Suppose that v1=(2,1,0,3), v2=(3,-1,5,2), and v3=(-1,0,2,1). Which of the following vectors are in span ? Explain your work. A) (2, 3, -7, 3) B) (0, 0, 0 ,0) C) (1, 1, 1, 1) D) (-4, 6, -13, 4)
Suppose that v1=(2,1,0,3), v2=(3,-1,5,2), and v3=(-1,0,2,1). Which of the following vectors are in span ? Explain your work. A) (2, 3, -7, 3) B) (0, 0, 0 ,0) C) (1, 1, 1, 1) D) (-4, 6, -13, 4)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Question:
Suppose that v1=(2,1,0,3), v2=(3,-1,5,2), and v3=(-1,0,2,1). Which of the following
A) (2, 3, -7, 3)
B) (0, 0, 0 ,0)
C) (1, 1, 1, 1)
D) (-4, 6, -13, 4)
My Comments:
I've attempted this problem by setting the vectors equal to the multiple choice answers. I placed them in augmented form and took the reduced row echelon form. However, so far A and B are coming out equal making me conclude that A and B are in span. Am I solving this correctly? Did I get it right? (I've attached my work as well.)
![(.... CONTINUED)
A.
DA=
B.
V₁ = (2, 1, 0₂3), V₂ = (3₁-1₁ 5₂2), V₂ = (-1₂ 0₂ 2₂1)
D
D
2 3
1
95 2
32
1
✓
D
D
(..)
DIC=
2x + 3y
2 = 2
2 (2) + 3(-1) - (-1)=2
4 + (-3) + 1 = 2
1 + 1 = 2
2 = 2²
B=
-1
N
3
2
2 3 -1 O
O S 2
1
↓
• 2x+3y-z = 0
2(0)+ 3 (0)-6=0
0=0
2 3
1
S
3 2
-1
O
2
-
3
-7
3
· 2x + 3y - 221
► 2 (0) + 3(0)-0=1
071
6
RREF(A) =
1 x - y = 3
2-(-1)=3
6 → RREF(B) =
6
3 =3 ✓
0-0
•x-y=0
0=00
1
1 → RREF(C)=
J
• X-y = 1
0-0 =1
071
(CONTINUED
0
O
A
O
0 O
1
O
6 6
O
0 1 O
O 0
0
ON
O
1
100
0
Sy + 22 =-7
S(-1) + 2(²-1) = -7
-5 +-2=-7
-72-7✓
• Sy +22=0
5(0) + 2(0) = 0
0 =0°
O
1
O
Q
5y-2z=1
(5(0)-2(0)=1
6
6
0+x=0
9 x ²0
02:0
1] → 01
NEXT
0x = 6
⇒ y = 0
01
2
N
O
→x=2
→ Y = -1
Z=-1
+ 3x + 3y + 2 =3
3 (2) +2(-1)+(²1) = 3
6 + (-2) + (-4)=3,
1²6 =3 = 3
6-3
3 (0) +2(0) +0 = 0
0 =0✓
3x + 2y + z = 1
3(0) +2(0) + (0) = 1
01
PAGE....)
12
13](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F48ae96ba-8296-4f48-84d9-a65ba8f4727e%2F9d4a5a78-5c29-487a-9ee0-69bce8d0c455%2F1mjtgvl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(.... CONTINUED)
A.
DA=
B.
V₁ = (2, 1, 0₂3), V₂ = (3₁-1₁ 5₂2), V₂ = (-1₂ 0₂ 2₂1)
D
D
2 3
1
95 2
32
1
✓
D
D
(..)
DIC=
2x + 3y
2 = 2
2 (2) + 3(-1) - (-1)=2
4 + (-3) + 1 = 2
1 + 1 = 2
2 = 2²
B=
-1
N
3
2
2 3 -1 O
O S 2
1
↓
• 2x+3y-z = 0
2(0)+ 3 (0)-6=0
0=0
2 3
1
S
3 2
-1
O
2
-
3
-7
3
· 2x + 3y - 221
► 2 (0) + 3(0)-0=1
071
6
RREF(A) =
1 x - y = 3
2-(-1)=3
6 → RREF(B) =
6
3 =3 ✓
0-0
•x-y=0
0=00
1
1 → RREF(C)=
J
• X-y = 1
0-0 =1
071
(CONTINUED
0
O
A
O
0 O
1
O
6 6
O
0 1 O
O 0
0
ON
O
1
100
0
Sy + 22 =-7
S(-1) + 2(²-1) = -7
-5 +-2=-7
-72-7✓
• Sy +22=0
5(0) + 2(0) = 0
0 =0°
O
1
O
Q
5y-2z=1
(5(0)-2(0)=1
6
6
0+x=0
9 x ²0
02:0
1] → 01
NEXT
0x = 6
⇒ y = 0
01
2
N
O
→x=2
→ Y = -1
Z=-1
+ 3x + 3y + 2 =3
3 (2) +2(-1)+(²1) = 3
6 + (-2) + (-4)=3,
1²6 =3 = 3
6-3
3 (0) +2(0) +0 = 0
0 =0✓
3x + 2y + z = 1
3(0) +2(0) + (0) = 1
01
PAGE....)
12
13
Transcribed Image Text:EX
D.)
▷
▷
D
▷
▷
▷
D
▷
D
▷
(.... CONTINUED)
▷
▷
D
D=
1
0
3
3-
nd
• 2x + 3y - 2 =4
2 (0) + 3(0) -0 =4
074
S
-1
0
2
1
2
4
1
~13 → RREP(D) = 0
6
-4
• x-y = -13
D
6-0=-13
07-13
CHOICES "A" AND
THEY BOTH
HAVE A
OF V₁, V₂, AND V₂, CHOILES
SO, THE ANSWER
CHOICES "C" AND "D" RESULT
LINEAR COMBINATIONS
of
TO
23
вотн
SOLUTION, THEY
A
0
IN
V₁ V ₂9 V₂
QUESTION
0
0
RESULY
Q
1
NO
AND
O
0
0
Sy +22= 6 • 3x +2y + z = -4
1
5 (0) + 2(0) = 6
3(0) + 2(0)+ 0 = -4
66
0 £-4
1
IN A SOLUTION. BECAUSE
ARE
BOTH
AND "B" ARE IN
THREE
⇒ x = 0
0⇒x=0
0-200
→ 01
1
ARE
IS
SOLUTION, SO THEY
IN SPAN.
LINEAR COMBINATIONS
NOT
SPAN,
11
11
A AND B.
ARE
NOT
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