Applied Math 245
Midterm Exam: Winter 2007
Answer the questions in the spaces provided on the question sheets. If you run out of room for an
answer continue on the back of the page. No notes, books, or other aids may be used on the exam.
Student Id:
Answer Key
1. (10 points)
2. (10 points)
3. (10 points)
4. (10 points)
5. (10 points)
6. (10 points)
7. (10 points)
8. (10 points)
9. (10 points)
10. (10 points)
Total (100 points)
Page 1 of 7
Applied Math 245
Midterm Exam: Winter 2007
1. (10 points) Determine whether each of the following statements is true or false. No justification is required.
(a) A linear system with fewer unknowns than equations must have infinitely many solutions or none.
Solution: False; the system x = 2, y = 3, x + y = 5 has a unique solution.
(b) If A and B are matrices of the same size, the formula rank(A + B) = rank(A) + rank(B) must hold.
Solution: False; let A = B = I2 for example.
x
(y + 1)2 − (y − 1)2
(c) The function T
=
represents a linear transformation.
y
(x − 3)2 − (x + 3)2
x
4y
0 4 x
Solution: True; simplify to see that T
=
=
.
y
−12x
−12 0 y
k
(d) The matrix
5
−2
is invertible for all real numbers k.
k −6
Solution: True; det(A) = k 2 − 6k + 10, which has no real solution.
 
1
(e) The vectors 0,
0
 
2
1,
0
 
3
2 form a basis for R3 .
1
Solution: True; by inspection the vectors are linearly independent.
Page 2 of 7
Applied Math 245
Midterm Exam: Winter 2007
x
2. (10 points) Find all solutions to the set of equations 1
3x1
+
x2
+ 3x2
= b1 .
= b2 Solution: By inspection, we see that the system is consistent only if b2 = 3b1 .
 
1

1

3. (10 points) Find all vectors in R4 that are perpendicular to the three vectors 
1,
1
 
1
2
 ,
3
4
 
1
9
 .
9
7
Solution: We need to find a vector ~
x such that ~
v1 · ~
x =~
v2 · ~
x =~
v3 · ~
x = 0. We can represent this as the
~
system A~
x = 0, where the rows of A are the three vectors given. We can easily compute:

1
rref 1
1
1 1
2 3
9 9
1
4
7
 
1
0
0 = 0
0
0
0
1
0

0
0.25 0
0 −1.50 0
1
2.25 0


−0.25
 1.50

Letting x4 = t, we see that any vector of the form ~
x =t
−2.25 is a solution.
1.00
Page 3 of 7
Applied Math 245
Midterm Exam: Winter 2007
4. (10 points) Let B −1 =
Solution: A = (AB)B
−1
1 2
1
and (AB)−1 =
3 5
2
= ((AB)
)
−1 −1
B
−1
3
. Find A.
5
1 3
=
2 5
−1 1
3
2
−5
=
5
2
3 1
−1 3
2
4
5
=
5
−1 −1
5. (10 points) Describe each of the following linear transformations as a well-known geometric transformation
combined with a scaling. Give the scaling factor in each case.
1 1
(a)
1 1
2
1 1
u1
u1 u2
=
, or equivalently
Solution: This matrix has the form of a projection: k
u1 u2
u22
1 1
" 1 #
2
√
1/k 1/k
u
u1
u1 u2
=
, where u12 + u22 = k1 + k1 = 1. Thus, k = 2 and u~ = 1 = √12 .
2
u2
u1 u2
u2
1/k 1/k
2
(b)
3 0
−1 3
1
Solution: This matrix has the form of a vertical shear: k
c
1 0
3/k
0
1
=
. Thus, k = 3 and c = − 3 .
c 1
−1/k 3/k
(c)
3
4
0
3
=
1
−1
0
, or equivalently
3
4
−3
a
b
3
4
Solution: This matrix has the form of a reflection: k
=
, or equivalently
b −a
4 −3
2
2
a
b
3/k
4/k
=
, where a2 + b2 = k3 + k4 = 1. Thus, k = 5.
b −a
4/k −3/k
Page 4 of 7
Applied Math 245
Midterm Exam: Winter 2007
6. (10 points) The color of light can be represented as a vector
 
