MAT 1341A, Fall 2017
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MAT 1341A, Fall 2017
1. Let U = {(x, y, z) e R3 l x + y + z = 0}. Which one of the following statements is true? (1)
mark (X) the correct answer:
A U is not a subspace of R3
B U is a subspace of R3 and dim U = 3
C U is a subspace of R3 and {(1, 1, -2), (0, 1, -1)} is a basis of U
D U is a line in R3 with direction vector (1, 1, 1)
E U is a subspace of R3 and {(1, 0, -1), (-1, 0, 1)} is a basis of U
F U is a plane in R3 with normal vector (-1, 2, -1)
Solution: As U is a plain which contains (0, 0, 0), it is a linear subspace of R3 of dimension 2. So one excludes A , B and D . Since {(1, 0, -1), (-1, 0, 1)} is dependent, it is not a basis
of U , so one excludes E . The normal vector to U is collinear to (1, 1, 1), so one excludes .
Hence, the correct answer is .
2. Which two of the following statements are true? (1)
I. {1, sin2 x, cos2 x} is a linearly independent set of vectors in the vector space of all real- valued functions F (R) = {f l f : R → R}
II. A homogeneous system of linear equations is always consistent
III. If A and B are 3 × 3 matrices, and both A and B are invertible, then their product AB is also an invertible matrix
IV. If u and v are linearly dependent vectors in R3 , then dim(Span{u, v}) = 2.
mark (X) the correct answer:
I. and II.
I. and III.
I. and IV.
II. and III.
II. and IV.
III. and IV.
Solution: I. Since sin2 x +cos2 x = 1, these vectors (functions) are linearly dependent.
Which of the following statements are true? (1)
I. If rank(A) = 1, there is just one parameter in the general solution of the system Ax = 0
II. If rank(A) = 1, there are n - 1 parameters in the general solution of the system Ax = 0
III. If A is invertible, the homogeneous system Ax = 0 has infinitely many solutions
IV. If the system Ax = 0 has infinitely many solutions, then rank(A) < n
mark (X) the correct answer:
A I. only
B II. only
C I. and III.
D II. and III.
E I. and IV.
F II. and IV.
Solution: Using the formula rank(A) + #parameters = #columns, we obtain that if rank(A) = 1, then #parameters = n - 1, so II. is correct. Also IV. is correct by the same formula. Hence, the correct answer is .
4. Find the |
polar |
form of the complex number |
1-^3i i-1 . |
(see the last page for the table of trigonometric functions) (1) |
mark (X) the correct answer:
A ^2(cos(-7π/12) + i sin(-7π/12))
B ^2(cos(5π/12) + i sin(5π/12))
C ^2(cos(-π/12) + i sin(-π/12))
^2(cos(π/12) + i sin(π/12))
E ^2(cos(-5π/12) + i sin(-5π/12))
F ^2(cos(11π/12) + i sin(11π/12))
Solution: We have
1 - ^3i - i ei(-T/3)
i - 1 -2(1) +2(1)i 1^2 ei(3T/4)
= ^2ei(-T/3-3T/4) = ^2ei(-13T/12) = ^2ei(11T/12)
The correct answer is F .
5. Let B = 、 and consider the subset U = {A e M2×2 (R) l BA = -AB}.
Which one of the following statements is true? (1)
mark (X) the correct answer:
A U is not a subspace of the vector space of 2 × 2 matrices M2×2 (R)
B U is a subspace of M2×2 (R), and dim U = 0
C U is a subspace of M2×2 (R), and dim U = 1
D U is a subspace of M2×2 (R), and dim U = 2
E U is a subspace of M2×2 (R), and dim U = 3
F U is a subspace of M2×2 (R), and dim U = 4
Solution: Suppose A = ╱c(a) d(b)、. Then BA = 、 ╱ c(a) d(b)、 = 、. On the other hand -AB = ╱c(a) d(b)、 = 、. Hence, AB = -BA would imply that
c = 0, a = -a and b - d = a.
So U = { ╱0(0) b(b)、 l b e R} is a subspace of M2×2 (R) of dimension 1.
The correct answer is .
6. If A = .(╱) .(、), then the third row of A-1 is:
mark (X) the correct answer:
(0 1 0)
(1 0 2)
(1 0 - 1)
(0 0 1)
(1 0 1)
(-1 1 1)
Solution: Let (x1 , x2 , x3 ) denote the third row of A-1 . Since A-1 A = I, we obtain
.(╱)**x1 **x2 **x3.(、) .(╱) .(、) = .(╱) .(、)
The correct answer is .
7. For a non-homogeneous system of 13 equations in 15 unknowns, answer the following three questions (Yes/No): (1)
● Can the system be inconsistent?
● Can the system have a unique solution?
● Can the system have infinitely many solutions?
mark (X) the correct answer:
A Yes, Yes, Yes
B Yes, Yes, No
C No, Yes, Yes
D Yes, No, Yes
E No, No, Yes
F Yes, No, No
Solution: The answers to the first and the third questions are YES.
If system has a unique solution, then the number of leading ones in its REF (< 13) must
coincide to the number of unknowns (15).
So the correct answer is D .
8. Let A be a square n × n matrix.
Which one of the statements below is not equivalent to the statement
“The columns of A are linearly independent”
mark (X) the correct answer:
The rows of A are linearly independent
Tank(A) = n
det(A) 1
The rows of A form a basis of Rn
The homogeneous system Ax = 0 has a unique solution
A is invertible
Solution: The correct answer is .
