Tuesday October 11 START: 13:10 DURATION: 110 mins

University of Toronto

Faculty of Applied Science & Engineering

MIDTERM EXAMINATION I
MAT188H1F
Linear Algebra

EXAMINERS: D. Burbulla, S. Cohen, D. Fusca, F. Lopez, M. Palasciano, M. Pugh, B. Schachter, S. Uppal

Last Name (PRINT):

Given Name(s) (PRINT):

Student NUMBER:

Student SIGNATURE:

EMAIL @mail.utoronto.ca:

Instructions.

1. There are 55 possible marks to be earned in this exam. The examination booklet contains a total of 11 pages. It is
your responsibility to ensure that no pages are missing from your examination. DO NOT DETACH ANY PAGES FROM
YOUR EXAMINATION.
2. DO NOT WRITE ON THE QR CODE AT THE TOP RIGHT-HAND CORNER OF EVERY PAGE OF YOUR EXAM-
INATION.
3. For the full answer questions, WRITE YOUR SOLUTIONS ON THE FRONT OF THE QUESTION PAGES THEM-
SELVES. THE BACK OF EVERY PAGE WILL NOT BE SCANNED AND SEEN BY THE GRADERS.
4. Ensure that your solutions are LEGIBLE.

5. No aids of any kind are permitted. CALCULATORS AND OTHER ELECTRONIC DEVICES (INCLUDING PHONES)
ARE NOT PERMITTED.
6. Have your student card ready for inspection.
7. There are no part marks for Multiple Choice (MC) questions.

8. You may use the two blank pages at the end for rough work. The last two pages of the examination WILL NOT BE
MARKED unless you clearly indicate otherwise on the question pages.
9. For the full answer questions, show all of your work and justify your answers but do not include extraneous information.

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Part I - Multiple Choice. Clearly indicate your answer to each question by circling your choice. Each question is worth 2 marks.

For each question, choose the BEST option from the given options.

1. Which of the following matrices are in row-echelon form?

0 0 1
(i) 0 0 1
0 0 1

1 1 0
(ii) 0 1 1
0 0 1

1 0 0
(iii) 0 0 0
0 0 1

(A) (ii) only

(B) (iii) only
(C) (i) only
(D) (ii) and (iii) only
(E) (i) and (ii) only

Answer: A
Solution:
i) This matrix is not in row-echelon form as there is a leading one in each row but each leading 1 is not to the right of the
one above it.
ii) This matrix is in row echelon form.
iii) This matrix is not in row-echelon form as there is a row full of 0s that is not at the bottom of the matrix.

2. Let x, y, z be non-zero vectors in R3 . Which of the following statements are TRUE?

(i) If x y = x z then y = z.
(ii) (x y) z = x (y z).
(iii) ||x y|| = 0 if and only if x and y are parallel.

(A) (ii) only

(B) (iii) only
(C) (i) and (iii) only
(D) (ii) and (iii) only
(E) none of (i), (ii), or (iii)

Answer: B
Solution:
0 1 0
i) This is not true. As a counterexample, we can take x = 0, y = 0 and x = 1. Since ||x|| = 0, x v (where v is
0 0 0

0
any vector in R3 ) will be equal to 0, in particular when v = y or z, but y 6= z.
0

2

1 0 0
ii) This is not true. As a counterexample, we can take x = y = 0 and z = 1. Then (x y) z = 0 but
0 0 0
0
x (y z) = 1.
0
iii) This is true. We know that ||x y|| = ||x||||y|| sin where is the angle between x and y. This being equal to 0 would
imply that either x or y is the zero vector (in which case they are parallel as one is a multiple of another), or the sine of the
angle between them is 0, meaning that angle is either 0 or , meaning that one of the vectors is a (possible negative) multiple
of another.

   
c c+2
3. For what value(s) of c is the set , linearly dependent?
1 c
(A) c = 0
(B) c = 1
(C) c = 2
(D) (A) and (B) only
(E) (B) and (C) only

Answer: E
Solution:
For a set of 2 vectors to be linearly dependent, one must be a scalar  multiple
  of the other.
0 2
A) This value of c does not work. When c = 0, the vectors are and which are clearly not scalar multiples. Thus, the
1 0
vectors are linearly independent.    
1 1
B) This value of c works. When c = 1, the vectors are and and one vector is -1 times the other. Thus, the
1 1
vectors are linearly dependent.    
2 4
C) This value of c works. When c = 2, the vectors are and and the second vector is twice the first. Thus, the vectors
1 2
are linearly dependent.
This portion of the page is left blank for your rough work, if necessary. Nothing written in this space will be graded or considered.

3
Part I - Multiple Choice. Clearly indicate your answer to each question by circling your choice. Each question is worth 2 marks.

For each question, choose the BEST option from the given options.

4. Let x, y, z be vectors in R3 . Which of the following statements are TRUE?

(i) If the set S = {x, y, z} is linearly independent, then the set {x, y} is linearly independent.

(ii) If the set S = {x, y, z} is linearly dependent, then the set {x, y} is linearly dependent.
(iii) If the three sets {x, y}, {y, z}, and {x, z} are linearly independent, then the set S = {x, y, z} is linearly independent.
(A) (i) only
(B) (ii) only
(C) (iii) only
(D) (i) and (ii) only
(E) (i) and (iii) only

Answer: A
Solution:
i) This is true.
Notice that if the set {x, y} is linearly dependent then it is possible to form the zero vector using a non-trivial combination
of x and y. The zero vector can still be formed by the same linear combination of x and y with z taken with a coefficient of 0
which would mean that S is linearly dependent too.
1 0
ii) This is false. As a counterexample, we can take x = z = 0 and y = 1. The set {x, y, z} is clearly linearly dependent
0 0
as x z = 0. But {x, y} is linearly independent as x and yare not scalar multiples.

1 0
iii) This is false. As a counterexample, we can take x = 0 and y = 1 and z = x + y. Then the three sets {x, y}, {y, z},
0 0
and {x, z} are linearly independent since no two x, y, z are scalar multiples of each other, but {x, y, z} is linearly dependent as
x + y z = 0.

1 1 0 1 1
5. Suppose that the augmented matrix of a system of linear equations has been reduced to 0 0 1 1 1 . Which of the
0 0 0 1 0
following statements describes the set of solutions to the system?

(A) infinitely many solutions with three parameters

(B) infinitely many solutions with two parameters
(C) infinitely many solutions with one parameter
(D) unique solution
(E) no solutions

Answer: C
Solution:
We interpret this matrix as the augmented matrix of the system

x1 + x2 + x4 = 1
x3 + x4 = 1
x4 = 0

4
We see that x2 is the only free variable which as assigned as a parameter in the general solution. Hence, there are infinitely
many solutions with one parameter.

This portion of the page is left blank for your rough work, if necessary. Nothing written in this space will be graded or considered.

5
Part II - Short Answer Questions. Write your solutions in the space provided below each question.

1. Let P be the plane 2x1 + x3 = 3.

(a) Find parametric equations of the line L that contains the point (1, 6, 0) and is orthogonal to P. [2 marks]

Answer:

x1 = 1 + 2t
x2 = 6
x3 = t

for any t R.

Solution:
2 2
Since the equation for P is 2x1 + 0x2 + 1x3 = 3, then the vector 0 is normal to the plane. Thus, any line with 0 as a
1 1
direction vector will be orthogonal to the plane. Since we are given the point that the line must contain, we can write the vector
equation of the line as
x1 1 2
x2 = 6 + t 0
x3 6 1
for any t R. Equating components gives the parametric equations.

Note: This question specifically asked for parametric equations so if your answer was the vector equation you were deducted
0.5 points.

(b) Find the (shortest) distance from the point (1, 6, 0) to P and determine the point on P closest to (1, 6, 0) . [6 marks]

Answer: Distance: 5 Point: (1, 6, 1)
Solution:
Let Q be the point (1, 6, 0). Take any point P on the plane. For simplicity we choose P to be (0, 0, 3). Then, the shortest
distance is

2
||projn P~Q|| = || 0 ||
1

= 5

To find the point on the plane nearest to the point (-1, 6, 0), we simply subtract (-2, 0, -1) from (-1, 6, 0) to get (1, 6, 1). You
may find drawing a picture will help you visualize the solution.

6

1 2
2. (a) Suppose that the angle between the vectors 7 and 2 is given by cos() = 31 . Find c. [4 marks]
c 1
47
Answer: c = - .
12
Solution:
1 2 1 2
We know that 7 2 = || 7 || || 2 || cos . Thus, we get that 12 + c = 50 + c2 so 144 + 24c + c2 = 50 + c2 . This
c 1 c 1
47
is equivalent to 144 + 24c = 50 which gives c = - .
12

2. (b) Does the plane containing the points P (4, 0, 5), Q(2, 2, 1), and R(1, 1, 2) pass through the origin? Support your answer.
[4 marks]

Answer: Yes.
Solution:
2 3 10
We start by finding the normal vector n to the plane. We know that n = (P~Q) (P~R) = 2 1 = 6 . This
4 3 8
means that the equation of the plane containing the three points is given by

10x1 + 6x2 + 8x3 = d

for some d. Since the point P (4, 0, 5) is on the plane, we have d = 10(4) + 6(0) + 8(5) = 0. Hence the equation of the plane is

10x1 + 6x2 + 8x3 = 0

which indeed passes through the origin since 10(0) + 6(0) + 8(0) = 0.

7
3. (a) Let {v1 , . . . , vk } be a set of vectors in Rn . Define span{v1 , . . . , vk }. [2 marks]

Word definition:

span{v1 , . . . , vk } is the set of all possible linear combinations of the vectors v1 , . . . , vk .

Math definition:

span{v1 , . . . , vk } = {c1 v1 + + ck vk | c1 , . . . , ck R}.

2 2 2 2 0
3. (b) Show that span 4 , 7 span 1 , 2 , 3 . Support your answer. [6 marks]
3 6 0 3 3

Solution:
2 2 2 2 0 2 2
To show that span 4 , 7 span 1 , 2 , 3 we must show that if x is a vector in span 4 , 7 , then
3 6 0 3 3 3 6

2 2 0 2 2
x is also in span 1 , 2 , 3 . In other words, if x can be written as a linear combination of 4, and 7, then x
0 3 3 3 6


2 2 0 2 2
can also be written as a linear combination of 1 , 2 , and 3. This will be true if each of the vectors 4 and 7 can
0 3 3 3 6
2 2 0
themselves be written as a linear combination of 1 , 2 , and 3.
0 3 3

2 2 2 2 2 2
Notice, for example, that 4 = 2 1 + 2 and 7 = 3 1 + 2 2 (other combinations are possible) so the result
3 0 3 6 0 3
follows.

2 2 2 2 0
In fact, though the question does not ask you to show this, span 4 , 7 = span 1 , 2 , 3 . Can you explain
3 6 0 3 3

why?

8
4. (a) Define what it means for a subset W of Rn to be a subspace of Rn . [3 marks]

Solution:
W is a subspace of Rn if

1. W is non-empty.

2. W is closed under addition. i.e. For all x, y W , x + y W .

3. W is closed under scalar multiplication. i.e. For all x W , and c R, cx W .

x1 x1 x2 x2 x3
4. (b) Consider the subset W = x2 | = and = of R3 . Determine whether W is a subspace of R3 . Support
2 3 3 4
x3

your answer. [5 marks]

Answer: W is a subspace of R3 .
Solution:
0
0 0 0 0
1. Notice that 0 = 0 is in W since = and = so that indeed W in non-empty.
2 3 3 4
0

x1 y1 x1 + y1
x1 + y1
2. We show that if x = x2 and y = y2 are in W , then x + y = x2 + y2 W . Now, x + y W iff =
2
x3 y3 x3 + y3
x2 + y2 x2 + y2 x3 + y3
and = . Since
3 3 4
x1 + y1 x1 y1 x2 y2 x2 + y2
= + = + = ,
2 2 2 3 3 3
and
x2 + y2 x2 y2 x3 y3 x3 + y3
= + = + =
3 3 3 4 4 4
we have that indeed x + y W . Notice that the second to last equality in each line above is true because x and y are in W .

x1 cx1
cx1 cx2 cx2 cx3
3. We show that if x = x2 W and c R, then cx = cx2 W . Now, cx W iff = and = . Since
2 3 3 4
x3 cx3
cx1 x1 x2 cx2
=c =c = ,
2 2 3 3
and
cx2 x2 x3 cx3
=c =c =
3 3 4 4
we have that indeed cx W . Notice that the second to last equality in each line above is true because x is in W .

9
5. Consider a linear system of equations whose augmented matrix is

1 1 3 2
1 2 4 3 .
1 3 a b

For what values of a and b will the system have:

(i) infinitely many solutions; (ii) a unique solution; (iii) no solution? [8 marks]
Solution:
We start by reducing the matrix to (check this!)

1 1 3 2
0 1 1 1 .
0 0 a5 b4

i) The system will have infinitely many solutions if the above matrix contains an entire row of zeros. i.e. if a = 5 and b = 4.

ii) The system will have a unique solution if the number of leading variables (pivots) = the total number of variables. This
happens if a 6= 5 (b can be any real number).
 
iii) The system will have no solution if the above matrix contains a row of the form 0 0 0 c , where c 6= 0. This
happens if a = 5 and b 6= 4.

10
6. Is it possible to find a 2 2 matrix whose rows are linearly independent but whose columns are linearly dependent? Prove
your answer. [5 marks]

Answer: No.
Solution:  
a b
We suppose such a matrix exists and arrive at a contradiction. Call this matrix A = .
c d
Case 1: None of the entries in the matrix are 0. Since the columns are linearly dependent, we get that b = ka and d = kc,
for some scalar k. Our matrix is then  
a ka
.
c kc

We notice that the rows of this matrix is linearly dependent as the second row is equal to ac times the first row. Thus, it is
not possible to find such a matrix in this case. Notice, however, that the above argument does not hold if a = 0!
   
0 b 0 0
Case 2: If a = 0, then A = . Since the columns are linearly dependent, it must be that b = 0 too. Hence, A = .
c d c d
But then the rows are linearly dependent too since the first row is 0 times the second.

11
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12
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13
 University of Toronto
MAT188H1F TERM TEST
WEDNESDAY, NOVEMBER 16, 2005, 5:10 PM
Duration: 50 minutes

Aids Allowed: Casio 260, Sharp 520 or Texas Instrument 30 calculator.

TOTAL MARKS: 45

1. [12 marks; 4 for each part] Find the following:

4 3
(a) the angle between the vectors ~u = 2 and ~v = 0

2 3
(b) the area of the triangle with vertices

P (1, 1, 1), Q(2, 3, 2), R(0, 2, 2)

(c) the shortest distance from the point P (1, 2, 0) to the plane with equation
3x + y 2z = 14.

2. [10 marks] The parts of this question are unrelated.

(a) [6 marks; 2 for each part] Write down the matrix for each of the following
transformations:
(i) a reflection in the line y = 2x

(ii) a rotation through
3
(iii) a projection on the line y = 3x
(b) [4 marks] Suppose T : R2 R2 is a linear transformation such that
" # " # " # " #
1 4 0 3
T = and T = .
0 5 1 4
" #
1 2
Find T .
3

1 1 1
0
3. [13 marks] Let A = 2 1
.
4 1 0
Find the eigenvalues and eigenvectors of A and use them to write down the
general solutions of the following system of differential equations
0

f1
= f1 f2 + f3
f20 = 2f2 f3 , where f1 , f2 and f3 are functions of x.
0

f3 = 4f1 + f2
 4.(a) [6 marks; 3 for each part] Decide if the following sets of vectors are linearly
independent or dependent.

1
1 1
(i) 2 , 2 , 1


3 1
1
1 3

1 3 1




0 2
1 2 1

(ii) , , , ,

3 0 1 1 1




4 2 0 2 5

4.(b) [4 marks] Find a spanning set for the following subspace of R4 :



s+t 


2s t



U =  s and t are real numbers .



s 




t


1
" #
9 1
ANSWERS: 1(a) 1(b) 38 1(c) 2(b)
6 2 14 2
" # " # " #
1 3 4 1 1 3 1 1 3
2(a)(i) Q2 = 2(a)(ii) R/3 = 2(a)(iii) P3 =
5 4 3 2 3 1 10 3 9

0 1 1
3. 1 = 1; X1 = 1 ; 2 = 1; X2 = 1 ; 3 = 3; X3 = 1

1 3 1

f1 (x)
x x 3x
f2 (x) = c1 X1 e + c2 X2 e + c3 X3 e , where c1 , c2 and c3 are arbitrary constants.

f3 (x)

1 1




2 1


4(a)(i) independent 4(a)(ii) dependent 4(b) ,

1 0





0 1

 University of Toronto
MAT188H1F TERM TEST
WEDNESDAY, OCTOBER 12, 2005, 5:10 PM
Duration: 50 minutes

Aids Allowed: Casio 260, Sharp 520 or Texas Instrument 30 calculator.

Instructions: Fill in the information on this page, and make sure this test contains
5 pages. Present your solutions in the space provided. Use the backs of pages if
you need more space. The value for each question is indicated in square brackets
beside each question number.
TOTAL MARKS: 45

1. [10 marks] Consider the system of linear equations

x1 + x2 x3 + x4 + x5 = 6
x2 + x3 x5 = 2
x1 + 2x2 + x4 = 8
3x1 + 2x2 4x3 + 3x4 + 4x5 = 16

Find the reduced row-echelon form of the augmented matrix for this system,
and then find the solution to the system.

2. [10 marks; 5 for each part] Find the following:

1 1 0
(a) the inverse of A = 0 1 2

1 2 3
3 1 0 0

1 1 7 1
(b) det
2 1 1 0


1 0 1 2

3. [11 marks] The parts of this question are unrelated.

(a) [5 marks] Use Cramers Rule to find x2 in the solution to the system of
equations
2x1 + x2 + x3 = 4
x1 + x2 + 4x3 = 2
x1 x2 + 5x3 = 2
(b) [6 marks] Find all values of a for which the system of equations

x + y + z = 1
x + ay = 3
6y + az = 8

has infinitely many solutions.

 4. [14 marks] Parts (a) and (b) of this question are unrelated.

(a) [4 marks] Suppose W, X, Y and Z are all n n matrices. Find X in terms

of W, Y and Z, if
2W + 3(X T + Y )1 = Z.
(b) [10 marks; 5 for each part] Suppose A and B are 5 5 matrices with
det A = 2 and det B = 1. Find the following:
(i) det(A2 B 3 AT B 2 )
(ii) det(A1 + adj A)

x1 4 + 2s t 2u
1 0 2 1 2 4

x2 2s+u

0 1 1 0 1 2

ANSWERS: 1.

;
x3 = s , where
0 0 0 0 0 0

x4
t
0 0 0 0 0 0

x5 u
s, t and u are parameters.

1 3 2
2.(a) 2

3 2
(b) 72 3.(a) x2 = 0 (b) a = 2
1 1 1
4.(a) X = (3(Z 2W )1 Y )T (b)(i) 8 (ii) 243/2
 University of Toronto
MAT188H1F Linear Algebra TERM TEST
Monday, November 22, 2004, 5:10 PM

1. [13 marks] Find the following:

2 1
(a) [4 marks] the angle between the vectors ~u = 1 and ~v = 0


1 1
(b) [5 marks] the area of the triangle with vertices

P (1, 1, 1), Q(2, 3, 5), R(0, 2, 2)

2 1
~
(c) [4 marks] the projection of ~v = 3 on d = 2

1 1

2. [12 marks; 4 marks for each part] The parts for this quesiton are unrelated.

(a) Find the (scalar) equation of the plane passing through the three points

A(2, 1, 1), B(3, 0, 2), C(1, 2, 4).

(b) Find the vector equation of the line of intersection of the two planes with
equations
x + y + z = 6 and 2x y 4z = 0.
(c) What is the shortest distance from the point P (1, 2, 0) to the line with
vector equation
x 1
y = t 2 ,

z 2
where t is a parameter.

