FINAL EXAM REVIEW -LINEAR ALGEBRA

Lê Xuân Đại, email: ytkadai@hcmut.edu.vn

HCMUT

Lê Xuân Đại Final Exam Review

[L.O.2] Using the following information, answer the questions
from 1 to 4.
When studying the market of two competing products A and B in a city
of 1409 people, it is found that after each month 15% of A’s users switch
to B and 16% stop using both, 16% of B’s users switch to use A and 14%
stop using both types of products, 10% of people who do not use both
products switch to A and 19% switch to B. Given that at this month,
there are 438, 514, and 457, people that use A, B and none of them,
respectively.

Lê Xuân Đại Final Exam Review

Question 1.

What is the transition matrix? (The 3 states should be in this order: A,

B, none
 of them)   
73/100 3/20 4/25 438 3/20 4/25
A.  17/100 33/50 7/50  B.  514 33/50 7/50 
1/10 19/100 7/10 457 19/100 7/10
   
73/100 17/100 1/10 73/100 438 4/25
C.  3/20 33/50 19/100  D.  17/100 514 7/50 
4/25 7/50 7/10 1/10 457 7/10

Lê Xuân Đại Final Exam Review

Question 1.

What is the transition matrix? (The 3 states should be in this order: A,

B, none
 of them)   
73/100 3/20 4/25 438 3/20 4/25
A.  17/100 33/50 7/50  B.  514 33/50 7/50 
1/10 19/100 7/10 457 19/100 7/10
   
73/100 17/100 1/10 73/100 438 4/25
C.  3/20 33/50 19/100  D.  17/100 514 7/50 
4/25 7/50 7/10 1/10 457 7/10
Solution. The correct answer is A
 73 3 4

100 20 25
The transition matrix is P =  17 33 7 
100 50 50
1 19 7
10 100 10

Lê Xuân Đại Final Exam Review

Question 2.

Find all eigenvalues

√ of the√transition matrix. √ √
509 13 291 13 13 13
A. 1; 200 − 200 ; − 200 + 200 B. 1; 109
200 − 200 √; 200 + 109
200
√ √ √
691 13 691 13 13 13
C. 1; − 200 − 200 ; − 200 + 200 D. 1; 909
200 − 200 ; 200 + 309
200

Lê Xuân Đại Final Exam Review

Question 2.

Find all eigenvalues

√ of the√transition matrix. √ √
509 13 291 13 13 13
A. 1; 200 − 200 ; − 200 + 200 B. 1; 109
200 − 200 √; 200 + 109
200
√ √ √
691 13 691 13 13 13
C. 1; − 200 − 200 ; − 200 + 200 D. 1; 909
200 − 200 ; 200 + 309
200
Solution. The correct answer is B
 73 3 4

100 20 25
17 33 7 
We have P =  100 50 50 . The eigenvalues λ of P satisfy the
1 19 7
10 100 10
characteristic equation:
det(P − λI) = 0
where I is the identity matrix.

Lê Xuân Đại Final Exam Review

We compute:
 73 3 4

100 − λ 20 25 2
det  17 33
−λ 7  = −λ3 + 209λ − 13867λ + 2967 = 0
100 50 50 100 10000 10000
1 19 7
10 100 10 −λ
√ √
109 13 13 109
λ1 = 1, λ2 = − , λ3 = +
200 200 200 200

Lê Xuân Đại Final Exam Review

Question 3.

Find an eigenvector corresponding to the smallest eigenvalue of the

transition
 matrix.
√   √ 
− 18 − √1813
11 25
18
√
− 13
18 
A. − + 13  B.  13 + 29 
 7  
18 18 18 18
1 0
 √   √ 
2 13 7 13
9 + 9 − +
 √13 2   11 √1813 
18
C. − D. − −
9 − 9 18 18
 
3 1

Lê Xuân Đại Final Exam Review

Question 3.

