First and last name:
Codename:
CLOSED book, calculators allowed. Remember that points are given for the steps
that you perform, not just the answer.
1. [10 points ] Solve the system of equations
2a +2b +2c = 0
a +2b
= 4
a
= ,1
by writing the corresponding augmented matrix and using elementary row
operations. I want to see you write the steps that you perform, e.g., R2 , 3R1,
to illustrate that you've understood the method. It's easy to see what the
answer should be, but I'd like to see that you understand the method.
Solution:
The augmented matrix is (2 points)
02
@1
1
2 2 0
2 0 4 A
1 0 0 ,1
First, to make the arithmetic easier it's convenient to ensure that there is a
1 in the top left so, e.g., R3 $ R1 to get
0 1 0 0 ,1 1
@1 2 0 4 A
2 2 2 0
Then R2 , R1 and R3 , 2R1 give
0 1 0 0 ,1 1
@0 2 0 5 A
0 2 2 2
(2 points for correct operations, 2 points for correct arithmetic). Finally, to
get a zero in the third row/second column, do R3 , R2 to get
01
@0
0 0 ,1
2 0 5
0 0 2 ,3
1
A
(1 point for correct operation, 1 point for correct arithmetic). Now it's in
echelon form, rewrite this as a system of equations (1 point):
a
= ,1
= 5
2c = ,3
2b
Therefore the answer is (a; b; c) = (,1; 5=2; ,3=2) (1 point).
1
2. [10 points ] Draw the graph of
r(x) = x2 x, 4
using the 6-step procedure introduced in lectures. Most of the points are for
your working and PRESENTATION, so explain what you're doing in each
step.
Solution:
(a) Not the translate of a simple function.
(b) Find y-intercepts (1 point): y = r(0) = 0 (1 point).
(c) i. Findx-intercepts (1 point): solve r(x) = 0 so x = 0 (1 point).
ii. Find vertical asymptotes (1 point): solve x2 , 4 = 0 so x = 2 (1
point).
(d) Find far left / far right behaviour:
1 = degree f < degree g = 2;
(1 point) so y = 0 is a horizontal asymptote (1 point).
(e) Plot one point between and beyond the zeroes and vert. asymptotes at
x = ,2; 0; 2. Thus, for example, calculate r(,3) = ,3=5, r(,1) = 1=3,
r(1) = ,1=3, r(3) = 3=5 (1 point).
(f) Now just draw the graph (1 point) as follows. From y = 0 at far left
it drops through (,3; ,3=5) down towards ,1 at the vert asymptote
x = ,2. On right hand side of x = ,2 it drops from +1 through
(,1; 1=3), then through the origin, then through (1; ,1=3) and down
towards ,1 at the vert asymptote x = 2. On right hand side of x = 2
it drops from +1 through (3; 3=5), then on towards y = 0 at the far
right.
3. [15 points ] Circle T or F according to whether you think the statement is True or False
(the rst three are worth 2 points each, the last three are worth 3 points
each):
Solution:
T
F
F
T
Any nonsquare matrix doesn't have an inverse.
01
The matrix @ 0
0 1 0
1 0 0
0 0 1 0
1
A is in row reduced echelon form.
For the rational function (x + 1)(xx , 10)2 , the partial fraction decomposition is of the form x +a 1 + (x ,b10)2 .
The graph of a rational function can cross one of its asymptotes.
2
F
Suppose A is an m n matrix, B is an n r matrix and AB = 0.
Then either A or B must have all entries equal to zero.
A circle and a straight line may intersect in a single point.
4. [10 points ] Write the partial fraction decomposition for the rational expression
T
4x2 , 1 :
2x(x + 1)2
[Hint: Eventually you'll have to solve a system of equations. Question 1
should be helpful!!! If it's not then move on to the next question now, but
return and check through your solution to this question.]
Solution:
The partial fraction decomposition is (3 points)
4x2 , 1 = a + b + c ;
2x(x + 1)2 2x x + 1 (x + 1)2
for some constants a; b; c. To work out what a; b; c are, put the fractions over
a common denomenator (1 point)
a + b + c = a(x + 1)2 + b(2x)(x + 1) + c(2x) :
2x x + 1 (x + 1)2
2x(x + 1)2
Multiply out the polynomials on the numerator to give (1 point)
ax2 + 2ax + a + 2bx2 + 2bx + 2cx :
2x(x + 1)2
Now gather the terms (1 point)
(a + 2b)x2 + (2a + 2b + 2c)x + a :
2x(x + 1)2
This is equal to the fraction in the question, namely (1 point)
4x2 , 1 = (a + 2b)x2 + (2a + 2b + 2c)x + a :
2x(x + 1)2
2x(x + 1)2
The coecients must be equal, so you get three equations (2 points):
2a +2b +2c = 0
a +2b
= 4
a
= ,1
This is the system from question 1, so you know the answer is (a; b; c) =
(,1; 5=2; ,3=2). Therefore (1 point)
4x2 , 1 = ,1 + 5=2 + ,3=2 :
2x(x + 1)2 2x x + 1 (x + 1)2
3
5. [10 points ]
(a) Find the inverse of the matrix
A=
3 ,3
,2 3
;
or show that A has no inverse. Using the formula from lectures will get
at most 3 marks out of 7 for this part. To gain full marks I want to see
you perform several ERO's on a matrix which you create. If you have
time, check your answer using matrix multiplication.
Solution: Augment the 2 2 matrix by the identity (2 points)
3 ,3 1 0
,2 3 0 1
Then reduce to row reduced echelon form (4 points): 31 R1 and 12 R2 give
Then R2 + R1 gives
1 ,1 1=3 0
,1 3=2 0 1=2
1 ,1 1=3
0
0 1=2 1=3 1=2
Then 2R2 gives
1 ,1 1=3
Then R1 + R2 gives
1
0
0 1 2=3 1
0 1 1
0 1 2=3 1
The identity appears on the left of this matrix, so A does have an inverse,
namely (1 point)
1 1
,
1
A = 2=3 1 :
(b) Using your answer to part (a), solve the system of equations
3x ,3y = 4
,2x +3y = 6 :
The system is a single matrix equation AX = B for the matrix
A as in part (a) above (1 point). Since A has an inverse, the solution is (2
Solution:
points)
X=
x
y
= A,1 B =
4
1 1
2=3 1
4 6
=
10
26=3 :