3. [13 marks] In this question all transformations are from R2 to R2 . Find the
following:

(a) [3 marks] the standard matrix, A, of a reflection in the line with equation
y = x.
(b) [3 marks] the standard matrix, B, of a projection onto the line with
equation y = x.
(c) [3 marks] T (~v ), where T is the reflection
! of part (a) followed by the
1
projection of part (b), and ~v = .
2
 ! !
90 cos 6 sin 6 1
(d) [4 marks] R (~u), where R = and ~u = .
sin 6 cos 6 0

4. [12 marks; 4 marks for each part] The parts of this question are unrelated.

1 1 1 0 0
(a) Find a basis for null(A) if A = 0 1 1 0 1


1 1 1 1 1
(b) Find a basis of R3 which is contained in the spanning set

1
1 0 1 0
2 , 3 , 5 , 1 , 4



1 1 2

1 2

a
(c) Show that the set, U, of all vectors b in R3 such that

c


1 1 a
dim span 1 , 0 , b = 2


1
1 c

is a subspace of R3 , and find its dimension.

5 7 7
ANSWERS: 1(a) (b) (c) d~
6 2 6

x 2 1
2(a) 2x + 3y + z = 8 (b) y = 4 + t 2 (c) 2

z 0 1
! ! ! !
0 1 1 1 1 1 1 1
3(a) (b) (c) (d)
1 0 2 1 1 2 1 0




1 0


1 1 1 1 1






4(a)
0 ,
1 (b) 2 , 3 , 1

1








0




1 1 1
1 0

1 1 h i
1 , 0 or U = null 1 0 1 ; dim(U ) = 2
4(c) U = span


1
1
 University of Toronto
MAT188H1F TERM TEST
MONDAY, OCTOBER 25, 5:10 PM, 2004
Duration: 50 minutes

1. [12 marks; 4 for each part] Find the following:

1 0 3
(a) the inverse of A = 1 2 1

0 2 3
2 3 1 1

4 1 1 0
(b) det

1 2 2 1


0 1 1 0
(c) an elementary matrix E such that B = EA if

1 3 1 2 1 3 1 2
A= 4

1 6 7 and B = 0 11 10 1


1 1 0 2 1 1 0 2

2. [13 marks] Consider the system of linear equations AX = B, with

x1
1 1 1 1 1 1

x2

3 2 2 2 1 2


A= ,X =

x3 and B =
1 1 1 0 0 0


x4
1 0 0 1 0 1

x5

(a) [4 marks] Find the reduced row-echelon form of the augmented matrix
(A|B) of this system.
(b) [2 marks] What is the rank of the augmented matrix?
(c) [1 mark] How many parameters are required to solve the system?
(d) [4 marks] What is the solution to the system AX = B?
(e) [2 marks] What are the basic solutions to the corresponding homogeneous
system of equations AX = O?

3. [12 marks; 4 for each part] Find the following:

(a) all values of a for which the homogeneous system of equations

2x1 + x3 = 0
ax1 + x2 = 0
15x1 + ax2 + ax3 = 0

has non-trivial solutions.

 !
1 2
(b) the characteristic polynomial of the matrix A =
2 1
(c) det(A2 B 3 AT B 5 ), if A and B are n n matrices with

det A = 1 and det B = 2.

1 1 0
4. [13 marks] Let A = 1 0 1

0 1 1
Find a diagonal matrix D and an invertible matrix P such that D = P 1 AP.

2 3 3 1 0 0
3/2 3/2 2 1(b) 2 1(c) 4 1 0
ANSWERS: 1(a)

1 1 1 0 0 1

1 0 0 0 1 0

0 1 1 0 1 0
2(a) 2(b) 3 2(c) 2

0 0 0 1 1 1


0 0 0 0 0 0

t 0 1
s t 1 1



2(d) X =
s , where s and t are parameters. 2(e)
1 ,
0


1t


0
1

t 0 1

3(a) a = 5 or 3 3(b) CA (x) = x2 2x 3 3(c) 4

1 0 0 1 1 1
4. D = 0 2 0 , P = 0 1 2

0 0 1 1 1 1
 MAT 188H1F Linear Algebra TERM TEST
Tuesday, November 11, 2003 Duration: 50 minutes
Only aids permitted: a Casio 260, Sharp 520, or Texas Instrument 30 calculator.

TOTAL MARKS: 45

1. [9 marks] Parts (a) and (b) are unrelated.

0 3 2 1

1 4 1 1
(a) [5 marks] Find det
0 2 1 1


1 3 5 6

1 1 1
(b) [4 marks] Find all values of a for which the matrix A =

1 a 0
is
0 6 a
not invertible.

2. [8 marks; 4 marks for each part] Parts (a) and (b) are unrelated.

(a) Use Cramers

Rule to find the value of x in the solution to the system of
x + 2y 3z =
4
equations y + z = 0

2x + y z = 0
(b) Let A and B be 3 3 matrices such that det A = 2 and det B = 1. Find
the value of  
det A2 B 6 A1 B T

3. [8 marks] Let S = {(1, 0, 1, 1), (2, 1, 1, 1), (1, 1, 0, 2), (5, 1, 4, 2)}.
Let W = span (S) . Find a basis for each of W and W .

4. [6 marks; 2 marks each] Find the standard matrices of the following linear
operators on R2 .

(a) S is a reflection in the line y = x.

(b) T is a dilation with scaling factor 3.
(c) U is an orthogonal projection onto the line y = x.

1 0 0
5. [8 marks] Find the eigenvalues of A = 0 0 1

and a basis for each
0 1 0
eigenspace of A.
 6. [6 marks; one mark each] Indicate whether each of the following statements is
True or False. There is no need to justify your answer.

(a) {(1, 2, 0), (1, 1, 2), (3, 3, 5), (1, 3, 7)} is a linearly independent set of vec-
tors in R3 .
(b) {(1, 2, 0), (1, 2, 0)} is a linearly independent set of vectors in R3 .
(c) If A is an orthogonal n n matrix, then det A = 1.
(d) If = 0 is an eigenvalue of A, then the row vectors of A are linearly
independent.
(e) If T : R2 R2 is a reflection in the line y = mx, then T is onto.
(f) If A is a symmetric matrix, then ker (TA ) = (ran (TA )) .

ANSWERS: 1(a) 4 1(b) a = 3 or 2 2.(a) x = 1 2(b) 2

3. A basis for W is {(1, 0, 1, 1), (0, 1, 1, 3)}; for W , {(1, 1, 1, 0), (1, 3, 0, 1)}
! ! !
0 1 3 0 1/2 1/2
4.(a) 4(b) 4(c)
1 0 0 3 1/2 1/2
5. 1 = 1; basis for eigenspace is {(1, 0, 0), (0, 1, 1)}
2 = 1; basis for eigenspace is {(0, 1, 1)}
6(a) False (b) False (c) True (d) False (e) True (f) True
 MAT 188H1F Linear Algebra TERM TEST
Friday, November 14, 2003 Duration: 50 minutes
Only aids permitted: a Casio 260, Sharp 520, or Texas Instrument 30 calculator.

TOTAL MARKS: 45

1. [9 marks] Parts (a) and (b) are unrelated.

4 1 1 1

3 0 2 1
(a) [5 marks] Find det
2 0 1 1


3 1 5 6

1 a 0
(b) [4 marks] Find all values of a for which the matrix A = 1 1 1 is

0 6 a
not invertible.

2. [8 marks; 4 marks for each part] Parts (a) and (b) are unrelated.

(a) Use Cramers

Rule to find the value of z in the solution to the system of
x + 2y 3z =
4
equations y + z = 0

2x + y z = 0
(b) Let A and B be 3 3 matrices such that det A = 2 and det B = 1. Find
the value of  
det 2A3 B T A1

3. [8 marks] Let S = {(1, 0, 1, 1), (2, 1, 1, 1), (1, 1, 0, 2), (5, 1, 4, 2)}.
Let W = span (S) . Find a basis for each of W and W .

4. [6 marks; 2 marks each] Find the standard matrices of the following linear
operators on R2 .

(a) S is a rotation of 45 degrees, counterclockwise, about the origin.

(b) T is a dilation with scaling factor 2.
(c) U is an orthogonal projection onto the line y = x.

2 0 0
5. [8 marks] Find the eigenvalues of A = 0 0 4

and a basis for each
0 1 0
eigenspace of A.
 6. [6 marks; one mark each] Indicate whether each of the following statements is
True or False. There is no need to justify your answer.

(a) {(1, 2, 0), (1, 1, 2), (3, 3, 5), (1, 3, 7)} is a spanning set of vectors for R3 .
(b) {(1, 2, 0), (3, 5, 0)} is a spanning set of vectors for R3 .
(c) If A is a 3 3 matrix such that AT = A, then det A = 0.
(d) If A is an nn matrix such that row (A) = col (A), then A is a symmetric
matrix.
(e) The composition of two reflections about lines through the origin of R2
ia a rotation about the origin.
(f) If u and v are non-zero vectors in R3 , then {u, v, u v} is a basis for R3 .

ANSWERS: 1(a) 4 1(b) a = 3 or 2 2.(a) z = 1 2(b) 32

3. A basis for W is {(1, 0, 1, 1), (0, 1, 1, 3)}; for W , {(1, 1, 1, 0), (1, 3, 0, 1)}
! ! !
1/2 1/ 2 2 0 1/2 1/2
4.(a) 4(b) 4(c)
1/ 2 1/ 2 0 2 1/2 1/2
5. 1 = 2; basis for eigenspace is {(1, 0, 0), (0, 2, 1)}
2 = 2; basis for eigenspace is {(0, 2, 1)}
6(a) True (b) False (c) True (d) False (e) True (f) False
 University of Toronto
MAT 188H1F Linear Algebra TERM TEST
Tuesday, October 7, 2003
Duration: 50 minutes
Only aids permitted: a Casio 260, Sharp 520, or Texas Instrument 30 calculator.

TOTAL MARKS: 45

1. [8 marks] Find the reduced row echelon form of the augmented matrix of the
system of equations

x1 + 3x2 + x3 x4 = 2
x2 x3 + 2x4 = 3
3x1 + 13x2 x3 + 5x4 = 6

and then solve the system.

2. [8 marks; 4 marks for each part] Find the following:

(a) the cosine of the angle between the two vectors

u = (1, 2, 3, 0) and v = (5, 1, 2, 2).

(b) the equation of the plane that is parallel to the vector (2, 1, 1) and
contains the line with vector equation

(x, y, z) = (0, 1, 0) + t(1, 1, 1),

where t is a parameter. Put your answer in the form ax + by + cz = d.

1 0 1
3. [7 marks] Find the inverse of the matrix A = 0 1 1 and use it to solve

1 1 0
the system of equations

x + z = 4
y + z = 8
x + y = 6

4. [7 marks] Find the following:

(a) [4 marks] two elementary matrices E and F such that B = F EA, if

1 4 1 1 4 1
A = 2 3 1 and B = 0 5 3

1 1 2 0 3 3
 (b) [3 marks] conditions on b1 , b2 and b3 that ensure the following system is
consistent.
x + y + z = b1
x y + z = b2
5x + y + 5z = b3

5. [8 marks; 4 marks each.] Let

! ! ! !
2 2 0 1 4 1 3 1
A= ,B = ,C = ,D = .
1 3 1 2 1 2 1 1

(a) Find
  1
(AB)1 AC 1 D1 C 1 D1
by first simplifying it as much as possible.
(b) Solve for X if
2B + (A X)T = C.

6. [7 marks]
!
2 2
(a) [4 marks] Let A = . Show that A2 5A + 4I = O.
1 3
(b) [3 marks] Suppose A is any square matrix such that A2 5A + 4I = O.
Find a formula for A1 in terms of A and I.

x1 11 4s + 7t


1 0 4 7 11 x2 3 + s 2t
ANSWERS: 1. 0 1 1 2 3
; = , where s

x3 s
0 0 0 0 0
x4 t
and t are parameters.

2.(a) 1/ 476 2(b) 2x 3y + z = 3

1 1 1 x 4 5
1
3. A1 = 1 1 1
1 ; y = A 8 = 1

2

1 1 1 z 6 9

1 0 0 1 0 0
4.(a) E = 2 1 0 ; F = 0 1 0 4(b) b3 = 2b2 + 3b1

0 0 1 1 0 1
! !
2 1 2 1 1 5
5.(a) 5(b) 6(b) A1 = A + I
1 0 2 5 4 4
 University of Toronto
MAT 188H1F Linear Algebra TERM TEST
Friday, October 10, 2003
Duration: 50 minutes
Only aids permitted: a Casio 260, Sharp 520, or Texas Instrument 30 calculator.

TOTAL MARKS: 45
1. [8 marks] Find the reduced row echelon form of the augmented matrix of the
system of equations
x1 + 3x2 + x3 x4 = 1
x2 x3 + 2x4 = 2
3x1 + 13x2 x3 + 5x4 = 5
and then solve the system.
2. [8 marks; 4 marks for each part] Find the following:
(a) the cosine of the angle between the two vectors
u = (1, 2, 3, 0) and v = (2, 1, 2, 5).

(b) the equation of the plane that contains the three points with coordinates
(1, 1, 2), (0, 2, 2) and (3, 1, 1).
Put your answer in the form ax + by + cz = d.

1 2 3
3. [7 marks] Find the inverse of the matrix A = 2 5 3 and use it to solve

1 0 8
the system of equations
x + 2y + 3z = 1
2x + 5y + 3z = 1
x + 8z = 1

4. [7 marks] Find the following:

(a) [4 marks] two elementary matrices E and F such that B = F EA, if

2 4 6 1 2 3
A = 3 4 1 and B = 0 2 10

1 1 2 1 1 2

(b) [3 marks] conditions on b1 , b2 and b3 that ensure the following system is

consistent.
x + 2y + z = b1
x + y z = b2
x + 3y + 3z = b3
 5. [8 marks; 4 marks each.] Let
! ! ! !
2 2 0 1 4 1 3 1
A= ,B = ,C = ,D = .
1 3 1 2 1 2 1 1

(a) Find
 1
CB 1 (CD) (AD)1 A2
by first simplifying it as much as possible.
(b) Solve for X if
2C + (X A)T = D.

6. [7 marks]
!
2 2
(a) [4 marks] Let A = . Show that A2 5A + 4I = O.
1 3
(b) [3 marks] Suppose A is any square matrix such that A2 5A + 4I = O.
Find a formula for A3 in terms of A and I.

x1 7 4s + 7t


1 0 4 7 7 x2 2 + s 2t
ANSWERS: 1. 0 1 1 2 2
; = , where s

x3 s
0 0 0 0 0
x4 t
and t are parameters.

2.(a) 5/ 119 2(b) x + y + 2z = 6

40 16 9 x 1 47
3. A1 1
= 13 5 3 ; y = A 1 = 15

5 2 1 z 1 6

1/2 0 0 1 0 0
4.(a) E = 0 1 0 ; F = 3 1 0 4(b) b3 = 2b1 b2

0 0 1 0 0 1
! !
1 3 3 5
5.(a) 5(b) 6(b) A3 = 21A 20I
0 4 0 0
 University of Toronto
Faculty of Engineering
MAT 188H1F TERM TEST
FRIDAY, NOVEMBER 16, 2001, 9:10 AM
Duration: 50 minutes

Aids Allowed: A non-programmable calculator, to be supplied by student.

Instructions: Fill in the information on this page, and make sure this test contains
4 pages. Present your solutions in the space provided. Use the back of the preceding
page if you need more space. The value for each question is indicated in square
brackets beside each question number.
TOTAL MARKS: 40

1. [15 marks] Assume that the row-reduced echelon form of the matrix
2 4 1 5 0 1 2 0 3 0

1 2 0 3 2 0 0 1 1 0
A= is R =

.

0 0 1 1 1 0 0 0 0 1


1 2 3 6 1 0 0 0 0 0
Find the dimension of, and a basis for, each of the following subspaces:
(a) the row space of A.
(b) the column space of A.
(c) the solution space of Ax = 0, where x is in R5 .
2. [10 marks] Let W be the subset of M2,2 consisting of all 2 2 matrices, the sum
of whose entries is zero.
(a) Show that W is a subspace of M2,2 .
(b) Find a basis for W, and its dimension.
3. [15 marks; 3 marks each] Determine if each of the following statements is True
or False, and give a brief justification for your choice.
(a) The set {(2, 0, 1, 1), (0, 1, 0, 5), (6, 2, 3, 7)} is a linearly independent set
in R4 .
(b) The subset W of M3,3 consisting of all 3 3 non-invertible matrices is a
subspace of M3,3 .
(c) If the set {u, v, w} is a basis of a vector space V, then so is the set
{u, u + v, u + v + w}.
(d) The set of polynomials {1 + x, 1 x, 2 + x2 , x3 + x4 } is a spanning set of
P4 .
(e) Any seven matrices in M2,3 must be linearly dependent.
 University of Toronto
Faculty of Engineering
MAT 188H1F TERM TEST
FRIDAY, NOVEMBER 16, 2001, 11:10 AM
Duration: 50 minutes
Aids Allowed: A non-programmable calculator, to be supplied by student.
Instructions: Fill in the information on this page, and make sure this test contains
4 pages. Present your solutions in the space provided. Use the back of the preceding
page if you need more space. The value for each question is indicated in square
brackets beside each question number.
TOTAL MARKS: 40

1. [15 marks] Assume that the row-reduced echelon form of the matrix
2 4 1 5 0 1 2 0 3 0

0 0 1 1 1 0 0 1 1 0
A= is R =

.

1 2 3 6 1 0 0 0 0 1


1 2 0 3 2 0 0 0 0 0
Find the dimension of, and a basis for, each of the following subspaces:
(a) the row space of A.
(b) the column space of A.
(c) the solution space of Ax = 0, where x is in R5 .
2. [10 marks] Let W be the subset of P3 consisting of all polynomials p(x) in P3
such that p(1) = 0.
(a) Show that W is a subspace of P3 .
(b) Find a basis for W, and its dimension.
3. [15 marks; 3 marks each] Determine if each of the following statements is True
or False, and give a brief justification for your choice.
(a) The set {(2, 0, 1, 1), (10, 1, 5, 10), (0, 1, 0, 5)} is a linearly independent
set in R4 .
(b) The plane in R3 with equation x + 2y + 3z = 4 is a subspace of R3 .
(c) The dimension of the subspace in R4 consisting of all vectors of the form
(a + b, 2a + 2b, 0, a b) is 2.
(d) The set of matrices
( ! ! ! !)
1 0 1 0 0 1 0 1
, , ,
0 1 0 1 1 0 1 0
is a spanning set of M2,2 .
(e) Any five cubic polynomials must be linearly dependent.
 University of Toronto
Faculty of Engineering
MAT 188H1F TERM TEST
FRIDAY, NOVEMBER 15, 11:10 AM, 2002
Duration: 50 minutes

Aids Allowed: A non-programmable calculator, to be supplied by student.

Instructions: Fill in the information on this page, and make sure this test contains
4 pages. Present your solutions in the space provided. Use the back of the preceding
page if you need more space. The value for each question is indicated in square
brackets beside each question number. TOTAL MARKS: 40

1. [14 marks] Assume that the row-reduced echelon form of the matrix

1 1 4 4 3 1 0 1 5 0

1 1 2 6 3 0 1 3 1 0
A= is R = .

2 0 2 10 0 0 0 0 0 1
1 2 7 3 2 0 0 0 0 0

Find the following:

(a) [4 marks]
(i) the rank of A. Answer:
(ii) the dimension of the row space of A. Answer:
(iii) the dimension of the column space of A. Answer:
(iv) the dimension of the solution space of
the system Ax = 0. Answer:
(b) [3 marks] a basis for the row space of A.
(c) [3 marks] a basis for the column space of A.
(d) [4 marks] a basis for the solution space of the system Ax = 0.