Find an eigenvector corresponding to the smallest eigenvalue of the

transition
 matrix.
√   √ 
− 18 − √1813
11 25
18
√
− 13
18 
A. − + 13  B.  13 + 29 
 7  
18 18 18 18
1 0
 √   √ 
2 13 7 13
9 + 9 − +
 √13 2   11 √1813 
18
C. − D. − −
9 − 9 18 18
 
3 1
Solution. The correct answer is D
√
109 13
λmin = −
200 200

Lê Xuân Đại Final Exam Review

The eigenvector satisfies the equation:

P v = λmin v
√ 
13 37 3 4  
200 + 200 √ 20 25 0
17 13 23 7
+ = 0
   
 100 200 200 √ 50
1 19 13
+ 31 0
10 100 200 200
An eigenvector v corresponding to λmin is:
 √ 
7
− 18 + √1813
v = − 11 − 13 
 
18 18
1

Lê Xuân Đại Final Exam Review

Question 4.

Find the number of users of A after a long time (round the answer to the
nearest integer).
A. 470 B. 106 C. 252 D. 440

Lê Xuân Đại Final Exam Review

Question 4.

Find the number of users of A after a long time (round the answer to the
nearest integer).
A. 470 B. 106 C. 252 D. 440
Solution. The correct answer is A
 T
We assume the steady-state distribution vector is s = s1 s2 s3
The steady-state satisfies the equation:

P · s = s, with s1 + s2 + s3 = 1

Solving this system, we obtain:

1 1 1
s1 = , s2 = , s3 =
3 3 3

Lê Xuân Đại Final Exam Review

Given the total number of users is 1409, the number of users of group A
after a long time is:
1
round(1409 × ) = 470
3

Lê Xuân Đại Final Exam Review

[L.O.1] Using the following information, answer the questions
from 5 to 7.
Suppose that u = (−4; 3; 4; −5)T is an eigenvector of the matrix A
corresponding to λu = −3, and v = (−3; 2; 3; 3)T is an eigenvector of A
corresponding to λv = 3. Given that A is invertible.

Lê Xuân Đại Final Exam Review

Question 5.

Find the component 2 of the vector A3 u.

A. −81 B. −86 C. −76 D. −84

Lê Xuân Đại Final Exam Review

Question 5.

Find the component 2 of the vector A3 u.

A. −81 B. −86 C. −76 D. −84
Solution. The correct answer is A
We have  
108
 −81 
A3 u = λ3u .u = 
−108 .


135
Therefore, the component 2 of the vector A3 u is −81.

Lê Xuân Đại Final Exam Review

Question 6.

Find the component 4 of the vector A(5u + 2v).

A. 88 B. 96 C. 93 D. 98

Lê Xuân Đại Final Exam Review

Question 6.

Find the component 4 of the vector A(5u + 2v).

A. 88 B. 96 C. 93 D. 98
Solution. The correct answer is C
We have  
42
−33
−42 .
A(5u + 2v) = 5.λu .u + 2.λv .v =  

93
Therefore, the component 4 of the vector A(5u + 2v) is 93.

Lê Xuân Đại Final Exam Review

Question 7.

Find the component 3 of the vector A−1 (4u + 4v).

A. − 19
3 B. − 37 C. 23 D. − 43

Lê Xuân Đại Final Exam Review

Question 7.

Find the component 3 of the vector A−1 (4u + 4v).

A. − 19
3 B. − 37 C. 23 D. − 43
Solution. The correct answer is D
We have  4 
3
4 4 − 4 
A−1 (4u + 4v) = ×u+ ×v = 3
− 4  .
λu λv 3
32
3

Therefore, the component 3 of the vector A−1 (4u + 4v) is − 43 .

