MAT 1341C — Final Examination
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MAT 1341C — Final Examination
1. Let B = ┐ and consider the subset U = nA ÷ M22 (R) : BA = AB{. Which of the following statements is true?
Cross (X) the box for the correct answer:
U is not a subspace of the vector space M22 (R) of 2 × 2 matrices U is a subspace of M22 (R), and dim(U) = 0
U is a subspace of M22 (R), and dim(U) = 1
U is a subspace of M22 (R), and dim(U) = 2 — Correct
U is a subspace of M22 (R), and dim(U) = 3
U is a subspace of M22 (R), and dim(U) = 4
Solution: Let A = ┌c(a)
and
d(b)┐ . Then
BA = ┐ ┌c(a) d(b)┐ = ┐
AB = ┌c(a) d(b)┐ = ┐ .
Hence, if AB = BA, then c = 0 and b _ d = _a. Thus
U = ┐ : a, b ÷ R} = span {┌0(1) 1(0)┐ , ┌0(0) 1(1)┐} ,
and U is a subspace of M22 (R) of dimension 2.
2. Which two of the following statements are true? [1pt]
I. Every homogeneous system of linear equations has infinitely many solutions.
II. If nu, v, w{ is a linearly independent set of vectors in R3 , then nu, v, w{ is a basis for R3 .
III. If A and B are invertible matrices, then (AB)6| = A6|B 6| .
IV. n2 sin x, cos 2x{ is a linearly independent set of vectors in the vector space of all real-valued functions.
Cross (X) the box for the correct answer:
I. and II.
I. and III.
I. and IV.
II. and III.
II. and IV. — Correct
III. and IV.3. Which of the following are subspaces of R3 ?
U = n(x, y, z) ÷ R3 : x _ 2y + z = 0{
V = n(x, y, z) ÷ R3 : x _ 5z = 0{
W = n(x, y, x + 1) ÷ R3 : x, y ÷ R{
X = n(x + y, y, x _ 2y) ÷ R3 : x, y ÷ R{
Cross (X) the box for the correct answer:
A Only U and V
B Only U and W
Only W and X
D Only U , V and W
E Only U , V and X — Correct
F Only U , W and X
Solution: Observe that W is not a subspace as it does not contain the zero vector. U and V each represent a plane through the origin, so they are subspaces, while X = spann(1, 0, 1), (1, 1, _2){.
4. Consider a non-homogeneous system of 13 linear equations in 9 following three Yes/No questions (in order):
· Can this system have a unique solution?
· Can this system be inconsistent?
· Can this system have infinitely many solutions?
Cross (X) the box for the correct answer:
Yes, Yes, Yes — Correct
Yes, Yes, No
C No, Yes, Yes
D Yes, No, Yes
No, No, Yes
F Yes, No, No
variables,
and answer the [1pt]
5. Consider the invertible matrix A = . The second row of A6| is: [1pt]
Cross (X) the box for the correct answer:
[0 1 0] — Correct
[1 0 1]
[0 0 1]
[_1 0 1]
[1 0 2]
[1 0 _ 1]
Solution: Using the matrix inverse algorithm:
┌┐ ┌┐ ┌┐
Since we know A is invertible and the second row of the augmented matrix will not change any more during row reduction, we can conclude that the second row of A6| is [0 1 0].
6. Let A be a 6 × 6 matrix. Which of the following statements are true? [1pt]
I. If rank(A) = 2, there are 2 parameters in the general solution of the system Ax = 0.
II. If rank(A) = 2, there are 4 parameters in the general solution of the system Ax = 0.
III. If rank(A) = 4, then the system Ax = 0 has infinitely many solutions.
IV. If rank(A) = 4, then the system Ax = 0 has a unique solution.
Cross (X) the box for the correct answer:
I. only
II. only
I. and III.
I. and IV.
II. and III. — Correct
II. and IV.
7. Find the polar form of the complex number . ^626i .
Hint: use trigonometric formulas on Page ??.
Cross (X) the box for the correct answer:
3(cos + i sin )
3(cos _ i sin )
(cos + i sin ) — Correct (cos _ i sin )
cos + i sin
cos + i sin
Solution: We have
= 2++^2^2 = T = . = ei
8. We know that vectors u| = (0, 4, 2), u2 = (1, _1, 2) and u3 = (5, 1, _2) form an orthogonal basis for R3 . Hence any vector v ÷ R3 can be expressed in a unique way as a linear combination v = a|u| + a2u2 + a3u3 , where a| , a2 , a3 ÷ R are the coordinates of v with respect to the basis nu| , u2 , u3 {.