2. [14 marks] Determine if each of the following statements is True or False, and
give a brief justification for your choice.

(a) [2 marks] {(1, 2, 3, 0), (0, 2, 3, 4)} is a spanning set of R4 .

(b) [2 marks] {(1, 3, 4), (6, 5, 2), (1, 7, 3), (4, 3, 6)} is a linearly independent
subset of R3 .
(c) [2 marks] {(1, 3, 4), (1, 0, 0), (0, 1, 0), (0, 0, 1)} is a spanning set of R3 .
(d) [3 marks] The plane with equation 2x + y 4z = 2 is a subspace of R3 .
(e) [3 marks] {(1, 3, 4), (6, 5, 2), (4, 1, 6)} is a linearly independent subset
of R3 .
 (f) [2 marks] The subspace of R4 consisting of all vectors of the form

(a + b, 2a + 2b, a b, 4a + 4b)

has dimension 2.

3. [12 marks; 4 marks each] Determine if the given set W is a subspace of the given
vector space V :

(a) W is the set of all 22 matrices the sum of whose entries is zero; V = M2,2 .
(b) W is the set of all 2 2 matrices whose determinant is zero; V = M2,2 .
(c) W is the subset of P2 consisting of all polynomials p satisfying p(1) = 0;
V = P2 .

ANSWERS: 1(a)(i) 3 1(a)(ii) 3 1(a)(iii) 3 1(a)(iv) 2

1(b) any 3 independent rows of A or R will do. For example,

{(1, 1, 4, 4, 3), (1, 1, 2, 6, 3), (1, 2, 7, 3, 2)} or {(1, 0, 1, 5, 0), (0, 1, 3, 1, 0), (0, 0, 0, 0, 1)}

1(c)
1 1 3




1 1
3
, ,

2 0 0




1 2 2

1(d)



1 5


3 1






1 ,
0







0
1


0 0

2.(a) F (b) F (c) T (d) F (e) F (f) F

3(a) W is a subspace (b) W is not a subspace (c) W is a subspace
 University of Toronto
Faculty of Engineering
MAT 188H1F TERM TEST
FRIDAY, NOVEMBER 15, 12:10 PM, 2002
Duration: 50 minutes

Aids Allowed: A non-programmable calculator, to be supplied by student.

Instructions: Fill in the information on this page, and make sure this test contains
4 pages. Present your solutions in the space provided. Use the back of the preceding
page if you need more space. The value for each question is indicated in square
brackets beside each question number. TOTAL MARKS: 40

1. [14 marks] Assume that the row-reduced echelon form of the matrix

1 1 4 4 3 1 0 1 5 0

1 1 2 6 3 0 1 3 1 0
A= is R = .

2 0 2 10 0 0 0 0 0 1
1 2 7 3 2 0 0 0 0 0

Find the following:

(a) [4 marks]
(i) the rank of A. Answer:
(ii) the dimension of the row space of A. Answer:
(iii) the dimension of the column space of A. Answer:
(iv) the dimension of the solution space of
the system Ax = 0. Answer:
(b) [3 marks] a basis for the row space of A.
(c) [3 marks] a basis for the column space of A.
(d) [4 marks] a basis for the solution space of the system Ax = 0.

2. [14 marks] Determine if each of the following statements is True or False, and
give a brief justification for your choice.

(a) [2 marks] {(1, 2, 3, 0), (0, 2, 3, 4)} is a spanning set of R4 .

(b) [2 marks] {(1, 3, 4), (6, 5, 2), (1, 7, 3), (4, 3, 6)} is a linearly independent
subset of R3 .
(c) [2 marks] {(1, 3, 4), (1, 0, 0), (0, 1, 0), (0, 0, 1)} is a spanning set of R3 .
(d) [3 marks] The plane with equation 2x + y 4z = 0 is a subspace of R3 .
(e) [3 marks] {(1, 3, 4), (6, 5, 2), (4, 1, 6)} is a linearly independent subset
of R3 .
 (f) [2 marks] The subspace of R4 consisting of all vectors of the form

(a + 2c, b + d, 2a + 4c, b d)

has dimension 2.

3. [12 marks; 4 marks each] Determine if the given set W is a subspace of the given
vector space V :

(a) W is the set of all 2 2 matrices whose entries on the main diagonal add
up to zero; V = M2,2 .
(b) W is the set of 2 2 invertible matrices; V = M2,2 .
(c) W is the set of polynomials with degree less than or equal to three, the
sum of whose coefficients is zero; V = P3 .

ANSWERS: 1(a)(i) 3 1(a)(ii) 3 1(a)(iii) 3 1(a)(iv) 2

1(b) any 3 independent rows of A or R will do. For example,

{(1, 1, 4, 4, 3), (1, 1, 2, 6, 3), (1, 2, 7, 3, 2)} or {(1, 0, 1, 5, 0), (0, 1, 3, 1, 0), (0, 0, 0, 0, 1)}

1(c)
1 1 3




1 1
3
, ,

2 0 0




1 2 2

1(d)



1 5


3 1






1 ,
0







0
1


0 0

2.(a) F (b) F (c) T (d) T (e) F (f) T

3(a) W is a subspace (b) W is not a subspace (c) W is a subspace
 Tuesday November 15 START: 13:10 DURATION: 110 mins

University of Toronto

Faculty of Applied Science & Engineering

MIDTERM EXAMINATION II
MAT188H1F
Linear Algebra

EXAMINERS: D. Burbulla, S. Cohen, D. Fusca, F. Lopez, M. Palasciano, M. Pugh, B. Schachter, S. Uppal

Last Name (PRINT):

Given Name(s) (PRINT):

Student NUMBER:

Student SIGNATURE:

EMAIL @mail.utoronto.ca:

Instructions.

1. There are 58 possible marks to be earned in this exam. The examination booklet contains a total of 11 pages. It is
your responsibility to ensure that no pages are missing from your examination. DO NOT DETACH ANY PAGES FROM
YOUR EXAMINATION.
2. DO NOT WRITE ON THE QR CODE AT THE TOP RIGHT-HAND CORNER OF EVERY PAGE OF YOUR EXAM-
INATION.
3. For the full answer questions, WRITE YOUR SOLUTIONS ON THE FRONT OF THE QUESTION PAGES THEM-
SELVES. THE BACK OF EVERY PAGE WILL NOT BE SCANNED AND SEEN BY THE GRADERS.
4. Ensure that your solutions are LEGIBLE.

5. No aids of any kind are permitted. CALCULATORS AND OTHER ELECTRONIC DEVICES (INCLUDING PHONES)
ARE NOT PERMITTED.
6. Have your student card ready for inspection.
7. There are no part marks for Multiple Choice (MC) questions.

8. You may use the two blank pages at the end for rough work. The last two pages of the examination WILL NOT BE
MARKED unless you clearly indicate otherwise on the question pages.
9. For the full answer questions, show all of your work and justify your answers but do not include extraneous information.

1
Part I - Multiple Choice. Clearly indicate your answer to each question by circling your choice. Each question is worth 2 marks.

For each question, choose the BEST option from the given options.

0 0 1 0 1 1
1. Let T : R3 R3 be the linear transformation such that T 1 = 2, T 1 = 0, and T 0 = 0.
1 0 0 1 0 0

0
Determine T 1.
2
1
(A) 2
1

1
(B) 1
1

1
(C) 4
1

1
(D) 1
0

0
(E) 0
0

Answer: C
Solution:
0 1 1
We start by noting that the three argument vectors, 1, 1, and 0, are linearly independent as we can use a linear
1 0 0

0 0 1 1
combination of them to write the three unit vectors. This means that we can express 1 as a1 + b1 + c0. We will find
2 1 0 0

0 0 1 1
what a, b, and c are later, but for now, we know that T 1 = aT 1 + bT 1 + cT 0 since T is a linear
2 1 0 0
transformation. Thus,
0 0 0 1 0 0 1 a
T 1 = a2 + b0 + c0 = 2 0 0 b
2 0 1 0 0 1 0 c
. Also, from before we know that
0 0 1 1 0 1 1 a
1 = a1 + b1 + c0 = 1 1 0 b
2 1 0 0 1 0 0 c
-1
a 0 1 1 0
. Thus, we can find b by evaluating 1 1 0 1.
c 1 0 0 2

2
 -1
0 1 1
To find 1 1 0 , we adjoin the identity matrix and do row operations.
1 0 0

0 1 1 1 0 0
1 1 0 0 1 0
1 0 0 0 0 1

1 0 0 0 0 1
Swap rows 1 and 3. 1 1 0 0 1 0
0 1 1 1 0 0

1 0 0 0 0 1
Subtract row 1 from row 2. 0 1 0 0 1 -1
0 1 1 1 0 0

1 0 0 0 0 1
Subtract row 2 from row 3. 0 1 0 0 1 -1
0 0 1 1 -1 1

-1
0 1 1 0 0 1 a 0 0 1 0 2
Thus, we get 1 1 0 = 0 1 -1, which means that b = 0 1 -1 1 = -1.
0 0 1 -1 1
1 c 1 -1 1 2 1
0 0 0 1 2 1
Therefore, T 1 = 2 0 0 -1 = 4 and the answer is C.
2 0 1 0 1 -1

3
        
a a + 2b 3 c
2. Let T : R2 R2 be the linear transformation defined by T = . If T 1 = , find c + d.
b a+b 1 d
(A) -1
(B) 0
(C) 3
(D) 1
(E) 2

Answer: D
Solution:          
a a + 2b a 1 2 a
We know that if T = then T = .
b a+b b 1 1 b
           
3 c 3 c 1 2 c
Thus if T -1 = , then =T = .
1 d 1 d 1 1 d
   -1  
c 1 2 3
This means that = .
d 1 1 1
 -1
1 2
To find , we adjoin the identity matrix and do row operations.
1 1
 
1 2 1 0
1 1 0 1
 
1 2 1 0
Subtract row 1 from row 2.
0 -1 -1 1
 
1 2 1 0
Negate row 2.
0 1 1 -1
 
1 0 -1 2
Subtract twice row 2 from row 1.
0 1 1 -1

 -1          
1 2 -1 2 c -1 2 3 -1
Thus, we get = , which means that = = .
1 1 1 -1 d 1 -1 1 2
If c = -1 and d = 2 then c + d = 1, and so the answer is D.

4
Part I - Multiple Choice. Clearly indicate your answer to each question by circling your choice. Each question is worth 2 marks.

For each question, choose the BEST option from the given options.

3. Which of the following functions T : R2 R are linear transformations?

 
x
(i) T = 2x + 5y 3.
y
 
x
(ii) T = 2x + 5y.
y
 
x
(iii) T = |x|
y

(A) (i) only

(B) (ii) only
(C) (i) and (ii) only
(D) (iii) only
(E) (ii) and (iii) only

Answer: B
Solution:         
x x x 1
i) Does not satisfy T c = cT . For example, take = and c = 3.
y y y 2
 
    x
ii) Works because we can represent T by the matrix 2 5 since 2 5 = 2x + 5y and matrices are linear.
     y
x x
iii) Does not work as it does not satisfy T - = -T .
y y
Thus the answer is B.

5
4. Let A, B, and C be n n matrices. Which of the following statements are TRUE?

(i) If Ax = Bx for every x Rn , then A = B.

(ii) If A 6= 0 and AB = AC, then B = C.
(iii) Either the system Ax = b has a unique solution for every b Rn or there exist non-trivial solutions to the homogenous
system Ax = 0.

(A) (i) and (ii) only

(B) (ii) only
(C) (i) and (iii) only
(D) (i), (ii), and (iii)
(E) none of (i), (ii), or (iii)

Answer: C
Solution:
i) If A and B differ in the j th column then we can consider x=ej . The output of A x will be the j th column of A. The
output of B x will be the j th column of B. If these
 outputs
 are the
  same, the columns
 cannot differ.
1 0 0 0 0 0
ii) This is false. One counterexample is A = ,B= , and C = . Here, A 6= 0, the left and right sides are
  0 0 0 0 0 1
0 0
both , yet B 6= C.
0 0
iii) This is true. The forwards direction is true because if there were two different solutions to the equation we would have
Ax1 = b and Ax2 = b. Thus, Ax1 Ax2 = 0 so A(x1 x2 ) = 0. The backwards direction is true because we can just choose
b = 0 and we have found our solution.

6
5. Let u, v, w be non-zero, non-parallel vectors in R4 and let W = span{u, v u, u w, v + w}. Determine all possible values
of dim(W ).
(A) 4
(B) 1 or 2
(C) 2 or 3
(D) 1, 2, or 3
(E) 1, 2, 3, or 4

Answer: C
Solution:
We observe that since 2u + (v u) (u w) (v + w) = 0, then the set whose span makes up W is not linearly independent
and so we can remove one of the vectors from the set and W will be unchanged. This also makes sense as all 4 vectors in the
definition of W are linear combinations of only three vectors, u, v, and w. Thus, dim(W ) cannot be 4. Also, if u and v are not
parallel, then u and v u are not parallel so dim(W ) is at least 2. Thus, the only option that this leaves us C.

7
Part II - Short Answer Questions. Write your solutions in the space provided below each question.

3 6 1 5 1 1 0 0 1 1
1. Suppose you are given that the matrix A = 1 2 1 2 0 has row-echelon form R = 0 0 1 1 1.
2 4 0 3 1 0 0 0 0 0

(a) Determine rank(A) and nullity(A). [2 marks]

The rank of a matrix is equal to the dimension of the row space, since the row space is invariant under row operations, and the
dimension of the row space of R is clearly 2, rank(A)=2.

By the rank-nullity theorem, we know that since A has 5 columns, rank(A)+nullity(A)=5 and since rank(A)=2, nullity(A)=3.

(b) Find a basis for row(A). [2 marks]

Whenever we perform a row operation on A, the basis for the row space of the resulting matrix will still be a basis for the row
space of the initial matrix. Thus, any basis for row(R) will be a basis for row(A).


1 0
0 0



It is clear that an adequate basis for row(R) is 0 , 1 , which would also then be an adequate basis for row(A).



1 -1

1 1

(c) Find a basis for col(A). [2 marks]

Since the row rank of A is equal to the column rank of A, we need to choose two linearly
independent
columns for our basis.
3 5
This means choosing two columns that are not multiples of each other, which means 1, 2 suffice.
2 3

(d) Find a basis for null(A). [2 marks]

We need to solve Ax = 0. We can perform row operations on this system until it is of the form Rx = 0. Thus, we get two
constrains, that x1 = -x4 x5 and x3 = x4 x5 . If we set x2 = a, x4 = b, and x5 = c, we get that

x1 -b c 0 -1 -1
x2 a 1 0 0

x3 = b c = a 0 + b 1 + c -1 .

x4 b 0 1 0
x5 c 0 0 1


0 -1 -1

1 0 0


Thus, our basis for the null space of A is 0 , 1 , -1 as it is the basis to the solution space of the homogeneous equation

0 1 0




0 0 1

Ax = 0.

8
2. (a) Let W be a subspace of Rn . Define what it means for a set S of vectors in Rn to be a basis for W . [2 marks]

The definition of a basis of a subspace W is a set of vectors, S W such that

S is linearly independent, and
span(S)=W .

x1 1 3
2. (b) Consider the subspace W = x2 | x1 + x2 + x3 = 0 . Is the set S = 1 , 2 a basis for W ? Support your
x3 2 1

answer. [6 marks]

Answer: Yes.

Solution:

Notice that W is the set of solutions to the homogeneous equation x

1+ x2+
x3 =0.
Since the set of solutions requires two
1 1
parameters, the dimension of W is 2. Notice that, for example, the set 1 , 0 is a basis for W . Since the dimension
0 1

of W is 2, any two linearly independent vectors from W will be a basis for W (cf. Tutorial Problems 5, 3(b)).
Thevectors in
1 3
the set S are in W since -1+(-1)+2=0 and -3+2+1=0. They are also linearly independent since the matrix 1 2 has rank
2 1
2 so S is indeed a basis for W .

9
3. (a) Let W be a subspace of Rn . Define dim(W ). [2 marks]

The dimension of W is the number of vectors in every basis of W or the minimum number of vector required to span W .

1 1 c
3. (b) Let W = span 1 , c , 1 . For what value(s) of c is dim(W ) = 2. Support your answer. [6 marks]
c 1 1

Answer: c=-2

Solution:

1 1 c
We look for the value(s) of c such that A = 1 c 1 has rank 2.
c 1 1
Rearranging the matrix into its RREF form gives us the following.

1 1 c
1 c 1
c 1 1

1 1 c
Subtract row 1 from row 2. 0 c1 1c
c 1 1

1 1 c
Subtract c times row 1 from row 3. 0 c1 1c
0 1c 1 c2

1 1 c
Add row 2 to row 3. 0 c1 1c
2
0 0 2cc

At this point we can stop. We see that the rank of A may be 2 if the expression 2 c c2 = (1 c)(c + 2) is equal to 0. i.e.
c = 1 or c = 2. However, if c = 1 the rank of A is 1 and the dimension of W is 1. If c = 2, the rank of A is 2 and so the only
value of c for which the dimension of W is 2 is -2.

10

1 1 1
4. (a) Find the inverse of A = 2 3 0 . [4 marks]
1 1 2
We do this by adjoining the identity matrix and putting the matrix A into its RREF form. We get the following series of
operations.

1 -1 1 1 0 0
2 -3 0 0 1 0
-1 1 -2 0 0 1

1 -1 1 1 0 0
Subtract twice row 1 from row 2. 0 -1 -2 -2 1 0
-1 1 -2 0 0 1

1 -1 1 1 0 0
Add row 1 to row 3. 0 -1 -2 -2 1 0
0 0 -1 1 0 1

1 -1 1 1 0 0
Negate rows 2 and 3. 0 1 2 2 -1 0
0 0 1 -1 0 -1

1 -1 1 1 0 0
Subtract twice row 3 from row 2. 0 1 0 4 -1 2
0 0 1 -1 0 -1

1 -1 0 2 0 1
Subtract row 3 from row 1. 0 1 0 4 -1 2
0 0 1 -1 0 -1

1 0 0 6 -1 3
Add row 2 to row 1. 0 1 0 4 -1 2
0 0 1 -1 0 -1

6 -1 3
Thus the inverse of A is 4 -1 2 .
-1 0 -1

11

x1 3 1 2 y1 z1 1 1 1 y1 x1 z1
4. (b) Given x2 = 1 0 4 y2 and z2 = 2 3 0 y2 . Find a matrix C such that x2 = C z2 .
x3 2 1 0 y3 z3 1 1 2 y3 x3 z3
Suggestion: Use part (a). [4 marks]

z1 y1 z1 y1
We see that z2 = A y2 (where A is the same as in part a). Thus, we get that A-1 z2 = y2 . This immediately tells us
z3 y3 z3 y3
x1 3 1 2 z1 3 1 2 6 -1 3 z1
that x2 = 1 0 4 A-1 z2 = 1 0 4 4 -1 2 z2 .
x3 2 1 0 z3 2 1 0 -1 0 -1 z3

3 1 2 6 -1 3 12 -2 5
Hence, C = 1 0 4 4 -1 2 = 2 -1 -1.
2 1 0 -1 0 -1 16 -3 8

12

5. Define a linear mapping
 T : R2 R2 by the following rule: T (x) is the result offirst
 rotating
  x counter-clockwise
   by
 4 and
1 1 0 1 1 0
then multiplying by A = . Let S be the unit square in R2 with vertices x = ,y = ,z = , and w = .
0 1 0 0 1 1

(a) Find a matrix B such that T (x) = Bx for every x R2 . [2 marks] " #
2
cos 4
- sin
-22



The rotation matrix for 4 counterclockwise is given by 4
= 2 .
sin 4 cos
4
2 2
2 2" "
# #
2
-22
2 0

From this, we get that to find T , we first apply the rotation, followed by A, so B = A 2
2 2
= 2 2 .
2 2 2 2

(b) Find and sketch, as accurately as possible, the image of S under T . [6 marks]
  " #  
0 2 2
We apply the matrix B to each vertex given in the problem and find that the vertices get mapped to , 2 , , and
0 2
2
 
0
, respectively. The square, then, gets mapped to the parallelogram with vertices at these points.
2
2

13
6. The diagram below is a map of a downtown area of a city. Each street is one-way in the direction of the respective arrows.
The numbers represent the average number of cars per minute that enter or leave a given street at 6:00pm. The variables also
represent the average numbers of cars per minute. Of course, unless there is an accident, the total number of cars entering any
intersection must equal the total number leaving. Thus, at the intersection of West and North street we have x + y = 50.

