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Math 125 – Exam 3 – Version 1
November 29, 2006
60 points possible
·
¸
4 α
1. Let A =
.
7 9
(a) (3pts) Find the adjoint of A.
Solution: For a 2 × 2 matrix, the adjoint of A is
·
¸t ·
¸
A11 A12
A11 A21
adj(A) =
=
.
A21 A22
A12 A22
Therefore,
·
(−1)1+1 M11 (−1)2+1 M21
adj(A) =
(−1)1+2 M12 (−1)2+2 M22
·
¸
(1)9 (−1)α
=
(−1)7 (1)4
¸
·
9 −α
=
−7 4
¸
(b) (2pts) Assuming that A is invertible, use the adjoint to determine the inverse of A.
Solution: Since A is invertible, we know that det (A) = 4 · 9 − α(7) = 36 − 7α is not equal
to zero. The adjoint construction of the inverse is
A−1 =
Therefore,
−1
A
1
adj(A).
det (A)
·
¸
1
9 −α
=
.
36 − 7α −7 4
2. Consider the points P1 = (4, 1, −3, −2) and P2 = (6, 5, −4, 8) in R4 .
(a) (2pts) Find the vector v that starts at P1 and ends at P2 .
Solution: To find the vector between them, we think of the points in space as being position
vectors. The vector that starts at P1 and ends at P2 is found by P~2 − P~1 .
~v = P~2 − P~1 = (6, 5, −4, 8) − (4, 1, −3, −2) = (2, 4, −1, 10).
(b) (3pts) Find the distance between P1 and P2 .
Solution: The distance between the two points is equivalent to the length of the vector
between the two points.
p
√
|~v | = (2)2 + (4)2 + (−1)2 + (10)2 = 121 = 11.
(c) (3pts) Find a vector equation of the line that contains both P1 and P2 .
Solution: Recall that a line ~l = (x1 , x2 , x3 , x4 ) is defined by any point on the line plus all
scalar multiples of a vector parallel to the line. Since ~v was formed from two points on the
line, it must be parallel to the line. Therefore,
~l = P~1 − t~v = (4, 1, −3, −2) + t(2, 4, −1, 10) for all t.
(d) (2pts) Does the point (0, −7, −1, −22) lie on the line? Clearly explain your answer.
Solution: If (0, −7, −1, −22) lies on the line, then there exists a value of t such that
(4, 1, −3, −2)+t(2, 4, −1, 10) = (0, −7, −1, −22). Solving for t, we see that there is a solution.
(4, 1, −3, −2) + t(2, 4, −1, 10) = (0, −7, −1, −22)
t(2, 4, −1, 10) = (0, −7, −1, −22) − (4, 1, −3, −2)
t(2, 4, −1, 10) = (−4, −8, 2, −20)
We see that equality happens when t = −2. Therefore, the point (0, −7, −1, −22) is on the
line.
3. (5pts) The Energysaver Window Company manufactures casement, awning, and double hung storm windows at the Skokie, Berwyn, and Cicero plants. The daily output for
the three plants is 600 casements, 900 awnings, and 1300 double hung windows at Skokie;
600 casements, 300 awnings, and 300 double hung windows at Berwyn; 200 casements, 400
awnings, and 600 double hung windows at Cicero. All plants are currently operating on an 8
hour work day. The Skokie plant, though large, is old and inefficient, and the company would
benefit from its closing. Is it possible to use the other plants to replace Skokie’s production
exactly? If so, how.
Solution: First, we represent the daily production at the different plants as vectors. Let
~ = (600, 900, 1300) represent Skokie, B
~ = (600, 300, 300) represent Berwyn, and C
~ =
S
~ can
(200, 400, 600) represent Cicero. In the language of vectors, this question is asking if S
~ and C
~ and, if so, does the linear combination make
be written as a linear combination of B
~ C|
~ S].
~
business sense. This is equivalent to solving the system [B

600 200
~ C|
~ S]
~ =  300 400
[B
300 600

1 0

0 1
rref ⇒
0 0

600
900 
1300

1/3
2 
0
~ is a linear combination of B
~ and C
~ and the coefficients
Since the system is consistent, S
are c1 = 1/3 and c2 = 2. Hence,
~ = 1B
~ + 2C.
~
S
3
Moreover, since the plants are currently operating on an 8 hour work day, this answer
makes business sense. To replace Skokie’s production (and not lose any of the current
production at Cicero and Berwyn), the company needs to run Berwyn and extra third of a
day and Cicero and extra 16 hours a day. Equivalently, run Berwyn 10.67 hours a day and
Cicero 24 hours a day.
4.
(a) (3pts) Define the null space of a matrix A.
Solution: The null space of a matrix A is the set of all solutions to the homogeneous system
A~x = ~0.