[1pt]
|
2
|
_
C |
D |
E |
F |
Solution: The Fourirer coefficient a2 is given by the formula
a2 = =
9. Let A be an n × n matrix.
Which one of the statements below is not equivalent to the statement
“The matrix A is invertible”?
Cross
(X) the box for the correct answer: The rows of A are linearly independent. det(A) 0.
The reduced row echelon form of A is the zero matrix. — Correct
The rank of A is n.
The columns of A form a basis for Rn .
The homogeneous system Ax = 0 has a unique solution.
10. Let A = be a matrix with det(A) = 2. (Here, a, b, c, d, e, f, g, h, i are real numbers.) [1pt]
Using elementary row and/or column operations, compute the determinant of the matrix
┌ ┐
( _2c _2f _2i )
Cross (X) the box for the correct answer:
A 12
B -12 — Correct
18
D -18
E 64
F -64
Solution:
det ┌
( _2c
e f hi(2)i = 3 . det b c(2)c e f(2)f hi(2)i
= (_2) . 3 . det ┌ ┐
( c f i )
= (_6) . det = (_6) . det = _12
11. Let P2 = nax2 + bx + c : a, b, c ÷ R{ be the vector space of polynomial functions of degree at most 2.
(a) Explain why B = n1 _ x, x + x2 , 1 _ x2 { is a basis for P2 . [1pt]
Solution: We first show that B is a linearly independent set. Suppose
a(1 _ x) + b(x + x2 ) + c(1 _ x2 ) = 0
for some a, b, c ÷ R. Then a + c = 0, _a + b = 0 and b _ c = 0. Solving this system of linear equations gives the unique solution a = b = c = 0. Hence B is linearly independent.
From class, we know that the dimension of P2 is 3, which is also the size of B . Hence B is a basis of P2 .
(b) Extend the basis B for P2 from (a) to a basis for P3 = nax3 +bx2 +cx+d : a, b, c, d ÷ R{.
Fully justify your answer.
[1pt]
ANSWER — an extended basis for P2 :
Solution: Adjoin x3 and show that the resulting set B\ = n1 _ x, x + x2 , 1 _ x2 , x3 { is linearly independent:
a(1 _ x) + b(x + x2 ) + c(1 _ x2 ) + dx3 = 0
yields the system of linear equations a + c = 0, _a + b = 0, b _ c = 0, and d = 0 with the unique solution a = b = c = d = 0. Hence B\ is linearly independent.
From class, we know that the dimension of P3 is 4, which is also the size of B\ . Hence B\ is a basis of P3 .
12. Let W = n(x, y, z, w) ÷ R4 : x _ z = 0 and y + 2z = 0{.
(a) Find a basis for the subspace W .
ANSWER:
Solution: We have
W = n(x, y, z, w) ÷ R4 : x = z and y = _2z{
= n(z, _2z, z, w) : z, w ÷ R{ = spann(1, _2, 1, 0), (0, 0, 0, 1){.
The set S = n(1, _2, 1, 0), (0, 0, 0, 1){ is clearly linearly independent, so it is a basis for W .
(b) Apply the Gram-Schmidt algorithm to find an orthogonal basis for the subspace [2pt] U = spann(1, 0, 1, 0), (0, 1, 0, 1), (1, 1, 0, 0){.
ANSWER:
Solution: Let u| = (1, 0, 1, 0), u2 = (0, 1, 0, 1) and u3 = (1, 1, 0, 0).
Let w| = u| be the first vector of our orthogonal basis. Since w| . u2 = 0 (w| and u2 are orthogonal), we can take as the second vector w2 = u2 . To find the last vector we use the Gram-Schmidt formula:
u3 . w| u3 . w2
Note that we can replace w3 by its scalar multiple (1, 1, _1, _1).
(c) Find the best approximation for the vector v = (0, 0, _3, 1) by a vector in the subspace
U from part (b).
[1pt]
ANSWER:
Solution: The best approximation of v by a vector in U is projU (v). We know
projU (v) = projw1 (v) + projw2 (v) + projw3 (v)
for any orthogonal basis nw| , w2 , w3 { for U .
From (b), we can take w| = (1, 0, 1, 0), w2 = (0, 1, 0, 1), w3 = (1, 1, _1, _1). Hence
13. Let A = ┌ 2(1) (2 |
2 1 2 |
2(2)┐ 1) |
(a) Find the (fully factored) characteristic polynomial of A.