Lê Xuân Đại Final Exam Review

Question 8.
In
D R3E, let the inner product be given
x, y = 5x1 y1 + 2x2 y2 + 4x3 y3 + 2x1 y2 + 2x2 y1 and
x = (5; 3; −1), y = (−2; −1; 3). Calculate the distance between x and y.
√ √ √ √
A. 458 B. 434 C. 453 D. 274

Lê Xuân Đại Final Exam Review

Question 8.
In
D R3E, let the inner product be given
x, y = 5x1 y1 + 2x2 y2 + 4x3 y3 + 2x1 y2 + 2x2 y1 and
x = (5; 3; −1), y = (−2; −1; 3). Calculate the distance between x and y.
√ √ √ √
A. 458 B. 434 C. 453 D. 274
Solution. The correct answer is C
 
5 2 0
We have A =  2 2 0 and x − y = (7; 4; −4). Therefore, the distance
0 0 4
between x and y is
q √ √
d(x, y) = (x − y)A(x − y)T = 453 = 453

Lê Xuân Đại Final Exam Review

Question 9.
Let f be the linear transformation f : R3 7→ R2 so that
f (x; y; z) = (−2x − y − z; −4x − 12y − 4z). Which vector belongs to
Kerf ?
A. (0; 0; 0) B. (−2; −1; 5) C. (−5; 2; 4) D. (1; −6; 2)

Lê Xuân Đại Final Exam Review

Question 9.
Let f be the linear transformation f : R3 7→ R2 so that
f (x; y; z) = (−2x − y − z; −4x − 12y − 4z). Which vector belongs to
Kerf ?
A. (0; 0; 0) B. (−2; −1; 5) C. (−5; 2; 4) D. (1; −6; 2)
Solution. The correct answer is B
n o
We have Kerf = (x; y; z) ∈ R3 : f (x; y; z) = (0; 0) ⇒ (x; y; z) =
(−2; −1; 5) ∈ Kerf.

Lê Xuân Đại Final Exam Review

Question 10.
Find all real values 
of m such that λ = 5 is an eigenvalue of
2 2 m
A =  −1 4 −5  .
0 5 5
A. −19 B. −20 C. −16 D. −15

Lê Xuân Đại Final Exam Review

Question 10.
Find all real values 
of m such that λ = 5 is an eigenvalue of
2 2 m
A =  −1 4 −5  .
0 5 5
A. −19 B. −20 C. −16 D. −15
Solution. The correct answer is D
λ = 5 is an eigenvalue of A if and only if
 
−3 2 m
det(A−λI) = 0 ⇔ det  −1 −1 −5  = 0 ⇔ −5m−75 = 0 ⇔ m = −15.
0 5 0

Lê Xuân Đại Final Exam Review

[L.O.1] Using the following information, answer the questions
from 11to 12. 
1 1 −1 3
Let A =  2 −5 3 0  and T be a linear transformation defined
−5 −5 −5 −2
by T (x) = AxT .

Lê Xuân Đại Final Exam Review

Question 11.

Find the third component of the image vector T (x) where x = (2; 0; 0; 0).

A. −7 B. −8 C. −10 D. −6

Lê Xuân Đại Final Exam Review

Question 11.

Find the third component of the image vector T (x) where x = (2; 0; 0; 0).

A. −7 B. −8 C. −10 D. −6
Solution. The correct answer is C
We have y = AxT , so
 
  2  
1 1 −1 3  0  2
y =  2 −5 3 0 .
 0
=
 4 
−5 −5 −5 −2 −10
0

Lê Xuân Đại Final Exam Review

Question 12.

Which
n of the following setsois a basis for the range of T.
A. (1; 2; −5), (1; −5; −5)
n o
B. (1; 1; −1; 3), (2; −5; 3; 0), (−5; −5; −5; −2)
n o
C. (1; 2; −5), (1; −5; −5), (−1; 3; −5)
n o
D. (1; 2; −5)

Lê Xuân Đại Final Exam Review

Question 12.