R
G  ,
B
where R = amount of red, G = amount of green, and B = amount of blue. The human eye and the brain
transform the incoming signal into the signal
 
I
L  ,
S
where
intensity
I
=
R+G+B
3
long-wave signal
L
=
R−G
R+G
B−
.
2
 
 
R
I



(a) Find the matrix P representing the transformation from G to L.
B
S
 1
1
1
short-wave signal S
3
3
Solution: P =  1
− 21
−1
− 21
=
3
0
1
(b) Consider a pair of yellow sunglasses for water sports that cuts out all blue light and passes all red and
green light. Find the 3 × 3 matrix A that represents the transformation incoming light undergoes as
it passes through the sunglasses.


1 0 0
Solution: A = 0 1 0
0 0 0
(c) Find the matrix for the composite transformation that light undergoes as it first passes through the
sunglasses and then the eye.
 1

1
0
3
3
Solution: P A =  1 −1 0
− 21 − 21 0
Page 5 of 7
Applied Math 245
Midterm Exam: Winter 2007

1
7. (10 points) Find vectors that span the kernel of A = 0
0
Solution: Reducing the matrix

1 2
0 1
0 0

3 4
2 3.
0 1
2
1
0
to reduced row-echelon form:




3 4
1 0 −1 −2
1
2 3 → 0 1
2
3 → 0
0 1
0 0
0
1
0
0 −1
1
2
0
0

0
0
1

 

1
1
−2
−2
 

Letting x3 = t, we see that any vector of the form t 
 1 is a solution. Thus, ker(A) = span  1.
0
0

1
0
8. (10 points) Express the kernel of A = 
0
0
2
0
0
0
0
1
0
0
0
0
1
0

0
0
 as the image of another matrix B.
0
0
3
2
1
0
Solution: This matrix is already in reduced row-echelon form and has three free variables. Letting x2 = r ,
x5 = s and x6 = t, we see that solutions are of the form:


 
 
 
      
−2r − 3s
−2
−3
0
−2
−3
0


 1
 0
0
 1  0 0
r


 
 
 
      
 −2s 
 
 
 
      

 = r  0 + s −2 + t 0 = span  0 , −2 , 0
 −s 
 0
−1
0
 0 −1 0


 
 
 
      


 0
 1
0
 0  1 0
s
t
0
0
Since the image of a matrix is simply the span of its

−2
 1

 0
B=
 0

 0
0
1
0
columns, we have:

−3 0
0 0

−2 0

−1 0

1 0
0
Page 6 of 7
1
0
1
Applied Math 245
Midterm Exam: Winter 2007

1
9. (10 points) Find a basis of the image of A = 1
1

1 1
2 5.
3 7
Solution: The first two columns are linearly independent, since the second is not a scalar multiple of the
first. The third column is not a linear combination of the first two, which can be seen from:

 
1 1 1
1
rref 1 2 5 = 0
1 3 7
0
     
1 
1
 1
Therefore, a basis for the image is 1 , 2 , 5 .


7
3
1
0
1
0

0
0
1
10. (10 points) Verify that the image of an n × m matrix A is a subspace of Rn .
Solution: We need to prove that im(A) meets the three necessary conditions:
1. A~0 = ~0 for any A. Therefore, ~0 is in im(A).
2. Let ~
x , y~ be in im(A). Then, by definition of the image, there exist vectors ~
v and w
~ such that A~
v =~
x
and Aw
~ = y~. We see that ~
x + y~ = A~
v + Aw
~ = A(~
v +w
~ ). Therefore, ~
x + y~ is in im(A).
3. Let ~
x be in im(A). Then, by definition of the image, there exists a vector ~
v such that A~
v =~
x . We
see that k~
x = kA~
v = A(k~
v ). Therefore, k~
x is in im(A).
Page 7 of 7