ANSWER:
Solution:
det(A _ λI) = det ┌ ┐
( 2 2 1 _ λ)
= (1 _ λ)╱ (1 _ λ)2 _ 4、_ 2 ╱2(1 _ λ) _ 4、+ 2 ╱4 _ 2(1 _ λ)、 = (1 _ λ)╱λ2 _ 2λ _ 3、_ 2 ╱ _ 2λ _ 2、+ 2 ╱2λ + 2、
= (1 _ λ)(λ + 1)(λ _ 3) + 4(λ + 1) + 4(λ + 1)
= (λ + 1)╱ _ λ2 + 4λ _ 3 + 8、
= (_1)(λ + 1)╱λ2 _ 4λ _ 5、= (_1)(λ + 1)2 (λ _ 5)
(b) Using the characteristic polynomial, explain why the eigenvalues of A are 5 and _1. [1/2pt]
Solution: The eigenvalues of A are the roots of the characteristic polynomial of A. Since the characteristic polynomial factors as (_1)(λ + 1)2 (λ _ 5), its roots are that are 5 (with
multiplicity 1) and _1 (with multiplicity 2).
(c) Find a basis for the eigenspace E5 = nv ÷ R3 : Av = 5v{. [1pt]
ANSWER:
Solution: Since E5 = Null(A _ 5I), we need to find a basis for the general solution of the homogeneous system [A _ 5I}0].
A _ 5I = 1 _22 5 1 2_2 5 1 22_ 5 = _224 2_24 22_4 ● 1_24 _222
┌ ┐ ┌┐ ┌┐
● ( ) ● ( ) ● ( )
2(1) ┐
_4)
Hence z = s, x = s, and y = s. From
E5 = n(s, s, s) : s ÷ R{ = spann(1, 1, 1){
we can see that n(1, 1, 1){ is a basis for E5 .
(d) Find a basis for the eigenspace E61 = nv ÷ R3 : Av = _v{. [1pt]
ANSWER:
Solution: Since E61 = Null(A _ (_1)I), we need to find a basis for the general solution
A + I = ┌ 1 2(+) 1 ( 2 |
2 1 + 1 2 |
1 + 1) = (2 |
2 2 2 |
2(2)┐ ┌0(1) 0(1) 2) ● (0 0 |
0(1)┐ 0) |
Hence y = s, z = t, x = _s _ t. From
E61 = n(_s _ t, s, t) : s, t ÷ R{ = spann(_1, 1, 0), (_1, 0, 1){
we can see that n(_1, 1, 0), (_1, 0, 1){ is a basis for E61 .
(e) Find an invertible matrix P and a diagonal matrix D such that P6|AP = D . [1/2pt]
ANSWER: P =
D =
Solution: Using the Diagonalization Theorem,
P = _101 _011 and D = 0_01 00_1
14. For each of the following statements, determine whether it is always true (T), or is possibly false (F). Write your response in the box following the statement.
Justify each answer:
· If you claim that the statement may be false, you must either give an explicit example
where it fails (a counterexample), or explain clearly why it fails (e.g. referring to theorems from class).
· If you claim that the statement is always true, you must give a clear explanation.
(a) The set B = n(1, 0, 2), (0, 5, 7){ is a basis for the subspace
U = spann(4, 0, 8), (1, 0, 2), (_1, 0, _2), (_2, 5, 3){
of R3 .
Answer (T/F):
Justification:
[1/2pts]
[1pt]
True |
┌ 4 0 8 ┐ ┌ 1 0 2┐ ┌ 1 0 2┐
A = ● ● =
(_2 5 3 ) ( 0 0 0) (0 0 0)
Since Row(A) = Row(), we conclude that the set B is indeed a basis for the subspace U .
(d) Let A be a 7 × 7 matrix with an eigenvalue equal to 0. Then A is invertible.
Answer (T/F): [1/2pts]
Justification: [1pt]
False |
(c) If U and W are subspaces of a vector space V , then their intersection U / W = nv ÷ V : v ÷ U and v ÷ W{
is also a subspace of V .
Answer (T/F):
Justification:
The intersection of subspaces U and W is closed under addition:
u, w ÷ U / W ~ ╱u + w ÷ U and u + w ÷ W、~ u + w ÷ U / W
and scalar multiplication:
u ÷ U / W, c ÷ R ~ ╱cu ÷ U and cu ÷ W、~ cu ÷ U / W,
and contains the zero vector. By the Subspace Test, it is a subspace.
(d) The matrix A = ┌ 1(2) 3(0)┐
2023-07-15