(a) Starting at the intersection of West and North Street, and continuing clockwise around the square, write a system of linear
equations that describes the traffic flow (assuming there are no accidents) and find all solutions to this system. You should find
there are infinitely many solutions with one parameter. Use w as your free variable. [5 marks]

In the order indicated in the problem, we write a system of equation by added all the incoming cars and subtracting all of the
leaving cars and setting that equal to 0. Thus we get

x + y 20 30 = 0
20 x w = 0
z + w 20 30 = 0
50 + 30 y z = 0

1 1 0 0 x 50
1 0 0 1
y = 20.

Therefore, we get that

0 0 1 1 z 50
0 1 1 0 w 80
1 0 0 1 x 20
0 1 0 -1 y 30
Row reducing this system of equations turns it into the system = .
0 0 1 1 z 50
0 0 0 0 w 0
Our solution is therefore x = 20 w, y = 30 + w, z = 50 w, and w is a free variable.

14
(b) Nefarious Construction Co. wants to close down West Street for six months but the City Council refused to grant the permit.
Explain why on the basis of the solutions to the system in part (a). [3 marks]

This has the same effect as setting x = 0, and by part a, this means that w = 20, so y = 50 and z = 30. This has the effect
of causing massive traffic jams along North Street. Before, when we had a smaller value for w, such as 10, there was 20% less
traffic among North Street and it could operate fine.

15
 Solutions to MAT188H1F - Linear Algebra - Fall 2014

Term Test 1 - September 30, 2014

Time allotted: 100 minutes. Aids permitted: Casio FX-991 or Sharp EL-520 calculator.

This test consists of 8 questions. Each question is worth 10 marks. Total Marks: 80

General Comments:

1. The range on every question was 0 to 10. Q4 and Q5 had the highest averages; Q2 and Q8 the lowest.

2. All the questions on this test, except for 2(b), were very similar to homework problems. In 2(b) the
usual process was reversed: given the solution what is the corresponding reduced augmented matrix?

3. In Q3 to Q8 you must explain your work fully to receive full marks. Many students simply wrote
down some matrices or equations with no indication of what they were doing, or why they were doing
it! The markers will not fill in the details for you; you must make it clear what you are doing.

Breakdown of Results: 962 students wrote this test. The marks ranged from 30% to 100%, and the
average was 78.4%. Some statistics on grade distributions are in the table on the left, and a histogram of
the marks (by decade) is on the right.

Grade % Decade %
90-100% 18.5%
A 50.3% 80-89% 31.8%
B 29.8% 70-79% 29.8%
C 14.2% 60-69% 14.2%
D 3.7% 50-59% 3.7%
F 1.9% 40-49% 1.3%
30-39% 0.6%
20-29% 0.0%
10-19% 0.0%
0-9% 0.0%
MAT188H1F Term Test 1

PART I : No explanation is necessary.

1. (avg: 7.6) Let

1 0

2 0
" # 1 2 8 4 1 2 8 4
1 0 2 16 3
A= , B = , C= 0 1 0 4 , D= 0 0 3 1

3 0 0 1 7 8 5
0 0 3 1 0 0 0 4
0 0

be the augmented matrices of four systems of linear equations. Answer the following questions. Note:
there can be more than one answer per part. For this question all correct choices count as +1 and
all incorrect choices count as 1.

(a) Which of the four above matrices are in echelon form? B, C, D

(b) Which of the four above matrices are in reduced echelon form? B

(c) Which of the four above matrices are augmented matrices

of an inconsistent linear system of equations? D

(d) Which of the four above matrices are augmented matrices

of a system of linear equations with a unique solution? A, C

(e) Which of the four above matrices are augmented matrices

of an echelon system of linear equations? B, C

(f) Which of the four above matrices are augmented matrices

of a triangular system of linear equations? C

Page 2 of 9 Continued...
MAT188H1F Term Test 1

2. (avg: 6.6) Each part is worth 5 marks.

(a) The general solution to a linear system of equations is

x1 = 4 + 6s1 5s2
x2 = s1
,
x3 = 9 + 3s2
x4 = s2

for parameters s1 , s2 . Express the solution as a linear combination of three vectors in R4 .

Solution:
x1 4 6 5

x 0 1 0
2
= + s1 + s2


x3 9 0 3

x4 0 0 1

(b) Find a vector b in R2 and a matrix A with four columns each of which is a vector in R2 , such
that the equation A x = b has solution

4 + 6s1 5s2

s1
x= .


9 + 3s2

s2

Solution: infinitely many possible answers. Easiest is to pick the reduced echelon matrix that
produces the given solution:

" # " #
1 6 0 5 4
A= , b=
0 0 1 3 9

Page 3 of 9 Continued...
MAT188H1F Term Test 1

PART II : Present COMPLETE solutions to the following questions in the space provided.

3. (avg: 8.4) Solve the system of linear equations

x1 + 2x2 + x3 + 3x4 = 0
x1 + x2 x3 2x4 = 2


2x + 3x3 + 5x4 = 5
1

by reducing its augmented matrix to reduced echelon form.

Solution: row reduce the augmented matrix.

1 2 1 3 0 1 2 1 3 0 1 2 1 3 0

0 1 2 5 2 0 1 2 5 2
1 1 1 2 2

2 0 3 5 5 0 4 5 11 5 0 0 3 9 3

1 2 0 0 1 1 0 0 2 1

0 1 0 1
0 1 0 1 0 0

0 0 1 3 1 0 0 1 3 1
This last matrix is the reduced echelon form of the augmented matrix of the given system of equations.
Let x4 = s be a parameter. Then

x1 1 2s 1 2

x s 0 1
2
= = + s


x3 1 3s 1 3

x4 s 0 1

Page 4 of 9 Continued...
MAT188H1F Term Test 1

4. (avg: 9.2) A total of 275 people attend a concert. Ticket prices are $12 for adults, $10 for seniors and
$8 for students. The total revenue was $3100. Determine how many adults, seniors and students
attended the concert, given that the number of seniors who attended was twice the number of
students.

Solution: let x, y, z be the number of adults, seniors and students, respectively, that attended. We
have
x + y + z = 275, 12x + 10y + 8z = 3100, y 2z = 0.

The augmented matrix for this system is

1 1 1 275 1 1 1 275 1 0 1 175 1 0 0 200

0 2 0 1 0 1 0 50 .
12 10 8 3100 4 200 2 100

0 1 2 0 0 1 2 0 0 0 4 100 0 0 1 25

So the unique solution is x = 200, y = 50 and z = 25. That is, 200 adults, 50 seniors, and 25 students
attended the concert.

Page 5 of 9 Continued...
MAT188H1F Term Test 1

5. (avg: 9.2) Balance the chemical reaction

C3 H8 + O2 CO2 + H2 O

which describes the oxidation of propane to produce carbon dioxide and water, by first setting up
and then solving an appropriate homogeneous system of equations.

Solution: suppose the balanced reaction looks like

x1 C3 H8 + x2 O2 x3 CO2 + x4 H2 O

Count the number of like atoms on each side of the reaction:

Carbon: 3x1 = x3 3x1 x3 = 0

Hydrogen: 8x1 = 2x4 4x1 x4 = 0
Oxygen: 2x2 = 2x3 + x4 2x2 2x3 x4 = 0

Reduce the augmented matrix for this homogeneous system:

3 0 1 0 0 3 0 1 0 0 3 0 1 0 0 4 0 0 1 0

4 0
0 1 0 0
0 4 3 0 0 2 2 1 0
0 4 0 5 0 .

0 2 2 1 0 0 2 2 1 0 0 0 4 3 0 0 0 4 3 0

Let x4 = s, then
s 5s 3s
x1 = , x2 = , x3 = .
4 4 4
To find the smallest whole integer solution, let s = 4, so that

x1 = 1, x2 = 5, x3 = 3.

That is, the balanced chemical reaction is

C3 H8 + 5O2 3CO2 + 4H2 O

Page 6 of 9 Continued...
MAT188H1F Term Test 1

6. (avg: 7.4) Find a set of vectors {u, v} in R4 that spans the solution set of the system of equations



x1 + 3x2 x3 + 2x4 = 0
3x1 + 9x2 11x3 + 14x4 = 0


2x 6x 6x + 4x = 0
1 2 3 4

Solution: first solve the system, for example by row reduction on the augmented matrix.

1 3 1 2 0 1 3 1 2 0 1 3 0 1 0

3
9 11 14 0 0 0 8 8 0 0
0 1 1 0 .
2 6 6 4 0 0 0 8 8 0 0 0 0 0 0

Let x2 = s1 , x3 = s2 be parameters. Then

x1 3s1 s2 3 1

x s1 1 0
2
= = s1 + s2 .


x3 s2 0
1

x4 s2 0 1

3 1

1 0
Thus every solution is a linear combination of and ; take


0

1

0 1

3 1

1 0
u= , v = .


0

1

0 1

Page 7 of 9 Continued...
MAT188H1F Term Test 1

7. (avg: 7.6) Find all values of h such that the set of vectors {a1 , a2 , a3 } spans R3 if

1 3 1

a1 =
1 , a2 = h , a3 = 0 .

1 4 1

Solution: need to find all values of h such that any vector in R3 can be written as a linear combination
of a1 , a2 , a3 . That is, the vector equation

a

x1 a1 + x2 a2 + x3 a3 =
b

c

must be consistent for every choice of a, b, c. Reduce the corresponding augmented matrix:

1 3 1 a 1 3 1 a 1 3 1 a

0 h 3 1 b a
1 h 0 b 0 7 2 c + a

1 4 1 c 0 7 2 c+a 0 h 3 1 b a

If h = 3 this matrix corresponds to a triangular system, which is always consistent. Now proceed
with h 6= 3:

1 3 1 a 1 3 1 a

0 7 2 c + a 0 7 2 c + a .

0 h 3 1 b a 0 0 2h + 1 (h 3)(c + a) 7(b a)

As long as 2h + 1 6= 0 this matrix corresponds to a triangular system which has a solution regardless
of a, b, c. Thus: the set of vectors {a1 , a2 , a3 } spans R3 if
1
h 6= .
2
1
If h = , then the augmented matrix is
2

1 3 1 a

0 7 2 c+a ,

0 0 0 (7/2)(c + a) 7(b a)

which represents an inconsistent system if (say) a = 1, b = 0, c = 1; that is, the bottom right
1
hand corner is 7 6= 0. Answer: h 6= .
2

Page 8 of 9 Continued...
MAT188H1F Term Test 1

8. (avg: 6.5) Indicate if the following statements are True or False, and give a brief explanation why.

(a) (2 marks) If a matrix has more rows than columns and is in echelon form then it must have at
N
least one row of zeros at the bottom. True False

Solution: let the number of rows of the matrix be n, let the number of columns be m. Then
n > m, and the number of leading entries in the echelon form of the matrix can be at most m,
ie one per column. So there are at least n m rows with no leading entries; they are zero rows.

(b) (2 marks) For every matrix A with columns that span Rn , A x = 0 has non-trivial solutions.
N
True False

1 0 0

0 1 0 .
Solution: not true for A =
0 0 1

2
1
1

N
(c) (2 marks) 3 is in the span of 0 , 0 .
True False



1 1 7

Solution: look at the middle components. s 0 + t 0 = 0 6= 3

(d) (2 marks) Every system of linear equations with more variables than equations must have
N
infinitely many solutions. True False

Solution: the system of two equations x1 + x2 + x3 = 1 and x1 + x2 + x3 = 2 in three variables

has no solutions, because it is obviously inconsistent.

(e) (2 marks) If a system of linear equations has the trivial solution then it must be a homogeneous
N
system of equations. True False

Solution: A x = b and x = 0 together imply b = 0, since A 0 = 0.

Page 9 of 9 The end.

 University of Toronto
FACULTY OF APPLIED SCIENCE AND ENGINEERING
Solutions to FINAL EXAMINATION, DECEMBER, 2012
First Year - CHE, CIV, CPE, ELE, ENG, IND, LME, MEC, MMS

MAT188H1F - LINEAR ALGEBRA

Exam Type: A

General Comments:

1. Most of this exam consisted of completely routine questions. In particular, questions

4, 5, 7 and 8 were almost identical to questions on previous exams.

2. The only question that could be considered non-routine is question 9. However, there
are many correct ways to do both parts (a) and (b), and many people got them. But
most students had no idea how to tackle part (b). Some students wrote things like
B = 3I = I or

1 0 0 1 0 0
~u~uT + ~v~v T + (~u ~v )(~u ~v )T = 0 + 1 + 0 = 0 1 0 !
0 0 1 0 0 1

Others wrote things like ~u ~uT = k~uk2 . In fact, ~u ~uT is a 3 3 matrix.

3. The range on every question was zero to perfect. Students did surprisingly well on the
True and False, but surprisingly poorly on the first question.

4. There were three perfect papers.

Breakdown of Results: 971 students wrote this exam. The marks ranged from 9% to
100%, and the average was 70.3%. Some statistics on grade distributions are in the table on
the left, and a histogram of the marks (by decade) is on the right.

Grade % Decade %
90-100% 6.6%
A 27.8% 80-89% 21.2%
B 29.0% 70-79% 29.0%
C 23.0% 60-69% 23.0%
D 12.2% 50-59% 12.2%
F 8.0% 40-49% 4.7%
30-39% 1.8%
20-29% 1.0%
10-19% 0.4%
0-9% 0.1%
1. [avg: 5.3/10] Find the following:

(a) [5 marks] a vector equation for the line of intersection common to the two planes
with equations x + y z = 5 and 2x + y + 2z = 2.

Solution: solve the system.

     
1 1 1 5 1 1 1 5 1 0 3 3

2 1 2 2 0 1 4 8 0 1 4 8

Let z = t be a parameter, then

x 3 3t 3 3
y = 8 + 4t = 8 + t 4 ,
z t 0 1

which is the vector equation of a line.

(b) [5 marks] the shortest distance between the two parallel lines L1 and L2

x = 1 + t x = 2 s
L1 : y = 0 t ; L2 : y = 3 + s
z = 1 + t z = 1 s

where s and t are parameters.

Solution: let the shortest distance be D. Let P be the point on L1 with coordi-
nates (1, 0, 1); let Q be the point on L2 with coordinates (2, 3, 1); let

1
~n = 1
1
2 2

be a direction vector for both lines. Then D2 + proj~n P Q = P Q , so

1 2
2
2
2
1
P Q ~n 3 2 1 = 10 4 = 26 ;

D2 = P Q ~n =

~n ~n
0
3 3 3
1

and r
26
D= .
3
2. [avg: 7.9/10]

(a) [5 marks] Find the area of the triangle P QR passing through the three points

P (1, 3, 1), Q(2, 2, 2), R(0, 2, 0).

Solution: let the area be A. Then

1 1 2
8

1 1
1
A = P Q P R = 1 1 = 0
= = 2.
2 2 2 2
1 1 2

n T o
(b) [5 marks] Let U = x y z | xyz = 0 .

1. Is U non-empty?

Solution: No. The vector [ 0 0 0 ]T is in U since (0)(0)(0) = 0.

2. Is U closed under scalar multiplication?

Solution: Yes. If the vector X = [ x y z ]T is in U, so is tX for any scalar t,

since (tx)(ty)(tz) = t3 xyz = t3 (0) = 0.

3. Is U closed under vector addition?

Solution: No. The vectors X = [ 1 0 0 ]T and Y = [ 0 1 1 ]T are in U but

the vector X + Y = [ 1 1 1 ]T is not in U, since (1)(1)(1) = 1 6= 0.

4. Is U a subspace of R3 ?

Solution: No, because U is not closed under addition.

3. [2 marks for each part; avg: 7.9/10] Decide if the following statements are True or False,
and give a brief, concise justification for your choice. Circle your choice.

(a) If = 0 is an eigenvalue of the n n matrix A then A is not invertible.

True or False

True: let a corresponding eigenvector be X; so X is non-zero and AX = 0X = O.

Thus the homogeneous system has a non-zero solution, which is equivalent to A
being non-invertible. OR: det(0I A) = 0 det(A) = 0, so A is not invertible.

1 1 1 2
(b) dim im 3 6 1 1 = 3 True or False
5 8 3 3

False: dim (im(A)) = r, where r is the rank of A. The matrix

1 1 1 2
3 6 1 1
5 8 3 3

has rank 2 because it only has 2 independent rows: R3 = R2 + 2R1 .

(c) If U is a subspace of R7 and dim U = 4 then dim U = 3. True or False

True: 7 = dim U + dim U dim U = 7 4 = 3.

1 1 1 0
(d) 1 is in the span of
3 , 0 , 3 .
True or False
2 2 1 1

False: in the spanning set, the first vector is the sum of the next two. So the
question reduces to, Are there scalars a and b such that

1 1 0
1 = a 0 + b 3 ?
2 1 1

This system is obviously inconsistent, since 1 + 1/3 6= 2.

(e) If {X1 , X2 , X3 } is linearly independent, then so is {X2 , X3 }. True or False

True: sX2 + tX3 = O 0X1 + sX2 + tX3 = O s = t = 0 since {X1 , X2 , X3 }

is linearly independent.
4. [10 marks; avg: 8.5/10] Given that the reduced row-echelon form of

1 0 2 1 1 1 0 2 1 0
1 2 0 1 0 0 1 1 0 0
A= is R = ,
1 2 4 1 2 0 0 0 0 1
2 1 3 2 2 0 0 0 0 0

state the rank of A, and find a basis for each of the following: the row space of A, the
column space of A, and the null space of A.

Solution: the rank of A is r = 3, the number of leading 1s in R.

U dim U description of basis vectors in basis

1 0 0

0 1 0




rowA r=3 three non-zero rows of R 2 , 1 , 0



1 0 0


0 0 1

three independent rows of A {R1 , R2 , R4 }

{R1 , R3 , R4 }
{R2 , R3 , R4 }

NB: R3 = R2 + 2R1

col A r=3 three independent columns of A {C1 , C2 , C5 }

NB: C3 = 2C1 + C2 ; C4 = C1

(not three independent columns of R) (colR 6= colA)

2 1
1 0




null A 5r =2 two basic solutions to RX = 0 1 , 0



0 1


0 0

5. [avg: 8.5/12] Let T : R2 R2 be the linear transformation defined by
   
x 2x + 5y
T = .
y 2x + y

(a) [6 marks] Draw the image under T of the unit square, and calculate its area.

Solution: the image of the unit square is the parallelogram determined by

       
1 2 0 5
T (~i) = T = and T (~j) = T = .
0 2 1 1

In the diagram to the right, the

unit square is in blue, and the
image of the unit square is in
green. Its area is
  
det 2 5  = |2 10| = 8.
 
 2 1 

 
x
(b) [6 marks] Find the formula for T R if R is a rotation of /6 clockwise
y
around the origin.

Solution: the matrix of R is

   
cos(/6) sin(/6) 1 3 1
= =B
sin(/6) cos(/6) 2 1 3

and the matrix of T is  

2 5
= A;
2 1
so
      
x x 1 23 5 2 + 5 3 x
T R = AB =
y y 2 2 31 2+ 3 y
 
1 (2 3 5) x + (2 + 5 3) y
= .
2 (2 3 1) x + (2 + 3) y
  
1 8 6
6. [avg: 8.8/12] Let Q = .
10 6 8

(a) [4 marks] Find the eigenvalues of Q.

