MATH 317: SPRING 2016
ANSWERS
EXAM 1
1. Let x = (x1 , x2 , . . . , xn ) and y = (y1 , y2 , . . . , yn ) be vectors in Rn .
(3pts)
(a) Define the dot product of x and y.
Answer:
x · y = x 1 y 1 + x2 y 2 + · · · + xn y n .
(3pts)
(b) What does it mean to say that x and y are parallel ?
Answer: The vectors x and y are parallel if they are scalar multiples of each
other. That is, there exists a real number c such that x = c y.
(4pts)
(c) What does it mean to say that x and y are orthogonal ?
Answer: The vectors x and y are orthogonal if their dot product is 0. That
is, x · y = 0.
(4pts)
(d) Suppose v1 , v2 , . . . , vk are also vectors in Rn . What does it mean to say that x is
a linear combination of the vectors v1 , v2 , . . . , vk ?
Answer: x is a linear combination of the vectors v1 , v2 , . . . , vk if there exist
scalars c1 , c2 , . . . , cn such that
x = c1 v 1 + c2 v 2 + · · · + cn v n .
(10pts) 2. Recall Prop. 2.1 of our text says that the dot product satisfies the following properties:
for all x, y, z ∈ Rn and c ∈ R,
1. x · y = y · x;
2. x · x = kxk2 ≥ 0, with equality if and only if x = 0;
3. (cx) · y = c(x · y);
4. x · (y + z) = x · y + x · z.
Prove that if x is orthogonal to each of the vectors v1 , v2 , . . . , vk , then x is orthogonal
to every linear combination of the vectors v1 , . . . , vk . (For full credit, identify places in
your proof where properties from the list above are used; to refer to these properties,
use the letters given above.)
Answer: Suppose x is orthogonal to each of the vectors v1 , . . . , vk . Then,
x · (c1 v1 + · · · + ck vk ) = x · (c1 v1 ) + · · · + x · (ck vk ) (∵ Prop. 2.1 (4))
= c1 (x · v1 ) + · · · + ck (x · vk ) (∵ Prop. 2.1 (1) and (3))
= c1 0 + · · · + ck 0
(∵ x · vi = 0 for all 1 ≤ i ≤ k)
= 0.
(5pts) 3. State a theorem about existence and uniqueness of solutions to the system Ax = b. You
may state more than one theorem if you wish, but quality is better than quantity. Only
write what you know is true and carefully state your assumptions. If you make a
broad statement that, in fact, only applies under a narrow set of conditions, and you
leave out those conditions, then you will not receive very much credit. (Use only the
space provided below.)
Answer: There is more than one good answer to this. Here’s one possibility, which
is probably the most helpful when answering the questions on the next page. (Other
possible answers are given on the last page below.)
Let A ∈ Rm×n . Then the equation Ax = b
1. is consistent for all b ∈ Rm when rank(A) = m (=number of rows of A);
2. has at most one solution when rank(A) = n (=number of columns of A).
Recall that rank(A) = m iff the echelon form of A has no rows consisting of all zeros;
rank(A) = n iff the echelon form of A has a pivot in every column.
Score for this page:
out of 15
4. For the given matrix, write down the letters corresponding to true statements in each
case. Select from the following statements:
A. For every b ∈ Rm , Ax = b is consistent.
B. For every b ∈ Rm , Ax = b is inconsistent.
C. For every b ∈ Rm , Ax = b has exactly one solution.
D. For every b ∈ Rm , Ax = b has infinitely many solutions.
E. There exists b ∈ Rm such that Ax = b is consistent.
F. There exists b ∈ Rm such that Ax = b is inconsistent.
G. There exists b ∈ Rm such that Ax = b has exactly one solution.
(4pts)
H. There exists b ∈ Rm such that Ax = b has infinitely many solutions.


1 0 0
(a) If the matrix A has reduced echelon form 0 1 0,
0 0 1
then which of the statements (a)–(h) above is true? (Select all that apply.)
A.
(4pts)
B.
C.
D.
E.
F.
G.
H.

1
0
(b) If the matrix A has reduced echelon form 
0
0
0
1
0
0

0
0
,
1
0
then which of the statements (a)–(h) above is true? (Select all that apply.)
A.
B.
C.
D.
E.
F.
G.
H.
(4pts)
1 0 1
(c) If the matrix A has reduced echelon form
,
0 1 2
then which of the statements (a)–(h) above is true? (Select all that apply.)
A.
(4pts)
B.
C.
D.
E.
F.
G.
H.

1
0
(d) If the matrix A has reduced echelon form 
0
0

0 −1 2 0
1
1
1 3 0 −2
,
0
0 0 1 −1
0
0 0 0
0
then which of the statements (a)–(h) above is true? (Select all that apply.)
A.
B.
C.
D.
E.
F.
G.
H.
Score for this page:
out of 16
5. Let A ∈ Rm×n , B ∈ Rn×m , C ∈ Rn×m , and b ∈ Rm . For each statement below, either
prove the claim or write FALSE and give a counter-example.
(5pts)
(a) Claim: If AB = Im , then a solution to Ax = b, if it exists, is unique.
Answer: FALSE. Let A = 1 1 , and suppose b = [0]. Then there are
infinitely many solutions to Ax = b, since every x = (x1 , x2 ) satisfying x1 +x2 =
0 is a solution.
(5pts)
(b) Claim: If CA = In , then a solution to Ax = b, if it exists, is unique.
Answer: Suppose Ax = b has a solution, call it x0 . Suppose x00 is another
solution, so that Ax00 = b. Then Ax0 = Ax00 , so BAx0 = BAx00 , so In x0 = In x00 ,
so x0 = x00 .
(5pts)
(c) Claim: If Ax = 0 has only the trivial solution x = 0, then for every b ∈ Rm there
is exactly one solution to Ax = b.
1
Answer: FALSE. Let A =
. Then the only solution to Ax = 0 is x = [0].
0
That is,
1 0
x1 =
if and only if x1 = 0.
0
0
In this case, if the equation Ax = b has a solution, then it is unique. However,
it is not true that Ax = b always has a unique solution,since
it may be incon 1
1
sistent. For example, continuing with the matrix A =
above, if b =
,
0
1
then Ax = b has no solution.
Score for this page:
out of 15


 
1 1
0
1



6. Let A = 0 1 −1 and b = 2.
1 1
1
3
(8pts)
(a) Find A−1 .
Answer:

A−1
(5pts)
(b) Use your answer to (a) to solve Ax = b.
Answer:
(2pts)

2 −1 −1
1
1
= −1
−1
0
1
 
−3

4
x=
2
(c) Use your answer to (b) to express b as a linear combination of the columns of A.
(Fill in the blanks with your answers.)
 
 
1
1
2 = −3 0 +
3
1
4
 
1
1 +
1

2

0
−1
1
Score for this page:
out of 15
Alternative answers to Question 3:
• Suppose the system Ax = b is consistent. Then it has a unique solution if and only if
the associated homogeneous system Ax = 0 has only the trivial solution. This happens
exactly when rank(A) is the number of columns of A.
• Let A ∈ Rn×n . The following are equivalent:
1. A is nonsingular.
2. Ax = 0 has only the trivial solution.
3. For every b ∈ Rn , the equation Ax = b has a solution (indeed, a unique solution).
Score for this page:
out of 0