Which
n of the following setsois a basis for the range of T.
A. (1; 2; −5), (1; −5; −5)
n o
B. (1; 1; −1; 3), (2; −5; 3; 0), (−5; −5; −5; −2)
n o
C. (1; 2; −5), (1; −5; −5), (−1; 3; −5)
n o
D. (1; 2; −5)
Solution. The correct answer is C
We have
T (1; 0; 0; 0) = A(1; 0; 0; 0)T = (1; 2; −5)T
T (0; 1; 0; 0) = A(0; 1; 0; 0)T = (1; −5; −5)T
T (0; 0; 1; 0) = A(0; 0; 1; 0)T = (−1; 3; −5)T
T (0; 0; 0; 1) = A(0; 0; 0; 1)T = (3; 0; −2)T

Lê Xuân Đại Final Exam Review

  
1 1 −1 3
Then, A =  2 −5 3 0  has rank(A) = 3. We can choose 3
−5 −5 −5 −2
columns from 4 columns as a basis for the range of T.

Lê Xuân Đại Final Exam Review

Question 13.
n o n o
Let E = (1; 1), (1; 0) and F = (5; 0), (−2; 5) be two bases of R2 and
f : R2 7→ R2 be the linear transformation defined by
f (x, y) = (3x − y; 5x − 3y) . Find the matrix AEF of f with respect to
the bases
 14 Eand F .  14
1 25
    
25 1 25 1 1 1
A. 2 B. 14 2 C. 14 D. 14
5 1 25 5 25 1 25 1

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Question 13.
n o n o
Let E = (1; 1), (1; 0) and F = (5; 0), (−2; 5) be two bases of R2 and
f : R2 7→ R2 be the linear transformation defined by
f (x, y) = (3x − y; 5x − 3y) . Find the matrix AEF of f with respect to
the bases
 14 Eand F .  14
1 25
    
25 1 25 1 1 1
A. 2 B. 14 2 C. 14 D. 14
5 1 25 5 25 1 25 1
Solution. The correct answer is A
   
1 1 5 −2
We have E = and F = .
1 0 0 5
 
3 −1
The matrix Ast of f with respect to the standard basis is Ast =
5 −3
The matrix AEF of f with respect to the bases E and F is
 14 
−1 25 1
AEF = F Ast E = 2
5 1

Lê Xuân Đại Final Exam Review

Question 14.
In
D R2E, let the inner product be given
x, y = 2x1 y1 + 5x2 y2 + 3x1 y2 + 3x2 y1 . Find m so that vector
u = (5; −2) is perpendicular to vector v = (−1; 2m).
A. 57 B. − 53 C. − 85 D. 25

Lê Xuân Đại Final Exam Review

Question 14.
In
D R2E, let the inner product be given
x, y = 2x1 y1 + 5x2 y2 + 3x1 y2 + 3x2 y1 . Find m so that vector
u = (5; −2) is perpendicular to vector v = (−1; 2m).
A. 57 B. − 53 C. − 85 D. 25
Solution. The correct answer is D
 
2 3
We have A = . Vector u = (5; −2) is perpendicular to vector
3 5
v = (−1; 2m) if and only if
  
D E
T
  2 3 −1 2
u, v = uAv = 5 −2 = 10m − 4 = 0 ⇒ m =
3 5 2m 5

Lê Xuân Đại Final Exam Review

[L.O.1] Using the following information, answer the questions
from 15 to 18.
In R2 with the standard inner product, let d be the line −2x − y = 0.

Lê Xuân Đại Final Exam Review

Question 15.

Find the matrix  of the reflection transformation f about the

 line d. 
− 54 − 35 − 54 − 35
 3 4 3 4
5 − 5 − 5 − 5
A. B. C. D.
− 53 − 45 − 54 − 45 − 53 3
5 − 4
5
3
5

Lê Xuân Đại Final Exam Review

Question 15.