Solution:
8 6

    
10 2 64 36
det(I Q) = det 6
10
8 = = 2 1.
10 10 100 100

So the eigenvalues of Q are 1 = 1 and 2 = 1.

(6) [8 marks] Find a basis for each eigenspace1 of Q and plot the eigenspaces of Q in
the plane, indicating which eigenspace corresponds to which eigenvalue of Q.

Solution:
1 + 45 35
     
9 3 3 1
E1 (Q) = null(IQ) = null = null = null ;
53 1 45 3 1 0 0

so could take  
1
X1 = .
3
4
53
     
1 + 5
1 3 1 3
E1 (Q) = null(IQ) = null = null = null ;
35 1 4
5
3 9 0 0
so could take  
3
X2 = .
1
Plot of Eignespaces:
E (Q)
y  1



X1
PPX2 
iPP
P 
PP
P
 PPP x
 PP


 E1 (Q)





Note: Q is the matrix of a reflection in the line y = 3x, so E1 (Q) is the axis of
reflection, and E1 (Q) must be the line orthogonal to the axis of reflection.

1
Recall: if A is an nn matrix, the eigenspace of A corresponding to is E (A) = {X Rn | AX = X}.
7. [avg: 10.4/12]

Let U = span X1 = [ 0 1 1 0 ]T , X2 = [ 1 0 0 1 ]T , X3 = [ 1 1 0 0 ]T ;
 T
let X = 1 1 0 1 . Find:

(a) [6 marks] an orthogonal basis of U.

Solution: since X1 X2 = 0 already, you only need to use the Gram-Schmidt

algorithm to find F3 . Take F1 = X1 , F2 = X2 , and

1
X3 X1 X3 X2 1 1 1 1 .

F3 = X3 X 1 X2 = X 3 + X1 X2 =
kX1 k2 kX2 k2 2 2 2 1
1

Optional: clear fractions and take

1
1
F3 =
1 .

Either way {F1 , F2 , F3 } is an orthogonal basis of U.

(b) [6 marks] projU (X).

Solution: using {F1 , F2 , F3 } with fractions cleared.

X F1 X F2 X F3
projU X = F 1 + F2 + F3
kF1 k2 kF2 k2 kF3 k2
1 1
= F1 + (0)F2 + F3
2 4
1
1 1 .

=
4 3

1

Cross-check/Alternate Solution: U = span{Y } with Y = [ 1 1 1 1 ]T .

Then

1
X Y 3 1 1 .

projU X = X projU (X) = X Y = X Y =
kY k2 4 4 3
1
8. [12 marks; avg: 9.8/12] Find an orthogonal matrix P and a diagonal matrix D such that
D = P T AP, if
1 0 1
A = 0 2 0 .
1 0 1

Step 1: Find the eigenvalues of A.

1 0 1  
1 1
det(I A) = det 0 2 0 = ( 2) det
1 1
1 0 1
= ( 2) ( 1)2 1 = ( 2)(2 2)

= ( 2)2

So the eigenvalues of A are 1 = 0 and 2 = 2.

Step 2: Find mutually orthogonal eigenvectors of A.

1 0 1 1 0 1 1
E0 (A) = null 0 2 0 = null 0 1 0 = span 0 .
1 0 1 0 0 0 1

1 0 1 1 0 1 1 0
E2 (A) = null 0 0 0 = null 0 0 0 = span 0 , 1 .
1 0 1 0 0 0 1 0

OR: use either E0 (A) = (E2 (A)) or E2 (A) = (E0 (A)) to simplify calculations. That
is, E2 (A) is the plane with equation x = z and E0 (A) is the line normal to the plane.

Step 3: Divide each eigenvector by its length to get an orthonormal basis of eigenvec-
tors, which are put into the columns of P. So

1/ 2 1/ 2 0 0 0 0
P = 0 0 1 and D = 0 2 0
1/ 2 1/ 2 0 0 0 2
9. [avg: 3.3/12] Suppose ~u and ~v are two orthogonal unit vectors in R3 .
 
Let A = ~u ~v ~u ~v and let B = ~u ~uT + ~v ~v T + (~u ~v ) (~u ~v )T .

(a) [6 marks] Explain why A must be an orthogonal matrix.

Solution 1: show that the columns of A form an orthonormal basis of R3 .

It is given that k~uk = k~v k = 1 and ~u ~v = 0; and ~u v is always orthogonal to
both ~u and ~v . So need only show that ~u ~v is also a unit vector:

k~u ~v k = k~ukk~v k sin(/2) = (1)(1)(1) = 1.

 T  
Solution 2: show that AT A = I: AT A = ~u ~v ~u ~v ~u ~v ~u ~v =

~uT   ~
u T
~
u ~
u T
~
v ~
u T
(~u ~
v )
~v T ~u ~v ~u ~v = ~v T ~u ~v T ~v ~v T (~u ~v )
(~u ~v )T (~u ~v )T ~u (~u ~v )T ~v (~u ~v )T (~u ~v )

~u ~u ~u ~v ~u (~u ~v ) k~uk2 0 0
= ~v ~u ~v ~v ~v (~u ~v ) = 0 k~v k2 0 = I.
2
(~u ~v ) ~u (~u ~v ) ~v (~u ~v ) (~u ~v ) 0 0 k~u ~v k

(b) [6 marks] Explain why B must be I, the 3 3 identity matrix.

~ = ~u ~v . By part (a), AAT = I, so

Solution 1: To save space, let w
T
 ~u
I = AAT = ~u ~v w ~ ~v T = ~u ~uT + ~v ~v T + w ~ T = B.

~w
~T
w

Solution 2: {~u, ~v , w}~ is an orthonormal basis. Apply the Expansion Theorem,

3
for ~x R : ~x = (~x ~u)~u +(~x ~v )~v +(~x w)
~ w~ = ~u~uT ~x +~v~v T ~x + w
~w~ T ~x = B~x B = I.
Solution 3: show ~u, ~v , w ~ are all eigenvectors of B with eigenvalue = 1.

B~u = (~u~uT + ~v~v T + w ~w ~ T )~u = ~u~uT ~u + ~v~v T ~u + w

~w~ T ~u
= ~u(~u ~u) + ~v (~v ~u) + w( ~ w~ ~u) = ~u(1) + ~v (0) + w(0)~ = ~u;
T T T T T T
B~v = (~u~u + ~v~v + w ~w ~ )~v = ~u~u ~v + ~v~v ~v + w ~w~ ~v
= ~u(~u ~v ) + ~v (~v ~v ) + w(
~ w~ ~v ) = ~u(0) + ~v (1) + w(0)~ = ~v ;
Bw
~ = (~u~uT + ~v~v T + w ~w ~ T )w~ = ~u~uT w ~ + ~v~v T w~ +w~w~Tw ~
= ~u(~u w)
~ + ~v (~v w) ~ + w( ~ w~ w)
~ = ~u(0) + ~v (0) + w(1) ~ = w. ~
 
Thus the matrix A = ~u ~v w ~ orthogonally diagonalizes B and

I = diag(1, 1, 1) = AT BA B = AIAT = I.
 University of Toronto
Solutions to MAT188H1F TERM TEST
of Tuesday, October 30, 2012
Duration: 100 minutes

Only aids permitted: Casio 260, Sharp 520, or Texas Instrument 30 calculator.
Instructions: Make sure this test contains 6 sheets with questions on both sides. Do
not tear any pages from this test. Present your solutions to all 11 questions in the space
provided. The value for each question is indicated in parantheses beside the question
number. Total Marks: 60

General Comments about the Test:

1. Since more than a third of the class got A, and almost two-thirds received at least
B, the results on this test are generally quite good. But if your test mark is much
less than 3 times your quiz mark, your performance is slipping, and you may be in
trouble, since things always seem much harder in Chapter 4.

2. It was quite amazing how many students could not solve Question 2 correctly.

3. Questions 1, 2, 5, 6, 7, 8 and 9 can all be double-checked; these questions should all

have been aced.

4. det(A + B) = det(A) + det(B) is false, but this was used by some students in some
of their calculations, for example in Questions 5 and 10(c).

5. There are still students putting = between reduced matrices, or using implication
() incorrectly. This is just throwing away marks.

Breakdown of Results: 987 students wrote this test. The marks ranged from 16.7% to
100%, and the average was 72.4%. Some statistics on grade distributions are in the table
on the left, and a histogram of the marks (by decade) is on the right.

Grade % Decade %
90-100% 11.7%
A 38.4% 80-89% 26.7%
B 24.6% 70-79% 24.6%
C 19.1% 60-69% 19.1%
D 9.8% 50-59% 9.8%
F 8.0% 40-49% 5.0%
30-39% 2.1%
20-29% 0.7%
10-19% 0.2%
0-9% 0.0%

2 2 1
1. [4 marks] Find the inverse of the matrix A = 2 1 0 .
1 0 0

Solution 1: use the adjoint formula. Since det(A) = (1)3 = 1,

1
A1 = adj(A) = adj(A).
det(A)

The 9 cofactors of A are:

     
1 0 2 0 2 1
C11 = det C12 = det C13 = det
0 0 1 0 1 0
=0 =0 = 1
     
2 1 2 1 2 2
C21 = det C22 = det C23 = det
0 0 1 0 1 0
=0 = 1 =2
     
2 1 2 1 2 2
C31 = det C32 = det C33 = det
1 0 2 0 2 1
= 1 = 2 =2

So
0 0 1
A1 = adj(A) = [Cij]T = 0 1 2 .
1 2 2

Solution 2: use the Gaussian algorithm.

2 2 1 1 0 0 1 0 0 0 0 1
(A|I) = 2 1 0 0 1 0 2 1 0 0 1 0
1 0 0 0 0 1 2 2 1 1 0 0

1 0 0 0 0 1 1 0 0 0 0 1
0 1 0 0 1 2 0 1
0 0 1 2
0 2 1 1 0 2 0 0 1 1 2 2

1 0 0 0 0 1
0 1 0 0 1 2 = (I|A1 )
0 0 1 1 2 2
2. [6 marks] Solve the following system of linear equations

x1 + x2 + 2x3 + x4 2x5 = 5
2x1 x2 + x3 x4 + x5 = 2
x1 + 4x2 + 5x3 + 4x4 7x5 = 13

by first finding the reduced row-echelon form of its augmented matrix.

Solution: reduce the augmented matrix:

1 1 2 1 2 5 1 1 2 1 2 5
2 1 1 1 1 2 0
3 3 3 5 8
1 4 5 4 7 13 0 3 3 3 5 8

1 1 2 1 2 5 1 0 1 0 1/3 7/3
0 1 1 1 5/3 8/3 0 1 1 1 5/3 8/3
0 0 0 0 0 0 0 0 0 0 0 0
Let x3 = s, x4 = t, x5 = u be parameters. Then

x1 7/3 s + u/3 7/3 1 0 1/3
x2 8/3 s t + 5u/3 8/3 1 1 5/3

x3 = s = 0 + s 1 + t 0 + u 0
.

x4 t 0 0 1 0
x5 u 0 0 0 1
| {z }
Not required; just neater.
3. [5 marks] Let
1 2 1 3
2 3 2 2
A=
1
.
2 0 3
2 1 2 1
Calculate det(A).

Solution: use properties of determinants to simplify your work.

1 2 1 3 1 4 1 6
2 4 1 6
3 2 2 2 7 2 8

det
1
= det
1 0 0 = (1)(1)4 det 7 2 8
2 0 3 0
3 2 7
2 1 2 1 2 3 2 7

4 1 6    
3 1 4 1 4
= det 1 0 4 = (1) det
= 5 det = (5)(3) = 15
5 5 1 1
5 0 5
4. [6 marks] Find all the values of c for which the system of equations

x1 + c x2 + c x3 = 2
x1 x2 + x3 = 4
c x1 + c x2 + x3 = 2

has no solution.

Solution: let A be the coefficient matrix. If det(A) 6= 0, then A is invertible and

the system will have a unique solution.

1 c c
det 1 1 1 = 1 + 3c2 2c = (3c + 1)(c 1).
c c 1

If c = 1, then the system is

x1 + x2 + x3 = 2
x1 x2 + x3 = 4
x1 + x2 + x3 = 2

which has infinitely many solutions, since the first and third equations are identical.
If c = 1/3, then the system is equivalent to the system

3x1 x2 x3 = 6
x1 x2 + x3 = 4
x1 + x2 3x3 = 6

which is inconsistent, since the second plus the third equation gives 2x1 2x3 = 2,
while the first plus the third equation gives 4x1 4x3 = 0. Answer: c = 1/3
5. [5 marks] Find the eigenvalues of the matrix

3 4 2
A = 1 2 2 .
1 5 5

Solution: need the characteristic polynomial of A.

Accept either det(I A) or det(A I). Working with the former:

3 4 2 2 2 0
det 1 + 2 2 = det 1 + 2 2
1 5 5 0 3 3

1 1 0
= ( 2)( 3) det 1 + 2 2
0 1 1

1 0 0
= ( 2)( 3) det 1 + 1 2
0 1 1
= ( 2)( 3)( + 1 2)
= ( 2)( 3)( 1);

so the eigenvalues of A are 1 = 1, 2 = 2, 3 = 3.

Alternate Approach: simply expanding det(I A) gives

det(I A) = 3 62 + 11 6

which has to be factored! If the expansion is incorrect then the factoring my be

very difficult!
6. [7 marks] Given that the eigenvalues of

0 4 6
A= 9 0 9
6 4 0

are 1 = 6 and 2 = 12, find an invertible matrix P and a diagonal matrix D such
that D = P 1 AP.

Solution: find the eigenvectors.

6 4 6 0 3 2 3 0 3 2 3 0
(6I A|O) = 9 6 9 0 3 2 3 0 0 0 0 0 ;
6 4 6 0 3 2 3 0 0 0 0 0

so there are two (basic) eigenvectors corresponding to 1 = 6. For example, any

two of
1 1 0 2
0 , 3 , 3 , 3 ,
1 1 2 0
or their multiples, will do.

12 4 6 0 6 2 3 0 6 2 3 0
(12I A|O) = 9 12 9 0 3 4 3 0 0 6 9 0
6 4 12 0 3 2 6 0 0 6 9 0

6 2 3 0 6 0 6 0 1 0 1 0
0 2 3 0 0 2 3 0 0 2 3 0 ;
0 0 0 0 0 0 0 0 0 0 0 0
so an eigenvector corresponding to 2 = 12 is

1 2
3/2 or 3 .
1 2

Then (for example) take

1 1 2 6 0 0
P = 0 3 3 , D = 0 6 0 .
1 1 2 0 0 12
| {z }
Columns of P must correspond to correct diagonal entries of D.
7. [6 marks] Show that the triangle P QR with vertices P (1, 1, 1), Q(2, 1, 0), R(4, 2, 2)
is a right angle triangle.

Solution 1:

21 1 41 3 42 2

PQ = 1 1 = 0 ; P R = 2 1 = 1 ; QR = 2 1 = 1

01 1 21 1 20 2

Observe that P Q QR = 2 + 0 2 = 0, so the sides P Q and QR are orthogonal.
That is, the interior angle at Q is /2.

Solution 2: Observe that

kP Qk2 + kQRk2 = 1 + 1 + 4 + 1 + 4 = 11 = kP Rk2 ,

which means the lengths of the three sides in P QR satisfy the Pythagorean The-
orem, whence the triangle is a right triangle.
8. [5 marks] If
1 3
~v = 2 and d~ = 1 ,
1 1
express ~v in the form ~v = ~v1 + ~v2 where ~v1 is parallel to d~ and ~v2 is orthogonal to d.
~

Solution: use the projection formula.

~ 12
~v d ~ 3 + 2 1 4 1
~v1 = projd~~v = d= = d~ = 4 .
d~ d~ 9+1+1 11 11
4

Then
1 12 1
1 1
~v2 = ~v ~v1 = 2 4 = 18 .
11 11
1 4 15
    
1 1
9. [6 marks] Given that X1 = and X2 = are (basic) eigenvectors of the
1 2
matrix  
1 2
A= ,
4 3
find all the solutions to the following system of differential equations:

f10 (x) = f1 (x) + 2f2 (x)

f20 (x) = 4f1 (x) + 3f2 (x)

where f1 and f2 are functions of x.

Solution: you can find the eigenvalues by using the definition of eigenvector:
   
1 1
AX1 = = = X1 1 = 1;
1 1
   
5 1
AX2 = =5 = 5X2 2 = 5.
10 2
Or you can find them directly:
 
1 2
det = 2 4 5 = ( 5)( + 1).
4 3

Either way, the general solution is F = c1 X1 e1 x + c2 X2 e2 x . Thus

     
f1 (x) 1 x 1
= c1 e + c2 e5x ;
f2 (x) 1 2

that is,
f1 (x) = c1 ex + c2 e5x
and
f2 (x) = c1 ex + 2 c2 e5x .
10. [6 marks] Indicate if the following statements are True or False, and give a brief
explanation why.

(a) [2 marks] If A is a square matrix such that adj(A) is the zero matrix, then
det(A) = 0. True False

Solution: True.

A adj(A) = det(A)I O = det(A)I

det(A) = 0

Alternate Solution: Every cofactor of A is zero, so the cofactor expansion

along any row or column of A must be zero; that is det(A) = 0.

(b) [2 marks] If A is a 3 3 matrix such that AT = A, then det(A) = 0.

True False

Solution: True.

AT = A det(AT ) = det(A)
det(A) = (1)3 det(A)
det(A) = det(A)
det(A) = 0

(c) [2 marks] If A is a 3 3 matrix such that A3 + 4A = 7I, then det(A) = 0.

True False

Solution: False.

A3 + 4A = 7I A(A2 + 4I) = 7I
det(A(A2 + 4I)) = det(7I)
det(A) det(A2 + 4I) = 73 6= 0
det(A) 6= 0
11. [4 marks] Suppose A is an invertible matrix. Indicate if the following statements are
True or False, and give a brief explanation why.

(a) [2 marks] A and A1 have the same eigenvalues. True False

Solution: False. Consider  

2 0
A= .
0 2
It has eigenvalue = 2, repeated. But its inverse is
 
1/2 0
A= ,
0 1/2

which has eigenvalue = 1/2, repeated, and 6= .

(b) [2 marks] A and A1 have the same eigenvectors. True False

Solution: True.
Recall that A is invertible if and only if = 0 is not an eigenvalue of A. Then
1
AX = X X = A1 (X) X = A1 X X = A1 X.

Thus X is an eigenvector of A if and only if X is an eigenvector of A1 .
 University of Toronto
FACULTY OF APPLIED SCIENCE AND ENGINEERING
Solutions to FINAL EXAMINATION, DECEMBER, 2012
First Year - CHE, CIV, CPE, ELE, ENG, IND, LME, MEC, MMS

MAT188H1F - LINEAR ALGEBRA

Exam Type: A

General Comments:

1. Most of this exam consisted of completely routine questions. In particular, questions

4, 5, 7 and 8 were almost identical to questions on previous exams.

2. The only question that could be considered non-routine is question 9. However, there
are many correct ways to do both parts (a) and (b), and many people got them. But
most students had no idea how to tackle part (b). Some students wrote things like
B = 3I = I or

1 0 0 1 0 0
~u~uT + ~v~v T + (~u ~v )(~u ~v )T = 0 + 1 + 0 = 0 1 0 !
0 0 1 0 0 1

Others wrote things like ~u ~uT = k~uk2 . In fact, ~u ~uT is a 3 3 matrix.

3. The range on every question was zero to perfect. Students did surprisingly well on the
True and False, but surprisingly poorly on the first question.

4. There were three perfect papers.

Breakdown of Results: 971 students wrote this exam. The marks ranged from 9% to
100%, and the average was 70.3%. Some statistics on grade distributions are in the table on
the left, and a histogram of the marks (by decade) is on the right.

