Q1: 1 2 3 1. Matrix A =| 2 3 4 is symmetric. 3 3 4 2 1 2. If { 1 1 } is a basis in R 2 3. { 0 is an orthogonal basis in R3 2 spans R2. 4 1) 1 5. {- is an orthonormal basis in R³ 0) V2 0 0 2 6. det 0 3 2 =7 1 4 4.
Q1: 1 2 3 1. Matrix A =| 2 3 4 is symmetric. 3 3 4 2 1 2. If { 1 1 } is a basis in R 2 3. { 0 is an orthogonal basis in R3 2 spans R2. 4 1) 1 5. {- is an orthonormal basis in R³ 0) V2 0 0 2 6. det 0 3 2 =7 1 4 4.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![Q1:
(1 2 3
1. Matrix A =| 2
3
4 is symmetric.
3 3
4
2
2
1
2. If { 1
1
1 } is a basis in R³
2
2
2
2
3. { 0
1
is an orthogonal basis in R³
1
4.
2
2
spans R².
1
5. {-
V2(0)2
1
is an orthonormal basis in R³
0 0 2
6. det 0
3 2 =7
1
4
Q2:
1. The Nullspace is defined only for square matrices.
2. C, and N, have the same dimension called rankA.
3. dim C,+dim R, = ncolumns of A
4. C = R,
5. If Basis(H) =0 then Basis(H-)=
1
6. If det A = 0 then dim N, = 0 ▪
Q3:
1. If (v,,v,) = 0 then || v ||° +|| v ||°=|| v, + v,||
2. (v,,v;) =|| v ||
3. If V = v, x v, then (V,v,)= (V,v,)=1.
4. || av ||= a || v |
V2
5. If (v,,v,) = 0 then v, || v,
6. if v, || V2 then v,×v, = 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc9009fbf-af52-4a96-9dc7-af8562671cf3%2Fc6531645-43f0-49a6-8116-0680bd9809ad%2Fazvkeh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Q1:
(1 2 3
1. Matrix A =| 2
3
4 is symmetric.
3 3
4
2
2
1
2. If { 1
1
1 } is a basis in R³
2
2
2
2
3. { 0
1
is an orthogonal basis in R³
1
4.
2
2
spans R².
1
5. {-
V2(0)2
1
is an orthonormal basis in R³
0 0 2
6. det 0
3 2 =7
1
4
Q2:
1. The Nullspace is defined only for square matrices.
2. C, and N, have the same dimension called rankA.
3. dim C,+dim R, = ncolumns of A
4. C = R,
5. If Basis(H) =0 then Basis(H-)=
1
6. If det A = 0 then dim N, = 0 ▪
Q3:
1. If (v,,v,) = 0 then || v ||° +|| v ||°=|| v, + v,||
2. (v,,v;) =|| v ||
3. If V = v, x v, then (V,v,)= (V,v,)=1.
4. || av ||= a || v |
V2
5. If (v,,v,) = 0 then v, || v,
6. if v, || V2 then v,×v, = 0
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