- (-1,2,1) 5. (3,1,1) yo (5,3,4) 18.mp (201²) ( 165/
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
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Plz last part only
Solution of ist three is given
Transcribed Image Text:- (-1,2,1)
5. (3,1,1)
y (5,3,4)
For which wales of L
(201²) (
19
6) Spon $1,9
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Sp- $1,95
![↑
Given - vector V₁ = {1,2,1}; v₂ = {3,₁1,1}; v₂ = { 5, 3, k}
To find for which values of 'k';
{V₁, V₂, V3} span R³.
V₂ € Spanv,, V₂3
Step 2: Calculation
Solution a) let A be 3x3 matrix with given vectors as columns.
(b)
Then A =
Since dim (R³) = 3
these vectors are linearly independent.
te
Now perform row operations on A and see vectors are L.I.
3 5
R3 R3-Re
- 3 5
0 7 13
_04 K+5
-1 3 5
R₁+R₂+3R₁
2
1
3
1 * ] R₂ + R₂ + R₂
13
K-17
This means that columns of A will be linearly independent if
7
k#1
Thus, the [v₁, V₂, Vs} span
2
ie
Augment
IF V₂ € span {V₁, V₂y then there exists constants & C₂ such
that
N₁ = GV₁ + G₂V₂
2
3
5
135
21 3
50;
1 3
So;
Step 3: Calculation 2
Thus ;
k
the given set of vectors will span R³ if
;
C₁
13 5
1 3
Solution
5 3
K
B-4
k
5
-
R₂ + R₂+2R₁
Rg Y Rz + Rp
8(A)
ⒸV₂ & span [V₁, V3}.
Here
(AIB) =
Then system has unique solution.
we have n=2
S (A) = 2
So for S(Alb) = 2; R = 17
7
3
for k=14
7
5
713
So the system Ac=b is consistent.
V₂ € span {V₁, V₂3 then k=1#
7
; perform row operations and solve for
G & C₂.
17-1K = 0
O
0 4 K+5
R₂ + R₂+2R₁
R₂ + R₂ + R
This is system Ac=b.
R₂ R₂ - 1/R2
0
- from
ⒸTF V₂ € span {V₁, V₂} then there exists constants ₁&₂ such
that V₂ = GV₁ + C₂ V3.
10-1
Now augment [Alb] : Perform now operation & solve for Cis
En in-number of variables
O
5 3
13 3
k+5 4
Now System S[Alb]= 5 [A] = 2 iff 17-7K = 0
13
je k = 17
4
7.
ها ترا
V₂ span {V₁, V₂] then k = 1 #
7
o t
This is system Ac=b.
TIE
5
13
O 0 K-17
F
R3-R₁ - (K+5) R₂
T
3
013 I
00
H 5
17-IK
13](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fed5788d2-a0b4-4eb1-998c-a4248e13736e%2F3696e0f9-6591-42f5-b470-ba3c2b1161d7%2Fi1bvqqh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:↑
Given - vector V₁ = {1,2,1}; v₂ = {3,₁1,1}; v₂ = { 5, 3, k}
To find for which values of 'k';
{V₁, V₂, V3} span R³.
V₂ € Spanv,, V₂3
Step 2: Calculation
Solution a) let A be 3x3 matrix with given vectors as columns.
(b)
Then A =
Since dim (R³) = 3
these vectors are linearly independent.
te
Now perform row operations on A and see vectors are L.I.
3 5
R3 R3-Re
- 3 5
0 7 13
_04 K+5
-1 3 5
R₁+R₂+3R₁
2
1
3
1 * ] R₂ + R₂ + R₂
13
K-17
This means that columns of A will be linearly independent if
7
k#1
Thus, the [v₁, V₂, Vs} span
2
ie
Augment
IF V₂ € span {V₁, V₂y then there exists constants & C₂ such
that
N₁ = GV₁ + G₂V₂
2
3
5
135
21 3
50;
1 3
So;
Step 3: Calculation 2
Thus ;
k
the given set of vectors will span R³ if
;
C₁
13 5
1 3
Solution
5 3
K
B-4
k
5
-
R₂ + R₂+2R₁
Rg Y Rz + Rp
8(A)
ⒸV₂ & span [V₁, V3}.
Here
(AIB) =
Then system has unique solution.
we have n=2
S (A) = 2
So for S(Alb) = 2; R = 17
7
3
for k=14
7
5
713
So the system Ac=b is consistent.
V₂ € span {V₁, V₂3 then k=1#
7
; perform row operations and solve for
G & C₂.
17-1K = 0
O
0 4 K+5
R₂ + R₂+2R₁
R₂ + R₂ + R
This is system Ac=b.
R₂ R₂ - 1/R2
0
- from
ⒸTF V₂ € span {V₁, V₂} then there exists constants ₁&₂ such
that V₂ = GV₁ + C₂ V3.
10-1
Now augment [Alb] : Perform now operation & solve for Cis
En in-number of variables
O
5 3
13 3
k+5 4
Now System S[Alb]= 5 [A] = 2 iff 17-7K = 0
13
je k = 17
4
7.
ها ترا
V₂ span {V₁, V₂] then k = 1 #
7
o t
This is system Ac=b.
TIE
5
13
O 0 K-17
F
R3-R₁ - (K+5) R₂
T
3
013 I
00
H 5
17-IK
13
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