Grade % Decade %
90-100% 6.6%
A 27.8% 80-89% 21.2%
B 29.0% 70-79% 29.0%
C 23.0% 60-69% 23.0%
D 12.2% 50-59% 12.2%
F 8.0% 40-49% 4.7%
30-39% 1.8%
20-29% 1.0%
10-19% 0.4%
0-9% 0.1%
1. [avg: 5.3/10] Find the following:

(a) [5 marks] a vector equation for the line of intersection common to the two planes
with equations x + y z = 5 and 2x + y + 2z = 2.

Solution: solve the system.

     
1 1 1 5 1 1 1 5 1 0 3 3

2 1 2 2 0 1 4 8 0 1 4 8

Let z = t be a parameter, then

x 3 3t 3 3
y = 8 + 4t = 8 + t 4 ,
z t 0 1

which is the vector equation of a line.

(b) [5 marks] the shortest distance between the two parallel lines L1 and L2

x = 1 + t x = 2 s
L1 : y = 0 t ; L2 : y = 3 + s
z = 1 + t z = 1 s

where s and t are parameters.

Solution: let the shortest distance be D. Let P be the point on L1 with coordi-
nates (1, 0, 1); let Q be the point on L2 with coordinates (2, 3, 1); let

1
~n = 1
1
2 2

be a direction vector for both lines. Then D2 + proj~n P Q = P Q , so

1 2
2
2
2
1
P Q ~n 3 2 1 = 10 4 = 26 ;

D2 = P Q ~n =

~n ~n
0
3 3 3
1

and r
26
D= .
3
2. [avg: 7.9/10]

(a) [5 marks] Find the area of the triangle P QR passing through the three points

P (1, 3, 1), Q(2, 2, 2), R(0, 2, 0).

Solution: let the area be A. Then

1 1 2
8

1 1
1
A = P Q P R = 1 1 = 0
= = 2.
2 2 2 2
1 1 2

n T o
(b) [5 marks] Let U = x y z | xyz = 0 .

1. Is U non-empty?

Solution: No. The vector [ 0 0 0 ]T is in U since (0)(0)(0) = 0.

2. Is U closed under scalar multiplication?

Solution: Yes. If the vector X = [ x y z ]T is in U, so is tX for any scalar t,

since (tx)(ty)(tz) = t3 xyz = t3 (0) = 0.

3. Is U closed under vector addition?

Solution: No. The vectors X = [ 1 0 0 ]T and Y = [ 0 1 1 ]T are in U but

the vector X + Y = [ 1 1 1 ]T is not in U, since (1)(1)(1) = 1 6= 0.

4. Is U a subspace of R3 ?

Solution: No, because U is not closed under addition.

3. [2 marks for each part; avg: 7.9/10] Decide if the following statements are True or False,
and give a brief, concise justification for your choice. Circle your choice.

(a) If = 0 is an eigenvalue of the n n matrix A then A is not invertible.

True or False

True: let a corresponding eigenvector be X; so X is non-zero and AX = 0X = O.

Thus the homogeneous system has a non-zero solution, which is equivalent to A
being non-invertible. OR: det(0I A) = 0 det(A) = 0, so A is not invertible.

1 1 1 2
(b) dim im 3 6 1 1 = 3 True or False
5 8 3 3

False: dim (im(A)) = r, where r is the rank of A. The matrix

1 1 1 2
3 6 1 1
5 8 3 3

has rank 2 because it only has 2 independent rows: R3 = R2 + 2R1 .

(c) If U is a subspace of R7 and dim U = 4 then dim U = 3. True or False

True: 7 = dim U + dim U dim U = 7 4 = 3.

1 1 1 0
(d) 1 is in the span of
3 , 0 , 3 .
True or False
2 2 1 1

False: in the spanning set, the first vector is the sum of the next two. So the
question reduces to, Are there scalars a and b such that

1 1 0
1 = a 0 + b 3 ?
2 1 1

This system is obviously inconsistent, since 1 + 1/3 6= 2.

(e) If {X1 , X2 , X3 } is linearly independent, then so is {X2 , X3 }. True or False

True: sX2 + tX3 = O 0X1 + sX2 + tX3 = O s = t = 0 since {X1 , X2 , X3 }

is linearly independent.
4. [10 marks; avg: 8.5/10] Given that the reduced row-echelon form of

1 0 2 1 1 1 0 2 1 0
1 2 0 1 0 0 1 1 0 0
A= is R = ,
1 2 4 1 2 0 0 0 0 1
2 1 3 2 2 0 0 0 0 0

state the rank of A, and find a basis for each of the following: the row space of A, the
column space of A, and the null space of A.

Solution: the rank of A is r = 3, the number of leading 1s in R.

U dim U description of basis vectors in basis

1 0 0

0 1 0




rowA r=3 three non-zero rows of R 2 , 1 , 0



1 0 0


0 0 1

three independent rows of A {R1 , R2 , R4 }

{R1 , R3 , R4 }
{R2 , R3 , R4 }

NB: R3 = R2 + 2R1

col A r=3 three independent columns of A {C1 , C2 , C5 }

NB: C3 = 2C1 + C2 ; C4 = C1

(not three independent columns of R) (colR 6= colA)

2 1
1 0




null A 5r =2 two basic solutions to RX = 0 1 , 0



0 1


0 0

5. [avg: 8.5/12] Let T : R2 R2 be the linear transformation defined by
   
x 2x + 5y
T = .
y 2x + y

(a) [6 marks] Draw the image under T of the unit square, and calculate its area.

Solution: the image of the unit square is the parallelogram determined by

       
1 2 0 5
T (~i) = T = and T (~j) = T = .
0 2 1 1

In the diagram to the right, the

unit square is in blue, and the
image of the unit square is in
green. Its area is
  
det 2 5  = |2 10| = 8.
 
 2 1 

 
x
(b) [6 marks] Find the formula for T R if R is a rotation of /6 clockwise
y
around the origin.

Solution: the matrix of R is

   
cos(/6) sin(/6) 1 3 1
= =B
sin(/6) cos(/6) 2 1 3

and the matrix of T is  

2 5
= A;
2 1
so
      
x x 1 23 5 2 + 5 3 x
T R = AB =
y y 2 2 31 2+ 3 y
 
1 (2 3 5) x + (2 + 5 3) y
= .
2 (2 3 1) x + (2 + 3) y
  
1 8 6
6. [avg: 8.8/12] Let Q = .
10 6 8

(a) [4 marks] Find the eigenvalues of Q.

Solution:
8 6

    
10 2 64 36
det(I Q) = det 6
10
8 = = 2 1.
10 10 100 100

So the eigenvalues of Q are 1 = 1 and 2 = 1.

(6) [8 marks] Find a basis for each eigenspace1 of Q and plot the eigenspaces of Q in
the plane, indicating which eigenspace corresponds to which eigenvalue of Q.

Solution:
1 + 45 35
     
9 3 3 1
E1 (Q) = null(IQ) = null = null = null ;
53 1 45 3 1 0 0

so could take  
1
X1 = .
3
4
53
     
1 + 5
1 3 1 3
E1 (Q) = null(IQ) = null = null = null ;
35 1 4
5
3 9 0 0
so could take  
3
X2 = .
1
Plot of Eignespaces:
E (Q)
y  1



X1
PPX2 
iPP
P 
PP
P
 PPP x
 PP


 E1 (Q)





Note: Q is the matrix of a reflection in the line y = 3x, so E1 (Q) is the axis of
reflection, and E1 (Q) must be the line orthogonal to the axis of reflection.

1
Recall: if A is an nn matrix, the eigenspace of A corresponding to is E (A) = {X Rn | AX = X}.
7. [avg: 10.4/12]

Let U = span X1 = [ 0 1 1 0 ]T , X2 = [ 1 0 0 1 ]T , X3 = [ 1 1 0 0 ]T ;
 T
let X = 1 1 0 1 . Find:

(a) [6 marks] an orthogonal basis of U.

Solution: since X1 X2 = 0 already, you only need to use the Gram-Schmidt

algorithm to find F3 . Take F1 = X1 , F2 = X2 , and

1
X3 X1 X3 X2 1 1 1 1 .

F3 = X3 X 1 X2 = X 3 + X1 X2 =
kX1 k2 kX2 k2 2 2 2 1
1

Optional: clear fractions and take

1
1
F3 =
1 .

Either way {F1 , F2 , F3 } is an orthogonal basis of U.

(b) [6 marks] projU (X).

Solution: using {F1 , F2 , F3 } with fractions cleared.

X F1 X F2 X F3
projU X = F 1 + F2 + F3
kF1 k2 kF2 k2 kF3 k2
1 1
= F1 + (0)F2 + F3
2 4
1
1 1 .

=
4 3

1

Cross-check/Alternate Solution: U = span{Y } with Y = [ 1 1 1 1 ]T .

Then

1
X Y 3 1 1 .

projU X = X projU (X) = X Y = X Y =
kY k2 4 4 3
1
8. [12 marks; avg: 9.8/12] Find an orthogonal matrix P and a diagonal matrix D such that
D = P T AP, if
1 0 1
A = 0 2 0 .
1 0 1

Step 1: Find the eigenvalues of A.

1 0 1  
1 1
det(I A) = det 0 2 0 = ( 2) det
1 1
1 0 1
= ( 2) ( 1)2 1 = ( 2)(2 2)

= ( 2)2

So the eigenvalues of A are 1 = 0 and 2 = 2.

Step 2: Find mutually orthogonal eigenvectors of A.

1 0 1 1 0 1 1
E0 (A) = null 0 2 0 = null 0 1 0 = span 0 .
1 0 1 0 0 0 1

1 0 1 1 0 1 1 0
E2 (A) = null 0 0 0 = null 0 0 0 = span 0 , 1 .
1 0 1 0 0 0 1 0

OR: use either E0 (A) = (E2 (A)) or E2 (A) = (E0 (A)) to simplify calculations. That
is, E2 (A) is the plane with equation x = z and E0 (A) is the line normal to the plane.

Step 3: Divide each eigenvector by its length to get an orthonormal basis of eigenvec-
tors, which are put into the columns of P. So

1/ 2 1/ 2 0 0 0 0
P = 0 0 1 and D = 0 2 0
1/ 2 1/ 2 0 0 0 2
9. [avg: 3.3/12] Suppose ~u and ~v are two orthogonal unit vectors in R3 .
 
Let A = ~u ~v ~u ~v and let B = ~u ~uT + ~v ~v T + (~u ~v ) (~u ~v )T .

(a) [6 marks] Explain why A must be an orthogonal matrix.

Solution 1: show that the columns of A form an orthonormal basis of R3 .

It is given that k~uk = k~v k = 1 and ~u ~v = 0; and ~u v is always orthogonal to
both ~u and ~v . So need only show that ~u ~v is also a unit vector:

k~u ~v k = k~ukk~v k sin(/2) = (1)(1)(1) = 1.

 T  
Solution 2: show that AT A = I: AT A = ~u ~v ~u ~v ~u ~v ~u ~v =

~uT   ~
u T
~
u ~
u T
~
v ~
u T
(~u ~
v )
~v T ~u ~v ~u ~v = ~v T ~u ~v T ~v ~v T (~u ~v )
(~u ~v )T (~u ~v )T ~u (~u ~v )T ~v (~u ~v )T (~u ~v )

~u ~u ~u ~v ~u (~u ~v ) k~uk2 0 0
= ~v ~u ~v ~v ~v (~u ~v ) = 0 k~v k2 0 = I.
2
(~u ~v ) ~u (~u ~v ) ~v (~u ~v ) (~u ~v ) 0 0 k~u ~v k

(b) [6 marks] Explain why B must be I, the 3 3 identity matrix.

~ = ~u ~v . By part (a), AAT = I, so

Solution 1: To save space, let w
T
 ~u
I = AAT = ~u ~v w ~ ~v T = ~u ~uT + ~v ~v T + w ~ T = B.

~w
~T
w

Solution 2: {~u, ~v , w}~ is an orthonormal basis. Apply the Expansion Theorem,

3
for ~x R : ~x = (~x ~u)~u +(~x ~v )~v +(~x w)
~ w~ = ~u~uT ~x +~v~v T ~x + w
~w~ T ~x = B~x B = I.
Solution 3: show ~u, ~v , w ~ are all eigenvectors of B with eigenvalue = 1.

B~u = (~u~uT + ~v~v T + w ~w ~ T )~u = ~u~uT ~u + ~v~v T ~u + w

~w~ T ~u
= ~u(~u ~u) + ~v (~v ~u) + w( ~ w~ ~u) = ~u(1) + ~v (0) + w(0)~ = ~u;
T T T T T T
B~v = (~u~u + ~v~v + w ~w ~ )~v = ~u~u ~v + ~v~v ~v + w ~w~ ~v
= ~u(~u ~v ) + ~v (~v ~v ) + w(
~ w~ ~v ) = ~u(0) + ~v (1) + w(0)~ = ~v ;
Bw
~ = (~u~uT + ~v~v T + w ~w ~ T )w~ = ~u~uT w ~ + ~v~v T w~ +w~w~Tw ~
= ~u(~u w)
~ + ~v (~v w) ~ + w( ~ w~ w)
~ = ~u(0) + ~v (0) + w(1) ~ = w. ~
 
Thus the matrix A = ~u ~v w ~ orthogonally diagonalizes B and

I = diag(1, 1, 1) = AT BA B = AIAT = I.
Here are the questions from the December 2005 Final Exam in MAT188H1F, with
answers. The first six questions are multiple choice.

1. If U is a subspace of R5 and dim U = 2, then dim U is

(a) 2
(b) 3
(c) 4
(d) 5

8

2 4 0 3




1
2 4 0 2


2. dim span , , , , =


1 2 2 6 4


0 0 1 1 2

(a) 2
(b) 3
(c) 4
(d) 5
" #
x
3. The best approximation Z = to a solution of the inconsistent system of
y
equations


x = 14
y = 14


2x + 3y = 0
is
" #
14
(a) Z =
14
" #
11
(b) Z =
16
" #
16
(c) Z =
11
" #
16
(d) Z =
11
4. What is the matrix of the transformation which is composed of a reflection in

the x-axis followed by a rotation through ?
2
" #
0 1
(a)
1 0
" #
1 0
(b)
0 1
" #
0 1
(c)
1 0
" #
0 1
(d)
1 0
" #
1 1 m
5. The eigenspaces of the projection matrix Pm = are
1 + m2 m m2
(" #) (" #)
1 m
(a) E1 (Pm ) = span and E1 (Pm ) = span
m 1
(" #) (" #)
1 m
(b) E0 (Pm ) = span and E1 (Pm ) = span
m 1
(" #) (" #)
m 1
(c) E0 (Pm ) = span and E1 (Pm ) = span
1 m
(" #) (" #)
m 1
(d) E1 (Pm ) = span and E0 (Pm ) = span
1 m
6. The equation of the plane passing through the point (x, y, z) = (1, 0, 1) and
perpendicular to the line [x y z]T = [2 3 4]T + t[2 1 3]T is
(a) 2x + 3y + 4z = 2
(b) 2x + y + 3z = 1
(c) 2x + 3y + 4z = 2
(d) 2x + y + 3z = 1
7. Suppose A is an n n matrix such that A2 = O. Explain clearly and concisely
why the following six statements about A are True.
(a) det(A) = 0
(b) (AT )2 = O
(c) (I A)1 = I + A
(d) If the system AX = B is consistent, then B is in nullA, where X and B
are n 1 matrices.
(e) The only eigenvalue of A is = 0.
(f) If A is diagonalizable then A = O
7. (continued) Suppose A is an n n matrix such that A2 = O. Explain clearly
and concisely why the following six statements about A are False.

(g) A = O
(h) adj(A) = O
(i) A is invertible
(j) colA = nullA
(k) imA = Rn
(l) dim(E0 (A)) = n

8. Given that
1 0 1 1 1 1 0 0 0 10

2 0 3 1 1 0 0 1 0 7
A= has reduced row-echelon form R =

;

1 0 0 4 2 0 0 0 1 2


0 0 1 4 1 0 0 0 0 0

state the rank of A, and then find a basis for each of the following: the row
space of A, the column space of A, and the null space of A.
n o
9. Let A be an n n matrix; let U = X in Rn | AX = AT X .

(a)[5 marks] Show that U is a subspace of Rn .

12 3 4 1 6
3 4 5 3 7



(b)[8 marks] Let A =
3 5 6 4 8 . Find a basis for U.


1 3 5 2 9

6 7 8 9 1
n o
10. Let U = span [ 0 1 0 0 ]T , [ 1 1 1 1 ]T , [ 1 2 2 0 ]T ;
h iT
let X = 1 1 0 1 . Find projU (X) and projU (X).

11. Find an orthogonal matrix P and a diagonal matrix D such that D = P T AP,
if
0 0 1
A= 0 1 0 .

1 0 0
ANSWERS: 1.(b) 2.(b) 3.(d) 4.(a) 5.(c) 6.(d)
7.(a) 0 = det(A2 ) = (det A)2 det A = 0
7.(b) (AT )2 = (A2 )T = OT = O.
7.(c) (I A)(I + A) = I A + A A2 = I O = I (I A)1 = I + A
7.(d) AX = B O = A2 X = AB B is in null(A)
7.(e) AX = X O = A2 X = A(X) = AX = 2 X 2 = 0 = 0
7.(f) D = P 1 AP D2 = P 1 A2 P = O 2i = 0 i = 0 D = O
A = P DP 1 = O OR use part (e) and start with D = O
" #
0 1
7.(g),(h) and (l): can use A = as a counterexample
0 0
7.(i), (j) and (k): can use A = O as a counterexample
8. rank of A is 3;
nh i h i h io
basis for row space of A is 1 0 0 0 10 , 0 0 1 0 7 , 0 0 0 1 2 ,
or any three independent rows of A

1

1 1




2 3
1

basis for column space of A is , , , or any three independent




1 0 4


0 1 4

columns of A




0 10


1 0






9.(a) U = null(A AT )

basis for the null space of A is
0 ,
7

2






0





0 1

0 0 1
4 2


1 0 0



1 6 1 0



9.(b) 0 , 0 , 0 10. projU (X) =

; projU (X) =

1 1


6 6


0
0
1

7 1



0 1 0

1 0 1

1 0 0 2 2
11. D = 0 1 0 ; P = 0 1 0


0 0 1 12 0 1
2
Here are the questions from the December 2004 Final Exam in MAT188H1F, with
answers.

1 2 3

1 2 3
1. [10 marks] Let A = .

2 3 1
1 1 2

(a) [3 marks] Find the reduced row-echelon form of AT .

(b) [2 marks] Find the basic solutions of the homogeneous system AT X = O,
where X is a 4 1 matrix.
(c) [2 marks] Let B be the matrix whose columns are the basic solutions of
AT X = O. Compute B T A.
(d) [3 marks] Circle the following formulas, if any, that are illustrated by the
above computations.
 
(i) null(AT ) = col(A) (ii) (row(A)) = null(A) (iii) null(AT ) = (col(A))

2. [10 marks] The parts of this question are unrelated.

1 a b
(a) [4 marks] Show that the matrix A = a 1 2
is invertible for any

b 0 1
real values of a and b.
(b) [6 marks] B is a 2 2 matrix such that
( !) ( !)
2 1
E2 (B) = span and E5 (B) = span .
1 1

Find B.

3. [10 marks; one mark for each part] Let A be an nn matrix. For each statement
below, decide if it is equivalent to the statement

A is invertible.

If it is, circle Yes at the right; if it isnt, circle No.

(a) = 0 is an eigenvalue of A. Yes No

(b) A7 is invertible. Yes No
(c) The rank of A is n. Yes No
(d) AAT is invertible. Yes No
(e) A is a product of elementary matrices. Yes No
(f) col(A) = Rn Yes No
 (g) null(A) = Rn Yes No
(h) A is similar to the identity matrix In Yes No
(i) O is not in the row space of A Yes No
(j) For any vector B in Rn , Yes No
the equation AX = B has a solution.