Find the matrix  of the reflection transformation f about the

 line d. 
− 54 − 35 − 54 − 35
 3 4 3 4
5 − 5 − 5 − 5
A. B. C. D.
− 53 − 45 − 54 − 45 − 53 3
5 − 4
5
3
5
Solution. The correct answer is D
We have f (1; −2) = (1; −2) and f (2; 1) = (−2; −1).
The matrix of the reflection transformation f about the line d is
−1
− 35 − 45
     
1 −2 1 2 1 −3 −4
Af = = =
−2 −1 −2 1 5 −4 3 − 45 3
5

Lê Xuân Đại Final Exam Review

Question 16.

Find the reflection of the triangle with vertices

A(3; −5), B(−2;
−4), C(−2; −1) about
the line d.
A. A′ − 275
; 2 , B ′ (2; 2) , C ′ 11 ; − 27 .
5 5

B. A′ 11 11 ′ − 4 ; 22 , C ′ − 27 ; 1 .


;
5 5 , B 5
5 5
C. A′ (1; 2) , B
′ − 45 ; 1 , C ′ (2; 2) .
D. A′ 11 27 ′ 22 ; − 4 , C ′ (2; 1) .


5 ;− 5 , B 5 5

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Question 16.

Find the reflection of the triangle with vertices

A(3; −5), B(−2;
−4), C(−2; −1) about
the line d.
A. A′ − 275
; 2 , B ′ (2; 2) , C ′ 11 ; − 27 .
5 5

B. A′ 11 11 ′ − 4 ; 22 , C ′ − 27 ; 1 .


;
5 5 , B 5
5 5
C. A′ (1; 2) , B
′ − 45 ; 1 , C ′ (2; 2) .
D. A′ 11 27 ′ 22 ; − 4 , C ′ (2; 1) .


5 ;− 5 , B 5 5
Solution. The correct answer is D
 
3 −2 −2
The matrix for the triangle ABC is A△ =
−5 −4 −1
The image of the reflection of the triangle is the triangle A′ B ′ C ′ :
 3
− 5 − 45
    11 22

3 −2 −2 5 5 2
Af × A△ = × =
− 45 3
5 −5 −4 −1 − 27
5 − 45 1

Therefore, A′ 11 27
, B′ 22 4
, C ′ (2; 1) .



5 ;− 5 5 ; −5

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Question 17.
π
Find the matrix of the rotation transformation g about the origin by 2
counter-clockwise.
       
0 −1 0 1 1 0 0 1
A. B. C. D.
1 0 −1 0 0 0 −1 1

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Question 17.
π
Find the matrix of the rotation transformation g about the origin by 2
counter-clockwise.
       
0 −1 0 1 1 0 0 1
A. B. C. D.
1 0 −1 0 0 0 −1 1
Solution. The correct answer is A

π
The matrix of the rotation transformation g about the origin by 2
counter-clockwise is  
0 −1
Ag =
1 0

Lê Xuân Đại Final Exam Review

Question 18.

Find the matrix

4 4
 of the composite
3 4
  3 3  f ◦ g.
transformation  4 3

−5 −5 −
A. 3 4 B. 53 53 C. 53 45 D. 35 5
4
5 − 5 5 5 5 5 5 5

Lê Xuân Đại Final Exam Review

Question 18.

Find the matrix

4 4
 of the composite
3 4
  3 3  f ◦ g.
transformation  4 3

−5 −5 −
A. 3 4 B. 53 53 C. 53 45 D. 35 5
4
5 − 5 5 5 5 5 5 5
Solution. The correct answer is D

The matrix of the composite transformation f ◦ g is

 3
− 5 − 45
    4 3

0 −1 −
Af ◦g = Af × Ag = × = 35 5 .
− 45 3
5 1 0 5
4
5

Lê Xuân Đại Final Exam Review

Question 19.
 
−3 4
Let A = . Find all eigenvalues of A.
1 −3
A. 0 or 0 B. 0 C. −5 or − 1 D. 0

Lê Xuân Đại Final Exam Review

Question 19.
 
