1. True or False (T/F) questions 1 -2 14 7 2 5 8 -2 -2 Let A = [a, a2 a3] and B = [b, b2 b3] = (Note: bị + b3 = 2b2.) %3D 3 -8 3 6 9 8 -4 -18 (a) T F: If u and v are both elements from Col A then so is u +v. (b) T F : Looking at the middle column, it's obvious that E Col A 16 (c) T E Nul A (d) T F : The matrix B is invertible. (e) T F: The vectors b1, b2 and bz are linear dependent. (f) T F: The dimension of the subspace W = Span (b, b2, b3) is 3. (g) T F : The vectors b1, b, and b; form a basis for R. (h) T F: Span (b, b2, b3) = Span (b,, b2) (i) T F: The dimension of the subspace Span (b1, b2, b3) is 2. (i) T F : rank(B) = 2. (k) T F: rank(A) = 4. (1) T F: Nul (A) C R. (m) T F: Col (A) CR*. (n) T F: Nul (B) is a subspace spanned by only 1 single vector. (o) T F: dim(P3) = 3. (p) T F : The set {r – 1, t-1,t+1, t² +1} is a spanning set for the vector space P3 (q) T F : 8. is in Col B 12 (r) T F: is an element of Col (A) (s) T F : is an element of Col (B) 3 (t) T F: -4 E Nul (B) (u) T F: nullity(B) = 2 (v) T F: The number of pivot columns of a matrix equals the dimension of its column space. w) T F: The dimension of Ps is 9. (x) T F: If W is a subspace of R' then its dimension is less than 3. (y) T F: If 5 vectors span the subspace W then dim(W) = 5. (z) T F: nullity(B) counts the number of vectors in a basis of Nul B.
1. True or False (T/F) questions 1 -2 14 7 2 5 8 -2 -2 Let A = [a, a2 a3] and B = [b, b2 b3] = (Note: bị + b3 = 2b2.) %3D 3 -8 3 6 9 8 -4 -18 (a) T F: If u and v are both elements from Col A then so is u +v. (b) T F : Looking at the middle column, it's obvious that E Col A 16 (c) T E Nul A (d) T F : The matrix B is invertible. (e) T F: The vectors b1, b2 and bz are linear dependent. (f) T F: The dimension of the subspace W = Span (b, b2, b3) is 3. (g) T F : The vectors b1, b, and b; form a basis for R. (h) T F: Span (b, b2, b3) = Span (b,, b2) (i) T F: The dimension of the subspace Span (b1, b2, b3) is 2. (i) T F : rank(B) = 2. (k) T F: rank(A) = 4. (1) T F: Nul (A) C R. (m) T F: Col (A) CR*. (n) T F: Nul (B) is a subspace spanned by only 1 single vector. (o) T F: dim(P3) = 3. (p) T F : The set {r – 1, t-1,t+1, t² +1} is a spanning set for the vector space P3 (q) T F : 8. is in Col B 12 (r) T F: is an element of Col (A) (s) T F : is an element of Col (B) 3 (t) T F: -4 E Nul (B) (u) T F: nullity(B) = 2 (v) T F: The number of pivot columns of a matrix equals the dimension of its column space. w) T F: The dimension of Ps is 9. (x) T F: If W is a subspace of R' then its dimension is less than 3. (y) T F: If 5 vectors span the subspace W then dim(W) = 5. (z) T F: nullity(B) counts the number of vectors in a basis of Nul B.
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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![1. True or False (T/F) questions
1 -2
14 7
2 5 8
-2 -2
Let A = [a, a2 a3]
and B = [b, b2 b3] =
(Note: bị + b3 = 2b2.)
%3D
3
-8
3 6 9
8 -4 -18
(a) T F: If u and v are both elements from Col A then so is u +v.
(b) T F : Looking at the middle column, it's obvious that
E Col A
16
(c) T
E Nul A
(d) T F : The matrix B is invertible.
(e) T F: The vectors b1, b2 and bz are linear dependent.
(f) T F: The dimension of the subspace W = Span (b, b2, b3) is 3.
(g) T F : The vectors b1, b, and b; form a basis for R.
(h) T F: Span (b, b2, b3) = Span (b,, b2)
(i) T F: The dimension of the subspace Span (b1, b2, b3) is 2.
(i) T F : rank(B) = 2.
(k) T F: rank(A) = 4.
(1) T F: Nul (A) C R.
(m) T F: Col (A) CR*.
(n) T F: Nul (B) is a subspace spanned by only 1 single vector.
(o) T F: dim(P3) = 3.
(p) T F : The set {r – 1, t-1,t+1, t² +1} is a spanning set for the vector space
P3
(q) T F :
8.
is in Col B
12
(r) T F:
is an element of Col (A)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F65713083-5849-4086-9415-4b0086144125%2F661deab3-e048-41fd-a440-14494b907316%2Fsvz9wbh.jpeg&w=3840&q=75)
Transcribed Image Text:1. True or False (T/F) questions
1 -2
14 7
2 5 8
-2 -2
Let A = [a, a2 a3]
and B = [b, b2 b3] =
(Note: bị + b3 = 2b2.)
%3D
3
-8
3 6 9
8 -4 -18
(a) T F: If u and v are both elements from Col A then so is u +v.
(b) T F : Looking at the middle column, it's obvious that
E Col A
16
(c) T
E Nul A
(d) T F : The matrix B is invertible.
(e) T F: The vectors b1, b2 and bz are linear dependent.
(f) T F: The dimension of the subspace W = Span (b, b2, b3) is 3.
(g) T F : The vectors b1, b, and b; form a basis for R.
(h) T F: Span (b, b2, b3) = Span (b,, b2)
(i) T F: The dimension of the subspace Span (b1, b2, b3) is 2.
(i) T F : rank(B) = 2.
(k) T F: rank(A) = 4.
(1) T F: Nul (A) C R.
(m) T F: Col (A) CR*.
(n) T F: Nul (B) is a subspace spanned by only 1 single vector.
(o) T F: dim(P3) = 3.
(p) T F : The set {r – 1, t-1,t+1, t² +1} is a spanning set for the vector space
P3
(q) T F :
8.
is in Col B
12
(r) T F:
is an element of Col (A)
Transcribed Image Text:(s) T F :
is an element of Col (B)
3
(t) T F:
-4
E Nul (B)
(u) T F: nullity(B) = 2
(v) T F: The number of pivot columns of a matrix equals the dimension of its
column space.
w) T F: The dimension of Ps is 9.
(x) T F: If W is a subspace of R' then its dimension is less than 3.
(y) T F: If 5 vectors span the subspace W then dim(W) = 5.
(z) T F: nullity(B) counts the number of vectors in a basis of Nul B.
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