9. Let A = .(、) for some numbers a, b, c, d, e, f, g, h, i be such that det(A) = 2.
Using the elementary row/column operations compute the determinant of the matrix
e + 5f
3d
-2f
mark (X) the correct answer:
-64
-18
-12
12
18
64
Solution: We apply elementary row operations to the matrix A: First, we switch the first and the second rows (the determinant changes its sign). Next, we multiply the first row by 3, we add to the second row the third row multiplied by 5 and, finally, we multiply the third row by -2. As a result we get a matrix of determinant 12 which is the transposed to the one in question.
The correct answer is .
10. The vectors u1 = (1, -1, 2), u2 = (-5, -1, 2), and u3 = (0, 2, 1) form an orthogonal basis of R3 . So that any vector v e R3 can be expressed in a unique way as a linear combination v = a1u1 + a2u2 + a3u3 , where a1 , a2 , a3 e R are coordinates of v with respect to the basis {u1 , u2 , u3 }. Find the coordinate a2 for v = (1, 0, 1). (1)
mark (X) the correct answer:
B
C
E
F
Solution: We have
a2 = = = -
So the correct answer is .
11. Let U = Span{(-1, 1, 1, 0), (1, 0, 1, 1), (2, 1, 4, 3), (0, 1, 2, 2)} in R4 . (a) Find a basis for U which is a subset of the given spanning set above.
ANSWER:
Solution: {(-1, 1, 1, 0), (1, 0, 1, 1), (0, 1, 2, 2)} is a basis.
(b) Extend the basis found in part (a) to a basis of R4 . (1)
ANSWER:
Solution: If we add (0, 0, 1, 0) to the basis above, we obtain a basis of R4 .
12. Let W = {(x, y, z, w) e R4 l y - z - w = 0}
(a) Find a basis for W .
ANSWER:
Solution: W = Null(A), where A = (0 1 - 1 - 1). We obtain
W = {(r, s + t, s, t) e R4 l r, s, t e R},
hence {(0, 1, 1, 0), (0, 1, 0, 1), (1, 0, 0, 0)} is a basis for W .
(b) Use the Gram-Schmidt algorithm to find an orthogonal basis for W . (1)
ANSWER:
Solution: {(0, 1, 1, 0), (0, 1, -1, 2), (1, 0, 0, 0)} is an orthogonal basis of W .
(c) Find the best approximation of (0, 1, 1, 1) by a vector in W .
ANSWER:
Solution: pTojW (0, 1, 1, 1) = (0, 4/3, 2/3, 2/3).
13. Let A = ╱
( 1 1
(a) Find the characteristic polynomial of A.
ANSWER:
Solution:
det(A - λI) = -1 -11 λ -1 1-1 λ -1 11- λ.(、) = -(1 + λ)3 + 1 + 1 + 3(1 + λ)
= -λ3 - 3λ2 + 4 = (λ + 2)2 (1 - λ)
(b) Using the characteristic polynomial explain why the eigenvalues of A are -2 and 1. (1/2)
Solution: The roots of the characteristic polynomial that are -2 and 1 are the eigenvalues of A.
(c) Find a basis of the eigenspace E-2 = {v e R3 l Av = -2v}. (1)
ANSWER:
Solution:
E-2 = ker(A + 2I) = ker .(、) = ker .(、) = {(-s - t, s, t) l s, t e R}
So that {(-1, 1, 0), (-1, 0, 1)} is a basis for E-2 .
(d) Find a basis of the eigenspace E1 = {v e R3 l Av = v}.
ANSWER:
Solution:
11-2.(、) = {(s, s, s) l s e R}
So that {(1, 1, 1)} is a basis for E1 .
(e) Find an invertible matrix P and a diagonal matrix D such that P-1 AP = D . (1/2)
ANSWER: P =
D =
Solution: P = ╱
( 0 1
.(、) and D = -002 0-02
.(、)
14. State whether each of the following is (always) true, or is (possibly) false, in the respective box. Provide the respective justification/example.
(a) If A is an 5×4 matrix and if a row echelon form of A has a row of zeros, then Tank(A) < 4
ANSWER (True/False):
Justification/example:
(Can have rank 4)
(b) The dimension of the null space of the 1 × 4 matrix (-1 0 1 2) is 3 ANSWER (True/False):
Justification/example:
True |
(1/2)
(1/2)
(c) Let z1 and z2 be two complex numbers which are not real numbers. Then their product z1 . z2 can not be a real number.
ANSWER (True/False):
Justification/example:
(i . i = -1)
(1/2)
(1/2)
(d) The function T : R2 → R2 defined by T (x, y) = (x, x + y) is a linear transformation.
ANSWER (True/False):
Justification/example:
True |
(1/2)
(1/2)
15. (BONUS question) Define a linear transformation T : R3 → R3 by T ( .(、)) = .(、)
(a) Find the standard matrix of T.
ANSWER: |
Solution: We have T (e1 ) = e1 - e3 , T (e2 ) = -e1 + e2 , T (e3 ) = -e2 + e3 . So AT = 10-1 -101 0-11.(、) .
(b) Find a basis for the image of T. (1) |
ANSWER: |
Solution:
2023-07-15