4. [10 marks] Indicate whether each of the following statements is True or False.
(Circle your choice.) It is not necessary to justify your choice, and there is no
penalty for an incorrect answer.

(a) True or False: 1 is the minimum distance between the two planes with
equations x + y + z = 1 and x + y + z = 0.
(b) True or False: The line of intersection of the two planes with equations
x+y +z = 3 and 2x3y +4z = 6 is parallel to the vector [ 7 2 5 ]T .
(c) True or False: If is an eigenvalue of the n n matrix A, then 2 + 4 is
an eigenvalue of the matrix A2 + 4In .

1 1 1
(d) True or False: An LU -factorization of the matrix A = 1 2 2 is

1 2 3
T
1 0 0 1 0 0
A= 1 1 0 1 1 0

1 1 1 1 1 1
(e) True or False: If E and F are both n n elementary matrices, then
(EF )1 = E 1 F 1 .

1
1 5
(f) True or False: 0 , 1 , 2 is an independent set in R3 .


2 3
12


1 3 9

(g) True or False: 1 , 1 , 1 is a spanning set for R3 .



2 3
12
(h) True or False: Every symmetric matrix is diagonalizable.
(i) True or False: If A is an m n matrix with rank equal to n and A = QR
is a QR-factorization of A, then AT A = RT R.
(j) True or False: If A is an n n matrix such that A3 A2 + A 2In = O,
then (A2 A + In )1 = 2A.

5. [10 marks] Given that the reduced row-echelon form of the matrix

1 0 1 0 1 1 0 1 0 0

1 2 3 4 1 0 1 2 2 0
A= is R =

2 2 6 4 2 0 0 0 0 1


3 4 11 8 4 0 0 0 0 0
 find a basis for each of the following: the row space of A, the column space of
A, and the null space of A.

6. [10 marks] Solve the system of differential equations

f10 (x) = f1 (x) + 5f2 (x)

f20 (x) = f1 (x) + 3f2 (x)

for f1 and f2 as functions of x, given the initial conditions f1 (0) = 1 and

f2 (0) = 1.
n o
7. [10 marks] Let U = span [ 1 1 2 1 ]T , [ 0 0 1 1 ]T , [ 0 1 2 3 ]T ;
h iT
let X = 1 1 0 1 . Find projU (X).

8. [10 marks] Find the least squares approximating line for the data points

(1, 1), (2, 2), (3, 2), (4, 3).

9. [10 marks] Find an orthogonal matrix P and a diagonal matrix D such that
D = P T AP, if
2 1 1
A = 1 2 1 .

1 1 2
You may assume that the eigenvalues of A are = 0 and = 3.

10. [10 marks; 5 marks for each part.] The parts of this question are unrelated.

(a) Suppose T : R2 R2 is a linear transformation such that

! ! ! !
1 1 2 0
T = and T =
1 0 3 1

Find ! !
x 1 x
T and T .
y y
(b) Let X and Y be any two vectors in Rn . Prove that
X and Y are orthogonal if and only if kX + Y k2 = kX Y k2 .
ANSWERS TO DEC 2004 EXAM:
1

1


1 1 0 1


!
1 0

0 0 0
0 0 1 1 1(b)
1(a) , 1(c) 1(d) (i) , (iii)

0 1 0 0 0


0 0 0 0


0 1

!
1 6
2(a) det A = 1 + (a + b)2 1, so det A 6= 0 2(b) B =
3 8
3(a) No (b) Yes (c) Yes (d) Yes (e) Yes (f) Yes (g) No (h) No (i) No (j) Yes
4(a) False (b) True (c) True (d) True (e) False (f) False (g) False (h) True (i)
True (j) False
nh i h i h io
5. basis for row(A) is 1 0 1 0 0 , 0 1 2 2 0 , 0 0 0 0 1

1 0 1




1 2
1
basis for column space of A is ,

,

2 2 2




3 4 4





1 0


2 2





basis for null space of A is
1 ,
0





0
1


0 0

2 5 2 1 h iT
6. f1 (x) = e4x + e2x ; f2 (x) = e4x e2x 7. 1 13 32 31
3 3 3 3

0 0 0 1/3 1/6 1/ 2
8. y = 0.5 + 0.6x 9. D = 0 3 0 ; P = 1/ 3 2/ 6
0


0 0 3 1/ 3 1/ 6 1/ 2
! ! ! !
x 3x 2y 1 x x + 2y
10(a) T = ;T =
y x + y y x + 3y
10(b)

kX + Y k2 = kX Y k2
(X + Y ) (X + Y ) = (X Y ) (X Y )
X X +Y X +X Y +Y Y = X X Y X X Y +Y Y
2X Y = 2X Y
X Y = 0
Here are the questions from the December 2003 Final Exam in MAT188H1F, with
answers.

1. [10 marks] Given that the reduced row echelon form of the matrix

1 1 2 1 10 1 1 0 0 1

1 1 3 2 7 0 0 1 0 2
A= is R =

2 2 1 2 14 0 0 0 1 7


3 3 4 1 2 0 0 0 0 0

find a basis for each of the following: the row space of A, the column space of
A, and the null space of A.

2. [12 marks] Solve the system of linear ordinary differential equations

y10 = y1 + 3y2
y20 = 4y1 + 2y2

for y1 and y2 as functions of t, given the initial conditions y1 (0) = 1 and

y2 (0) = 6.

3. [10 marks] Find an orthogonal basis for the subspace of R4 spanned by

{w1 = (1, 1, 3, 2), w2 = (1, 2, 0, 1), w3 = (0, 2, 1, 2)} .

4. [12 marks] Find the least squares line of best fit y = a + bx to the four data
points

(x1 , y1 ) = (2, 1), (x2 , y2 ) = (3, 2), (x3 , y3 ) = (5, 3), (x4 , y4 ) = (6, 4).

5. [12 marks] Find an orthogonal matrix P and a diagonal matrix D such that
D = P T AP, if
0 1 1
A = 1 0 1 .

1 1 0

6. [10 marks] Indicate whether each of the following statements is True or False.
(Circle your choice.) It is not necessary to justify your choice, and there is no
penalty for an incorrect answer.

(a) True or False: (6, 4, 5) is the point on the plane with equation x+y2z = 0
that is closest to the point (7, 5, 3).
(b) True or False: If A is a non-zero square matrix such that AT = kA, then
k = 1.
(c) True or False: If A is a square matrix such that A2 = A, then det A = 0.
(d) True or False: proj(1,0,2) (2, 1, 4) = (2, 0, 4)
 (e) True or False: {(1, 2, 0), (1, 1, 3), (2, 3, 3)} is a spanning set for R3 .
(f) True or False: If x is a least squares solution to the system Ax = b, then
Ax b is in the column space of A.
(g) True or False: {(1, 0, 1), (2, 1, 1), (0, 1, 1)} is a linearly independent set
of vectors in R3 .
(h) True or False: If A and B are n n diagonal matrices, then AB = BA.
(i) True or False: If A and B are similar matrices, then they have the same
eigenvalues and the same eigenspaces.
(j) True or False: If the nullity of the matrix A is the same as the nullity of
the matrix AT , then A must be a square matrix.

7. [34 marks] Consider the four matrices

0 0 1 0 1 0 1/2 0 1/2 1 0 0
A = 0 1 0 ,B = 0 0 1 ,C = 0 1 0 ,D = 0 0 1


1 0 0 1 0 0 1/2 0 1/2 0 1 0

(a) [2 marks] Which of the four matrices are symmetric?

(b) [2 marks] Which of the four matrices are NOT orthogonal?
(c) [2 marks] Which of the four matrices has trace 2?
(d) [4 marks] What is the determinant of each of the four matrices?
(e) [2 marks] Which of the four matrices does not have rank 3? What is its
rank?
(f) [8 marks] Find the characteristic polynomial of each of the four matrices.
Put your answers in factored form.
(g) [7 marks] Which of the four matrices is the standard matrix of an orthog-
onal projection? Find its kernel and range.
(h) [7 marks] Which of the four matrices is the standard matrix of a rotation?
Find its axis of rotation, and the angle of rotation.
ANSWERS: 1. basis for row space of A : {(1, 1, 0, 0, 1), (0, 0, 1, 0, 2), (0, 0, 0, 1, 7)}

1

1 2




1 3
2

basis for column space of A : , ,

2 1 2





3 4 1

1 1


1 0






basis for null space of A :
0 ,
2






0
7



0 1

2. y1 = 3e5t 2e2t ; y2 = 4e5t + 2e2t

6 9 3 3 1 1 1
    
3. (1, 1, 3, 2), , , , , , 0, , ; many other correct answers
5 5 5 5 3 3 3

1/ 2 1/6 1/ 3 1 0 0
4. y = 0.3 + 0.7x 5. P = 0 2/6 1/ 3 ; D = 0 1 0


1/ 2 1/ 6 1/ 3 0 0 2

6(a) T (b) T (c) F (d) T (e) F (f) F (g) F (h) T (i) F (j) T

7.(a) A, C (b) C (c) C (d) det(A) = 1; det(B) = 1; det(C) = 0; det(D) = 1

(e) rank of C is 2 (f) det(I A) = ( 1)2 ( + 1); det(I B) = ( 1)(2 + + 1)

det(I C) = ( 1)2 ; det(I D) = ( + 1)(2 + 1)

(g) null(C) = span {(1, 0, 1)} ; col(C) = span {(1, 0, 1), (0, 1, 0)}

OR could say: the range is the plane with equation x z = 0 and the kernel is the
line normal to the plane.

(h) axis of rotation is null(I B) = span {(1, 1, 1)}; angle is /3, or 120 degrees.
Here are the questions from the December 2002 Final Exam in MAT188H1F, with
answers.

1. [30 marks: 5 marks for each part] Find the following:

1 2 4
(a) the inverse of 0 1 1


1 0 2
1 1 2 3

1 0 1 0
(b) det

0 0 2 7


0 1 1 4

0 0 3
(c) the eigenvalues of the matrix 0 4 0

3 0 0
(d) the coordinate vector of p(x) = 1 + 4x + 5x2 with respect to the basis

B = {1 + x, 1 + x2 , x + x2 }

of P2 .

1 a 2+a
(e) the values of a for which the matrix a 4 4 is not invertible.

a 4 6
(f) the point on the plane with equation x + y + z = 2 closest to the point
(3, 2, 1).

2. [12 marks] Let W be the subspace of R4 consisting of all vectors of the form

(a + c, b + c, a + 2b + c, a b).

Find an orthonormal basis of W. (Use the usual dot product in R4 .)

3. [12 marks] Let W be the set of 3 3 matrices, A, satisfying the condition

AT = A.

(a) [6 marks] Show that W is a subspace of M3,3 .

(b) [6 marks] Find a basis for W and its dimension.

1 1 1
4. [12 marks] Let A = 1

1 1 . Find an orthogonal matrix P and a diagonal
1 1 1
matrix D such that D = P T AP.
5. [14 marks] Suppose A is a 3 3 invertible matrix with eigenvalues, 3, 1, and
1. Find the following:

(a) [4 marks] the eigenvalues of A1 .

(b) [5 marks] the eigenvalues of AT .
(c) [5 marks] the eigenvalues of Adj(A).

6. [10 marks; 2 marks for each part] Suppose u and v are two non-zero vectors
in R3 . What does each of the following conditions imply about the linear
independence or dependence of the set {u, v}?

(a) u = 3v
(b) au + bv = 0 a = b = 0
(c) u v = 0
(d) u v = 0
(e) {u, v, u v} spans R3

7. [10 marks: 5 marks for each part.] Let B = {w1 , w2 , . . . , wn } be an orthonormal

basis of an inner product space V. Prove the following:

(a) For any vectors u and v in V,

(u, v) = x y,

where x and y are the coordinate vectors of u and v, respectively, with

respect to the basis B.
(b) If xi is the coordinate vector of wi with respect to the basis B, then the
set {x1 , x2 , . . . , xn } is an orthonormal basis of Rn , with respect to the
usual dot product in Rn .

2 4 6 1
1
ANSWERS: 1(a) 1 2 1 1(b) 30 1(c) = 4, 3 1(d) 0

4

1 2 1 5
7 4 5
1(e) a = 2 1(f) ( . , )
3 3 3
( )
1 1 1
2. (1, 0, 1, 1), (1, 1, 1, 0), (1, 1, 0, 1) is one possible answer.
3 3 3


0 1 0 0 0 1 0 0 0
3(a) Use subspace test. 3(b) 1 0 0 , 0 0 0 , 0 0 1


1 0 0 0 1 0


0 0 0
1 1 1



2 6 3


0 0 0

1 1 1
4. P =

,D = 0 0 0

2 6 3
0 0 3
2 1

0

6 3
1
5(a) reciprocals: , 1, 1 5(b) same 5(c) 1, 3, 3
3
6(a) dependent 6(b) independent 6(c) independent 6(d) dependent
6(e) independent
n
X n
X
7(a) u = ai wi x = (a1 , a2 , . . . , an ); v = bj wj y = (b1 , b2 , . . . , bn )
i=1 j=1

Then:

Xn n
X
(u, v) = ai wi , bj wj
i=1 j=1
n
XX n
= ai bj (wi , wj ) , by distribution
i=1 j=1
Xn
= ai bi (wi , wi ) , since (wi , wj ) = 0, for i 6= j
i=1
Xn
= ai bi , since (wi , wi ) = 1
i=1
= xy

(
0 if i 6= j
7(b) Use part (a): xi xj = (wi , wj ) = . That is all.
1 if i = j
Here are the questions from the December 2000 Final Exam in MAT188H1F, with
answers. 403 students wrote this exam; marks ranged from 13% to 100%, with an
average mark of 62.9%

1. [15 marks: avg: 12.7] Find the following:

1 0 1 5

1 3 1 0
(a) (5 marks) det
2 1 0 2


0 1 1 4
(b) (5 marks) the standard equation of the plane passing through the three
points (1, 0, 2), (1, 2, 1) and (3, 0, 1).
1
1 2 4
(c) (5 marks) 0 1 1

1 0 2

2. [10 marks] avg: 7.7

1 3 4
(a) [5 marks] Find the LU decomposition of the matrix A = 1 2 1 .

1 0 5
(Do not use any row interchanges.)
(b) [5 marks] For s and t parameters, show that the two lines

x(s) = (3, 1, 2) + s(2, 1, 1) and x(t) = (1, 1, 1) + t(0, 1, 1)

do not intersect.

3. [10 marks; avg: 6.7] Let S be the subspace of R4 consisting of all vectors of the
form (a + b, a b, c, a + c), where a, b, and c are in R. Find an orthonormal
basis of S, relative to the usual dot product in R4 .

4. [20 marks: 2 marks for each part. Avg: 10.5] Indicate whether each of the
following statements is true (T) or false (F), and give a brief justification for
your choice:

(a) If A and B are any 2 2 matrices, then det(A + B) = det(A) + det(B).

(b) If 5 is an eigenvalue of the matrix A, then 125 is an eigenvalue of the
matrix A3 .

1 0 1 1 2 3
(c) B = 1 2 1 cannot be obtained from A = 4

5 6
by using
1 0 1 7 8 9
elementary row opertions.
(d) If A and B are both 2 2 matrices with a common eigenvalue 1, then
A + B also has eigenvalue 1.
(e) If A is a 7 7 matrix such that AT = A, then A is not invertible.
 1
(f) If A is a square matrix such that (A 2I)2 = O, then A1 = I A
4
(g) If A is an m n matrix, then every vector in the solution space of the
homogeneous system Ax = 0 is orthogonal to every vector in the row
space of A, with respect to the usual dot product in Rn .
(h) If A is an m n matrix, x is in Rn , and b is in Rm , then the equation
Ax = b has a solution x if and only if b is in the column space of A.
(i) If A is a square matrix which is not invertible, then 0 is an eigenvalue of
A.
(j) Any five 2 2 matrices must be linearly dependent.

5. [10 marks; avg: 5.2] For this question the inner product on P2 is defined to be

(p, q) = p(1)q(1) + p(0)q(0) + p(1)q(1).

(a) (5 marks) Verify that the set {1, x, 3x2 2} is an orthogonal basis of P2 .
(b) (5 marks) Find the coordinate vector of r(x) = x + x2 with respect to the
basis of part (a).

1 0 2
6. [15 marks; avg: 9.8] Let A = 0 2 0 . Find an orthogonal matrix P and

2 0 2
a diagonal matrix D such that D = P T AP.

7. [10 marks; avg: 4.1] Let S be the subset of M3,3 consisting of all 3 3 matrices
A such that the sum of the diagonal entries of A is zero.

(a) (5 marks) Prove that S is a subspace of M3,3 .

(b) (5 marks) Find a basis for S, and its dimension.

8. [10 marks; avg: 6.1] Let A be a 3 3 matrix such that

1 1 1 1 0 0
A 0 = 3 0 , A 1 = 2 1 , and A 1 = 4 1 .

1 1 2 2 1 1

4
Find A
3 .

5

2 4 6
1
ANSWERS: 1.(a) 9 (b) x y + 2z = 5 (c) 1 2 1
.
4
1 2 1

1 0 0 1 3 4
2.(a) L =

1 1 0 ;U =

0 1 5 .

1 3 1 0 0 16

1 1 1
3. { (1, 1, 0, 1), (1, 1, 0, 0), (1, 1, 3, 2)}
3 2 15

4. All are True, except for (a) and (d), which are both False.

5.(a) Show (1, x) = 0; (1, 3x2 2) = 0 and (x, 3x2 2) = 0 (b) (2/3, 1, 1/3)

1 2

0
3 0 0
5 5
6. 0
D= 2 0 ;P = 0 0 1
.


0 0 2

2 1
0

5 5

1 0 0
1 0 0 0 1 0 0 0 1 0 0 0
7.(b) basis = 0 1 0 , 0 0 0 , 0 , 0 0 0 , 0 0 1 ,
0 0

0 0 1


0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0
1 0 0 , 0 0 0 , 0 0 0 ; dimS =8


0 0 0 1 0 0 0 1 0

4 10
8. A 3 = 8

.
5 10
Here are the questions from the December 2001 Final Exam in MAT188H1F. 451
students wrote this exam. The marks ranged from 3% to 97%, and the average was
58.1%

1. [15 marks: 5 marks for each part] Find the following:

(a) parametric equations of the line of intersection of the two planes with
equations x + y z = 6 and 3x y + 3z = 4.
1 1 2 5

1 0 1 0
(b) det

5 1 0 2

0 1 1 4

1 2 4
(c) the adjoint of 0 1 1

1 0 2

1 2 7
2.(a) [5 marks] Find the LU decomposition of the matrix A = 1 5 1 . (Do

1 0 5
not use any row interchanges.)
2.(b) [5 marks] Let A and B
 be 33 matrices
 such that det A = 2 and det B = 4.
2 T 3 2
Find the value of det B A B A .

