℝ3.
( to { v1 b , ) v I 2 f
, the answer of the above question is yes, find the coordinates of u = (4 , 8 , −9) with v3 } respect SOLUTION
a) Dim(ℝ3) = 3, { v1 , v2, v3 } contains 3 vectors (1)
Assume that α1, α2, α3, α4 ∈ ℝ such that v1α1 + v2α2 + v3α3 = 0 1 1 0 0
 α1 [1] + α2 [1] + α3 [ 2 ] = [0] 2 0 −3 0 α1 + α2 = 0  {α1 + α2 + 2α3 = 0 2α1 − 3α3 = 0 1 1 0 0  ⟨1 1 2 |0⟩ 2 0 −3 0  ⟨10 10
02 |00⟩ 𝑅𝑅32 → −2𝑅→ −𝑅11++𝑅𝑅23 0 −2 −3 0 1 1 0 0  ⟨0 −2 −3|0⟩ 𝑅2 ↔ 𝑅3 lOMoARcPSD|359 747 69 0 0 2 0 α1 = 0
 {α2 = 0  The equation has a unique solution (0,0,0,0) α3 = 0
 { v1 , v2, v3 } are linearly independent (2)
From (1),(2)  { v1 , v2, v3 } forms a basis of ℝ3
b) Assume that α1, α2, α3, α4 ∈ ℝ such that v1α1 + v2α2 + v3α3 = u 1 1 0 4
 α1 [1] + α2 [1] + α3 [ 2 ] = [ 8 ] 2 0 −3 −9 α1 + α2 = 4  {α1 + α2 + 2α3 = 8 2α1 − 3α3 = −9 1 1 0 4  ⟨1 1 2 | 8 ⟩ 2 0 −3 −9  ⟨10 10
02 | 44 ⟩ 𝑅𝑅32→ −2𝑅→ −𝑅11++𝑅𝑅23 0 −2 −3 −17 1 1 0 4  ⟨0 −2 −3|−17⟩ 𝑅2 ↔ 𝑅3 0 0 2 4 −3 α1 = 2  {α2 = 112 α3 = 2 −3 11
Hence, 𝑢 = 2 𝑣1 + 2 𝑣2 + 2𝑣3
Therefore, 𝑢⁄{𝒗𝟏 , 𝒗2, 𝒗𝟑} = ( , , 2)
Document Outline

  • APPLIED LINEAR ALGEBRA
    • SOLUTION
    • SOLUTION (1)
    • SOLUTION (2)