Math 317: Linear Algebra
Exam 1
Fall 2015
Name:
*No notes or electronic devices. You must show all your work to receive full credit. When
justifying your answers, use only those techniques that we learned in class up to Section
2.2. Good luck!
1. Let u = (1, 2, 3) and v = (1, 1, 1).
(a) (10 points) Calculate 3u + 2v and u − v.
(b) (10 points) Calculate (3u + 2v) · (u − v), and calculate cos θ where θ is the
angle between 3u + 2v and u − v. You do not have to simplify your answer.
(c) (10 points) Find a vector w so that w is orthogonal to u.
(a) 3u + 2v = 3(1, 2, 3) + 2(1, 1, 1) = (3, 6, 9) + (2, 2, 2) = (5, 8, 11) and
u − v = (1, 2, 3) − (1, 1, 1) = (0, 1, 2).
(b) (3u + 2v) · (u − v) = (5, 8, 11) · (0, 1, 2) = 5(0) + 8(1) + 11(2) = 30.
(3u + 2v) · (u − v)
30
√
=√
.
2
2
k3u + 2vkku − vk
5 + 8 + 112 02 + 12 + 22
(c) w = (w1 , w2 , w3 ) is orthogonal to u = (1, 2, 3) if w · u = 0. This is
equivalent to saying that 1w1 + 2w2 + 3w3 = 0. Many choices work
here. For instance, picking w1 = 1, w2 = 1, w3 = −1 will do the job.
cos θ =
1
Math 317: Linear Algebra
Exam 1
Fall 2015
2. Let A be an n × 2 matrix with columns a1 and a2 in that order.
(a) ( 5 points) Find a vector
x
x= 1 ,
x2
such that Ax = 2a1 + 3a2 .
2
Recalling that Ax = x1 a1 + x2 a2 , we see that x =
is precisely the
3
vector that we are looking for.
(b)
i. (10 points) Suppose that Ax = 0 for every x ∈ R2 . Prove that A = 0.
That is, A is the zero matrix.
ii. (5 points) Suppose that B is an n × 2 matrix, and that Ax = Bx for every
x ∈ R2 . Prove that A = B. Hint: Use part(i) to help prove this claim.
1
2
i. Proof: Suppose that Ax = 0 for every x ∈ R . Then A
= 0.
0
1
By definition of matrix multiplication, A
= 1a1 +0a2 = a1 = 0.
0
0
So the first column of A is entirely 0. Similarly, A
= 0 =⇒
1
a2 = 0. So, the second column of A is entirely 0. Since A is a
n × 2 matrix, it only has two columns and hence A = 0.
ii. Proof: Suppose that B is an n × 2 matrix, and that Ax = Bx
for every x ∈ R2 . Then Ax = Bx =⇒ Ax − Bx = 0 =⇒
(A − B)x = 0 for all x ∈ R2 . From part i, this tells us that
A − B = 0 or A = B.
2
Math 317: Linear Algebra
Exam 1
Fall 2015
3. 25 points, 5 points each.
(a) Find a parametric equation for the line 2x1 − x2 = 0. Write your answer as
span(a) for some vector a.
(b) Let T1 : R2 → R2 be the transformation that projects x onto the line 2x1 −
x2 = 0 in R2 . Prove that T1 is linear.
(c) Give the standard matrix that represents T1 .
(d) Let T2 : R2 → R2 be the linear transformation that reflects x across the line
2x1 − x2 = 0. Give the standard matrix that represents T2 .
(e) Give the standard matrix that represents T1 ◦ T2 .
(a) 2x1 − x2 = 0
=⇒
x2 = 2x1 and so x = (x1 , x2 ) =
(x1 , 2x1 ) = x1 (1, 2) + (0, 0). Thus, the parametric line is given by
x = {(0, 0) + t(1, 2); t ∈ R} = span((1, 2)).
(b) Proof : The transformation that projects x onto the line 2x1 − x2 = 0
in R2 is given by T1 (x) = proja x where a is the vector that spans the
line 2x1 − x2 = 0, that is, a = (1, 2). We note that for any vectors
x, y ∈ R2 , we have that:
(x + y) · a
(x · a) + (y · a)
a=
a
T1 (x + y) = proja (x + y) =
2
kak
kak2
y·a
x·a
a
+
a = proja x + proja y = T1 (x) + T1 (y).
=
kak2
kak2
We also see that T1 (cx) = proja (cx) =
(cx) · a
x·a
a
=
c
a = cT1 (x).
kak2
kak2
Thus T1 is a linear transformation.
1
0
(c) We calculate T1
and T1
.
0
1
(1, 0) · (1, 2) 1
1
T1
=
2
0
k(1, 2)k2
1
1
= 2
2
1 +2 2
1/5
=
.
2/5
Similarly,
3
Math 317: Linear Algebra
Exam 1
Fall 2015
(0, 1) · (1, 2) 1
0
T1
=
1
2
k(1, 2)k2
2
1
= 2
2
1 +2 2
2/5
=
.
4/5
1/5 2/5
So the standard matrix for T1 is given by: A =
.
2/5 4/5
(d) To find the standard matrix for the reflection transformation, T2 , we
first recall that T2 (x) = 2Pl (x) − I(x) where Pl (x) is the linear transformation that projects x onto some line l through the origin, and
I(x) = x. But this Pl (x) is precisely what we worked with in parts (b)
and (c). Using the standard matrix that we obtained in (c) and using
the fact that the standard matrix for the identity operator is in fact
the identity matrix we obtain
1/5 2/5
1 0
−3/5 4/5
B = 2A − I = 2
−
=
,
2/5 4/5
0 1
4/5 3/5
as the standard matrix for T2 .
(e) We know that the action of the composition between two linear transformations is given through matrix multiplication, once we have the
standard matrices for T1 and T2 . Thus the standard matrix for T1 ◦ T2
is given by:
1/5 2/5
1/5 2/5 −3/5 4/5
AB =
=
.
2/5 4/5
4/5 3/5
2/5 4/5
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Math 317: Linear Algebra
Exam 1
Fall 2015
4. (a) (10 points) Find the reduced row echelon form and the rank of the following matrix.