3. [10 marks] Let S be the subspace of R4 consisting of all vectors of the form
(a c, a b, c, 2a + b + c), where a, b, and c are in R. Find an orthogonal basis
of S, relative to the usual dot product in R4 .
4. [20 marks: 2 marks for each part] Indicate whether each of the following state-
ments is true (T) or false (F), and give a brief justification for your choice:
(a) If E and F are any 3 3 elementary matrices, then EF = F E.
(b) If 3 is an eigenvalue of the square matrix A, then 27 is an eigenvalue of
the matrix A3 .
(c) If the 6 6 matrix B is obtained from the 6 6 matrix A by replacing
the third column of A with the sum of the second and fourth columns of
A, then det B = det A.
(d) If is an eigenvalue of the n n matrix A, then 2 + 1 is an eigenvalue
of A2 + I, where I is the n n identity matrix.
(e) The set of all n n symmetric matrices is a subspace of Mn,n .
(f) Suppose {v1 , v2 , . . . , vm } is an orthogonal basis of Rm , with respect to
the usual dot product, and A is the m m matrix with v1 , v2 , . . . , vm as
its columns. Then the rows of (14A)T form an orthogonal basis of Rm .
(g) If A and B are two 2 3 matrices such that Ax = 0 and Bx = 0 have
the same solution spaces, then A = B.
 (h) (3, 2, 4) is the coordinate vector of 3 + x 4x2 , relative to the ordered
basis {x + 1, x 1, 1 + x + x2 } of P2 .
(i) The value of y in the solution of the system of equations
2x + 4y + z = 6
x y + 3z = 3
x + y 4z = 3
is y = 0.
(j) dimPn = n + 1
5. [10 marks] For which values of k does the following system of equations
2x + ky z = 2
y + kz = 2
kx + y = 2
have
(a) no solutions?
(b) a unique solution?
(c) infinitely many solutions?
2 1 0 0

2 1 0 0
6. [15 marks] Let A = . Find an invertible matrix P and a diagonal

0 0 0 3
0 0 3 0
matrix D such that D = P 1 AP.
7. [10 marks] For this question let the inner product on R3 be defined by

(u, v) = 2u1 v1 + 4u2 v2 + u3 v3 .

Let S be the subset of vectors in R3 which are orthogonal to (1, 1, 2), with
respect to the above inner product.
(a) Show that S is a subspace of R3 .
(b) Find an orthonormal basis of S.
!
cos sin
8. [10 marks] Let A = .
sin cos

(a) Show that for any value of the matrix A is orthogonal.

(b) Show that for any vector v in R2 , ||Av|| = ||v||, with respect to the usual
dot product of R2 .
(c) Let P be any 2 2 orthogonal matrix with determinant equal to 1. Show
that P = A, for some value of .

2 4 6
Answers: 1.(a) x = 5/2 t; y = 7/2 + 3t; z = 2t (b) 44 (c) 1 2 1


1 2 1

1 0 0 1 2 7
2.(a) L =

1 1 0 ,U =

0 3 8 (b) 2

1 2/3 1 0 0 22/3

3. (Gram-Schmidt) {(1, 1, 0, 2), (1, 7, 0, 4), (12, 4, 11, 4)}; one of many possible
answers.

4.(a) F (b) T (c) F (d) T (e) T (f) T (g) F (h) T (i) T (j) T

5.(a) k = 1 (b) k 6= 0, k 6= 1, k 6= 1 (c) k = 0, k = 1

0 1 0 1 3 0 0 0

0 2 0 1 0 0 0 0
6. CA (x) = x(x + 3)(x 3)2 P = ,D =

1 0 1 0 0 0 3 0


1 0 1 0 0 0 0 3

7.(a) ((x, y, z), (1, 1, 2)) = 0 x2y +z = 0. So S is just a plane passing through
the origin, hence it is a subspace of R3 . (NB: you could use the subspace test and
all that, if you wanted to.)

7.(b) One possible answer: {(1/ 3, 0, 1/ 3), (2/ 60, 3/ 60, 4/ 60)}
! !
cos sin T cos sin
8.(a) A = ,A = . Show AAT = I.
sin cos sin cos

8.(b) (Compute!)

||(x cos + y sin , x sin + y cos )||

q
= x2 (cos2 + sin2 ) + y 2 (sin2 + cos2 )
2
= x + y2
= ||(x, y)||

!
a b
8.(c) Let P = . Use P P T = I and det P = 1 to solve for a, b, c and d.
c d
 MAT187 - Calculus II - Winter 2015

Term Test 1 - February 3, 2015

Time allotted: 100 minutes. Aids permitted: None.

Total marks: 50

Full Name:
Last First

Student Number:

Email: @mail.utoronto.ca

Instructions

DO NOT WRITE ON THE QR CODE AT THE TOP OF THE PAGES.

Please have your student card ready for inspection, turn off all cellular phones, and read all the
instructions carefully.

DO NOT start the test until instructed to do so.

This test contains 14 pages (including this title page). Make sure you have all of them.

You can use pages 1214 for rough work or to complete a question (Mark clearly).

DO NOT DETACH PAGES 1214.

GOOD LUCK!
PART I No explanation is necessary. (10 marks)

1. Consider the solid of revolution generated by revolving the region between two functions f (x) 6 g(x)
for x [a, b] around the xaxis. Then its volume is given by (circle one choice)
Z b Z b


2
(a) g(x) f (x) dx (c) g(x) f (x) dx
a a
Z b Z b
g(x)2 f (x)2 dx



(b) 2x g(x) f (x) dx (d)
a a

Z
2
2. Consider sin83 x cos83 x dx and make a substitution to obtain
0

Z
2
Z b
83 83
sin x cos x dx = f (u) du.
0 a

The substitution is

u=

a=

b=

3. On the integral of question 2, the integrand becomes

f (u) =

4. A radioactive material decayed by 10% in 50 years.

Its half-life is .

5. Let a > 0 and consider the region bounded by the graph of y = aeax and the xaxis on the interval
[0, ).

Its area is .
 4x2 2x2 + x
6. Consider the rational function . When using partial fractions, we write
(x + 1)(x 2)3 (x2 + 9)2
this function as a sum of the following terms (circle all that apply):

A D G J Mx + N
(a) (d) (g) (j) (m)
x (x + 1) (x 2) (x2 + 9) (x2 + 9)

B E H K Ox + P
(b) (e) (h) (k) (n)
x2 (x + 1)2 (x 2)2 (x2 + 9)2 (x2 + 9)2

C F I L Qx + R
(c) (f ) (i) (l) (o)
x3 (x + 1)3 (x 2)3 (x2 + 9)3 (x2 + 9)3

7. Consider two functions f (x) and g(x) satisfying 0 6 f (x) 6 g(x) for x (0, ).
Z Z
Assume that g(x) dx converges. Then f (x) dx
1 1

(a) converges (b) diverges (c) we cannot tell

8. Consider two functions f (x) and g(x) satisfying 0 6 f (x) 6 g(x) for x (0, ).
Z Z
Assume that g(x) dx diverges. Then f (x) dx
1 1

(a) converges (b) diverges (c) we cannot tell

Z b
9. Recall that when approximating the integral f (x) dx using the trapezoid rule, we make an error
a
K(ba)
of at most ET 6 2
12 (x) , where K = max |f 00 (x)| and x = ba
n .
x[a,b]
Z 1
2
To approximate the integral e(x ) dx with a maximum error of e
32 , I should choose
0

n> .

10. A free-hanging rope forms a catenary: a curve which satisfies

1
q
2
00
y (x) = 1 + y 0 (x) for x [b, b].
a
Assume that for this rope, y 0 (b) = y 0 (b) = 10
a. Then the length of the rope is
Z b q
2
L= 1 + y 0 (x) dx = .
b

(express the length as a number explicitly)

 PART II Justify your answers.

11. You are working at a biology lab with a population of bacteria which grows (10 marks)

proportionally to its population. Moreover, the population doubles its size every hour.

(a) Assuming that you start with P0 million bacteria, find a formula for the population of bacteria
after t hours.
(b) You start with 100 million bacteria and you have two containers. Each can hold 300 million
bacteria. Your job is to grow as many bacteria as you can in 2 hours.
What is the best way to divide the bacteria in the two containers? Justify your answer.
(Hint. This question is not hard)
12. Compute the following integrals. (10 marks)
Z b
(a) Let b, > 0. Calculate ex sin(x) dx.
0
 1

arcsin(x) 1 x2
Z
(b) Calculate
dx.
0 cos arcsin(x)

(Hint. Use a substitution)

13. Let u(t) be the temperature in o C at the Pearson airport t years after March 1, 2000. (10 marks)

Then the average temperature for the first decade (2000-2010) is

Z 10
1
Average temperature = u(t) dt.
10 0

(a) Let a < b. What is the average temperature from March 1 of the year 2000 + a to September 1
of the year 2000 + b?
(b) Assume that u(t) = 5 + 30et sin(2t). If this temperature pattern holds forever, what is the
limiting average temperature?
 p
14. Consider the function f (x) = . Consider the solid created by rotating this function around the
xp
xaxis over the interval [1, ).

(a) Calculate the volume of the solid. (7 marks)

(b) Find the value of p that minimizes the volume of this solid. (3 marks)
USE THIS PAGE TO CONTINUE OTHER QUESTIONS OR FOR ROUGH WORK.
USE THIS PAGE TO CONTINUE OTHER QUESTIONS OR FOR ROUGH WORK.
USE THIS PAGE TO CONTINUE OTHER QUESTIONS OR FOR ROUGH WORK.

The end.
 University of Toronto, Faculty of Applied Science and Engineering
MAT187H1S - Calculus II

Final Exam - April 24, 2015

o-Sousa, Y. Liokumovich,
Examiners: S. Cohen, C. Dodd, B. Galva
P. Milgram, D. Ojeda, L.-P. Thibault

Duration: 150 minutes. No Aids permitted.

Full Name:
Last First

Student ID:

Email: @mail.utoronto.ca

Instructions

DO NOT WRITE ON THE QR CODE AT THE TOP OF THE PAGES.

Please have your student card ready for inspection, turn o all cellular phones, and read all the
instructions carefully.

DO NOT start the test until instructed to do so.

This test contains 14 pages and a detached formula sheet. Make sure you have all of them.

You can use pages 1314 for rough work or to complete a question (Mark clearly).

DO NOT DETACH PAGES 1314.

GOOD LUCK!
 PART I. No explanation is necessary. (20 Marks)

2
For questions 1-4, consider the dierential equation for y(t):
6
6
6
6 y 0 + ty + t = 0.
6
6
6 1. The equilibrium solution is (circle one option)
6
6
6
6 (a) y(t) = 1 (c) y(t) = 1
6
6
6
6
6
6 (b) y(t) = 0 (d) There is no equilibrium solution
6
6
6 2. If a solution y(t) has a horizontal asymptote, then what is it?
6
6
6
6
6 y= .
6
6
6
6
6 3. Consider the solution y(t) which satisfies y(0) = 1. Complete the blanks:
6
6
6
6 y 0 (0) = ,
6
6
6
6
6
6
6 y 00 (0) = .
6
6
6
6
6 4. (Harder!) Consider the solution y(t) which satisfies y(0) = 1. What is the first value T > 0 for which
6
6 y(t) < 0 for all t > T ?
6
4
T = .

5. Consider two particles with positions r1 (t) = cos(t), 2t and r2 (t) = 1 + t2 , et .

(a) These particles collide and the paths intersect.

(b) The particles collide, but the paths dont intersect.

(c) The particles dont collide, but the paths intersect.

(d) The particles dont collide and the paths dont intersect.

Continued...
2 For questions 69, please consider the following power series
6
6 1
X
6 k+1
6 f (x) = xk .
6 4k
6 k=0
6
6
6
6
6
6
6
6 6. The radius of convergence for this series is R = .
6
6
6
6
6
6
6 Z
6
6 7. f (x) dx = (not as a power series).
6
6
6
6
6
6
6
6
6 8. f (x) = (not as a power series).
6
6
6
6
6
6
6 1
X
4 (k + 1)
9. ( 1)k = .
2k
k=0

10. The position of a particle is ~r(t) = x(t), y(t) as given in the

figure on the right. x
1 1
(a) The slope of the path of the particle at the point 2, 0 is
t
1
.
y
1
(b) At this point, the particle is moving to the t
1
left / right and up / down .

(circle the correct options)

Continued...
PART II. Answer the following questions. Justify your answers.

t3 2
11. Consider the curve ~r(t) = ,t ,1 2t . (20 Marks)
3

(a) (10 marks) Find the maximum curvature.

(Hint. Use the formula with the cross product)

Continued...
(b) (6 marks) Show that there is no minimum curvature.

(c) (4 marks) Consider the particles with positions r~1 (t) = ~r(t) and r~2 (t) = ~r(e3 t). Will these
particles collide? If so, when?

Continued...
12. Archer fish hunt by spitting a jet of water at a nearby flying insect. (20 Marks)
?

(a) (5 marks) Assume that the fishs position, as shown,

p p
is ( 1, 1) and the insects position is 3, 1 and the 3, 1

water surface is at y = 0.
When the fish sees the insect, water refraction changes ( 1, 1)
the angle of the light as in the figure on the right,
causing the fish to misjudge the insects position.

The fish sees the path to the insect as a straight line. What does the fish think is the position
of the insect?

Continued...
(b) (10 marks) The fish hunts the insect by spitting a jet p
3, 1
of water in the direction of the dotted line. Ignoring
the eects of water and air resistance, but considering
gravity, ~a = (0, g), how fast should the fish spit the ( 1, 1)
water to hit the insect? Assume all distances given
are in cm and the insect is not moving.

(c) (5 marks) How much time does the insect have to move out of the way?

Continued...
13. Consider a vector-valued function (20 Marks)
z ~r(b)
~r(t) = x(t), y(t), z(t) ,

for t 2 [a, b]. The graph of this function forms a curve. ~r(a)

Find a formula for the area between the curve and the
xy plane (as indicated in the figure). y

Hint. To make the exercise easier, you can follow the steps:

(a) Let a < t0 < t1 < b. Sketch the rectangle with coordinates

P0 = x(t0 ), y(t0 ), 0 , P1 = x(t0 ), y(t0 ), z(t1 ) , P2 = x(t1 ), y(t1 ), z(t1 ) , P3 = x(t1 ), y(t1 ), 0

(b) What is the area of the rectangle from (a)?

(c) Approximate the area (between the curve and the xy plane) with the area of n thin rectangles.

(d) Find a formula for the exact area between the curve and the xy plane by taking the limit as n ! 1.

Continued...
(Use this page to continue question 13)

Bonus. Test your formula. Use your formula to find the area for the function (4 Marks)

~r(t) = (t, t, t) for t 2 [0, 1] and confirm that it matches the formula for the area of a triangle.

Continued...
14. In this question we will investigate the shells of some land snails. (20 Marks)
750

500

(a) (8 marks) A certain snails shell has the shape of the

spiral
250

r = 2 ,
-1000 -750 -500 -250 0 250 500 750 1000

-250

where the magnitude of the angle in radians equals

-500

the number of days since the snail was hatched.

-750

19
If the variable r is in mm, how long will it take for the shells spiral to be mm long?
3

Continued...
(b) (8 marks) A dierent snails shell also forms a spiral. Each night, since its humid, the length
1 1
grows by cm, but during the day, since its dry the length decreases by cm, where n
2n 2n + 1
is the number of days since it hatched.
If the snail lives forever, how long will the shells spiral be?
P
(Hint. If you use notation, it should remind you of a Taylor series, with x set equal to 1)

Continued...
 (c) (4 marks) For the same snail as in (b), how many days will it take for the shells spiral to be
1
within 100 cm of the limiting size?

Bonus. Excluding this bonus question, what is your mark on this exam? 2 points (3 marks)

Continued...
 MAT197 - Calculus B - Winter 2014

Term Test 2 - February 27, 2014

Time allotted: 90 minutes. Aids permitted: None.

Full Name:
Last First

Student ID:

Email: @mail.utoronto.ca

Instructions

Write only on the front pages (with QR code on top).

ONLY THE FRONT PAGES WILL BE SCANNED. THE BACK PAGES

WILL NOT BE SEEN BY THE GRADERS.
DO NOT WRITE ON THE QR CODE AT THE TOP OF THE PAGES.

Please have your student card ready for inspection, turn off all cellular phones, and read all the
instructions carefully.

DO NOT start the test until instructed to do so.

This test contains 11 pages (including this title page). Make sure you have all of them.

You can use the back of pages for rough work.

GOOD LUCK!
MAT197 Term Test 2

PART I No explanation is necessary.

For questions 14, please fill in the blanks. (4 marks)

Z
1
1. What is the best substitution to calculate dx ?
17 + 16x + 4x2

u=

Z
x+2
2. What are two good substitutions to calculate dx ?
5 + 4x + x2

u= or u=

Z b
3. Using the Trapezoid Rule with n = 10 we obtain an error of 1 when computing the integral f (x) dx.
a
What is the smallest value for n to obtain an error of at most 106 ?

n=

4. During Cinco de Mayo, you shoot a bullet straight up into the sky at the speed of 500 m/s. The
altitude of the bullet y(t) at time t seconds after being shot satisfies the differential equation y 00 = g.
What is the velocity of the bullet when it hits the ground?

The velocity is

Page 2 of 11 Continued...
MAT197 Term Test 2

For questions 5-8, please choose the correct answer. (4 marks)

(x 2)2 (x 1)
5. Which terms appear in the partial fraction decomposition of ?
(x + 1)(x2 + 3)2 (x )2
(Select all that apply)

A D Gx + H
(a) (d) (g)
x1 x+1 x2 + 3

B E Ix + J
(b) (e) (h)
x2 x (x2 + 3)2

C F
(c) (f ) (i) Kx + L
(x 2)2 (x )2

6. Which of the following integrals is convergent?

Z Z 1
1
(a) x dx (c) dx
0 x
Z
1
Z
1
(b) dx (d) dx
1 x 1
5
x4

7. Which of the following integrals is divergent?

Z 1 Z
1
(a) dx (c) 0 dx
x4 0

Z Z 1
1
1
(b) dx (d) dx
1 + x2 0 x1+p2

Z
8. Which of the following substitutions is best to evaluate cos7 x sin4 x dx ?

(a) u = sin x (c) u = cos x

(b) u = tan x (d) u = sec x

Page 3 of 11 Continued...
MAT197 Term Test 2

PART II Justify your answers.

1. Evaluate (6 Marks)
3
x2
Z
5
dx.
0 9 25x2

Page 4 of 11 Continued...
MAT197 Term Test 2

2. Evaluate the integral (8 Marks)

x4 + 2x3 + 9x2 + 8x + 16
Z
dx.
x(x2 + 4)2

Page 5 of 11 Continued...
MAT197 Term Test 2

3. You are working for a biologist who is studying a new kind of virus that reproduces (6 Marks)

extremely quickly, with the population following the function

Z t
2
p(t) = ex dx,
0

1
where t is in days. The size of each virus is 10 m2 .

The biologist wants to make sure he has enough space for this virus to grow for 4 days.

(a) Using the Midpoint rule with n = 4, give an approximation of the necessary area.

Page 6 of 11 Continued...
MAT197 Term Test 2

(b) Recall the formula of the error EM we obtain when approximating an integral using the Midpoint
Rule:
b a
|EM | max f 00 (x) (x)2 .

x[a,b] 24
What is the smallest value of n that we can use to ensure that the error we are making is at
most 1?

Page 7 of 11 Continued...
MAT197 Term Test 2

4. Gravitational force decreases with the square of the distance, in particular we can write (6 Marks)
mM G
F = ,
y2
where y is the altitude of the object from the center of a planet of radius R and mass M , m is mass
of the object, and G is the gravitational constant.
1
To find the escape velocity ve , we know that the total kinetic energy of the object is 2 m ve2 , and that
it equals the work necessary to take the object from the surface of the planet (y = R) to infinity
(y = ).

(a) Find the escape velocity (in terms of M , G, and R).

Page 8 of 11 Continued...
MAT197 Term Test 2

(b) Assuming that c is the speed of light, what is the minimum radius R for a black-hole of mass
M ? (in terms of G)
(Hint. A black-hole is a celestial body where the escape velocity is at least the speed of light)

Page 9 of 11 Continued...
MAT197 Term Test 2

5. A bullet of mass m travelling upwards with initial velocity v0 is slowed by the force (6 Marks)

of gravity mg and by air resistance kv 2 , where k is a positive constant. As the bullet moves upward,
its velocity v satisfies
dv
m = (kv 2 + mg).
dt

(a) Find a formula for v(t).

 q 
mg
Hint. For simplicity, write a = k and dont forget the initial condition.

Page 10 of 11 Continued...
MAT197 Term Test 2

(b) Assuming m = k, and v0 = 100 g, find the maximum height the bullet will reach in the form
of an integral. Dont solve the integral.

Page 11 of 11 The end.