−3 4
Let A = . Find all eigenvalues of A.
1 −3
A. 0 or 0 B. 0 C. −5 or − 1 D. 0
Solution. The correct answer is C
The characteristic equation is

det(A − λI) = λ2 + 6λ + 5 = 0 ⇔ λ = −5 or − 1

Lê Xuân Đại Final Exam Review

[L.O.1] Using the following information, answer the questions
from 20 to 22.
In R3 we define the weighted inner product by the formula
D E
x, y = 4x1 y1 + 3x2 y2 + 3x3 y3 .

Let F be the subspace of R3ospanned by the set of vectors

n
u = (2; 2; 4); v = (−4; 0; 3)

Lê Xuân Đại Final Exam Review

Question 20.

One n
basis for the orthogonal complementnof F is
9 11

o 9 11

o
A. ; 1; − B. ; − ; 1
n 16 4

o n 16 4
  o
11 9


C. 1; 0; − 4 , 0; 1; 16 D. 2; 2; 4 , − 4; 0; 3

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Question 20.

One n
basis for the orthogonal complementnof F is
9 11

o 9 11

o
A. ; 1; − B. ; − ; 1
n 16 4

o n 16 4
  o
11 9


C. 1; 0; − 4 , 0; 1; 16 D. 2; 2; 4 , − 4; 0; 3
Solution. The correct answer is B
Let x= (x1 ; x2 ; x
3 ) be the vector of the orthogonal complement of F and
4 0 0
A =  0 3 0 . Then, we have
0 0 3
 D E
 x, u = xAuT = 0 
9 11
8x1 + 6x2 + 12 = 0
D E ⇔ ⇔ x1 = ; x2 = − .
 x, v = xAv T = 0 9 − 16x1 = 0 16 4

Therefore,
n oneobasis for the orthogonal complement of F is
9 11


16 ; − 4 1;

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Question 21.

The projection of z =
(0; −3; 1) onto F is

A. 1967 478 2892
B. − 333 97 17


575 ; 575 ; − 575
575 ; − 575 ; − 575

817
C. 575 ; − 1822 592
575 ; − 575 D. − 2058 1053 1742
575 ; 575 ; − 575

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Question 21.

The projection of z =
(0; −3; 1) onto F is

A. 1967 478 2892
B. − 333 97 17


575 ; 575 ; − 575
575 ; − 575 ; − 575

817
C. 575 ; − 1822 592
575 ; − 575 D. − 2058 1053 1742
575 ; 575 ; − 575
Solution. The correct answer is B
The projection of z = (0; −3; 1) onto F has a form λ1 u + λ2 v where
λ1 , λ2 are defined by the linear system
 D E D E D E
 z, u = λ1 u, u + λ2 u, v 
D E D E D E ⇔ 76λ1 + 4λ2 + 6 = 0
 z, v = λ1 u, v + λ2 v, v 4λ1 + 91λ2 − 9 = 0

97 59
⇔ λ1 = − ; λ2 = .
1150 575

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Therefore, the projection of z = (0; −3; 1) onto F is proj(z) = λ1 u+λ2 v =
     
97 59 333 97 17
= − × (2; 2; 4) + × (−4; 0; 3) = − ;− ;−
1150 575 575 575 575

Lê Xuân Đại Final Exam Review

Question 22.

The distance
√
from z = (0;√−3; 1) to F is √ √
404294
A. 115 B. 7411569 C. 7 7981
115 D. 2 107686
115

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Question 22.

The distance
√
from z = (0;√−3; 1) to F is √ √
404294
A. 115 B. 7411569 C. 7 7981
115 D. 2 107686
115
Solution. The correct answer is B
The distance from z = (0; −3; 1) to
F is ||g|| where
333 1628 592
g = z − proj(z) = 575 ; − 575 ; 575 .
Therefore, √
p
T
74 69
||g|| = gAg = .
115

Lê Xuân Đại Final Exam Review