1 0 0 3
 0 1 1 0 .
−1 2 1 −2


1
(b) (10 points) Find (if possible) a linear combination of the vectors v1 =  0 ,
−1
 
 
 
0
0
3
v2 = 1, v3 = 1, that gives b =  0 . Hint: Use your work in (a) to
2
1
−2
help you with this problem.
(a)


1 0 0 3
3 →R3
 0 1 1 0  R1 +R
−→
−1 2 1 −2

1 0 0
−R3 →R3

0 1 1
−→
0 0 1




1 0 0 3
1 0 0 3
+R3 →R3
0 1 1 0 −2R2−→
0 1 1 0
0 2 1 1
0 0 −1 1



3
1 0 0 3
−R3 +R2 →R2
0 1 0 1 
0
−→
−1
0 0 1 −1
There are three non-zero rows in the reduced row echelon form of A
and so the rank of A is 3.
(b) From (a), we know that
 
 
 
 
3
1
0
0
 0  = x1  0  + x2 1 + x3 1 ,
−2
−1
2
1
only if x1 = 3, x2 = 1, x3 = −1.
5
Math 317: Linear Algebra
Exam 1
Fall 2015
5. (15 points, 5 points each) Determine if the following statements are true or false. If
the statement is true, give a brief justification. If the statement is false, provide a
counterexample or explain why the statement is not true.
(a) Suppose that A is a 2 × 2 nonsingular matrix. Then rank(A2 ) = 2.
True. Since A is a 2 × 2 nonsingular matrix, then rank(A)=2. We
also know that the product of two nonsingular matrices is nonsingular.
Hence, A2 is nonsingular and it must also have rank 2 (as A2 is a 2 × 2
matrix).
(b) Suppose that A is an m × n matrix where m < n. The system Ax = b will
possess an infinite number of solutions.
False. This is true only if we know that the system is consistent. For
a counterexample, we give:
1 0 0 0
[A|b] =
.
0 0 0 1
Notice that 2 = m < n = 4, but the system has no solution.
(c) Suppose that A is an m × n matrix such that rank(A) = n. Then Ax = b is
consistent for every b ∈ Rm .
In general this is false. To see why, consider the following example:

1
0

[A|b] = 
0
0
0
0
1
0
0
0
0
0
1
0
0

0
0

0
.
1
2
Then the rank of A is 3 since all three columns have pivot entries.
Furthermore, the rank of A is n since A is a 5 × 3 matrix. However,
the associated linear system has no solution.
6