1
4 2 −3
(b) (5pts) Let A = −2 −9 8 1 . Find the null space of A. Write your answer as
1
3 14 −8
the span of a set of vectors.
Solution: To find the null space, we need to find the solutions to [A | ~0]. [A | ~0] in reduced
row echelon form is


1 0
50 −23
0
 0 1 −12
5
0 .
0 0
0
0
0
The coefficients that correspond to columns without leading ones are parameters. Let x3 = s
and x4 = t. Using these parameters, we find x1 = −50s + 23t and x2 = 12s − 5t. The null
space are all vectors of the form
nul(A) =
=
=
=
(−50s + 23t, 12s − 5t, s, t) for all s and t
(−50s, 12s, s, 0) + (23t, −5t, 0, t) for all s and t
s(−50, 12, 1, 0) + t(23, −5, 0, 1) for all s and t
span{(−50, 12, 1, 0), (23, −5, 0, 1)}
(c) (2pts) In what Euclidean space does the null space reside? Clearly explain your answer.
Solution: Since the vectors in the basis are 4-tuples, we know that these vectors are elements
of R4 .
(d) (2pts) What is the dimension of the null space? Clearly explain your answer.
Solution: From our work in part (b), we know that a basis for the null space is
{(−50, 12, 1, 0), (23, −5, 0, 1)}.
Since there are two vectors in the basis of the null space, the null space is a 2-dimensional
subspace of R4 .
5. Let v1 = (−3, 5, 1, −3), v2 = (4, −2, 3, −4), v3 = (2, 11, 9, −17), and v4 = (8, 4, 11, −18).
(a) (3pts) Define what it means for a set of vectors to be linearly dependent.
Solution: The vectors v1 , v2 , v3 , and v4 are linearly dependent if there exists constants c1 ,
c2 , c3 , and c4 not all zero, such that
c1 v1 + c2 v2 + c3 v3 + c4 v4 = ~0.
Equivalently, at least one vector in the set of four can be written as a linear combination
of the others.
(b) (2pts) Show that the set {v1 , v2 , v3 , v4 } is linearly dependent.
We want to show that there exists a linear combination of the set, without all zero scalars,
that forms the zero vector. Solving for the coefficients equates to solving the linear system
[v1 v2 v3 v4 |0]. Forming the augmented matrix,


−3
4
2
8
0
 5 −2
11
4
0 
.

 1
3
9
11
0 
−3 −4 −17 −18
0
The reduced row echelon form of this matrix is

1 0 0
0
 0 1 0 5/3

 0 0 1 2/3
0 0 0
0

0
0 
.
0 
0
Since the system has parametric solutions, the set is linearly dependent.
Since the linear system [v1 v2 v3 v4 |0] has a square coefficient matrix, we could have used
the alternative approach of showing that the coefficient matrix is invertible. The easiest way
would be to show that the det ([v1 v2 v3 v4 ]) 6= 0.
(c) (3pts) Find a dependency equation.
Solution: Solving for the coefficients in the above reduced row echelon form, we get c1 = 0,
5
2
c2 = − t, c3 = − t, and c4 = t. Therefore,
3
3
5
2
0 = − tv2 − tv3 + tv4 for all t.
3
3
To find a dependency equation, the easiest way is to let t = 1. This yields the dependency
equation,
5
2
0 = − v2 − v3 + v4 .
3
3
An equivalent (and easier answer) is found by viewing the above reduced row echelon
form matrix as a dependency table. The fourth column says
5
2
v4 = 0v1 + v2 + v3 .
3
3
(The same equation we found with our previous work.)
6. Let P3 be the collection of all real polynomials whose degree is less than or equal to 3.
In class, we showed that this is a vector space.
(a) (1pt) Give the general form of a cubic (third degree) polynomial.
Solution: p(x) = a0 + a1 x + a2 x2 + a3 x3 where a0 , a1 , a2 and a3 are any real numbers.
(b) (2pts) Describe the zero vector (the additive identity) of this vector space.
Solution: The zero vector is the polynomial 0(x) such that p(x) + 0(x) = p(x) and 0(x) +
p(x) = p(x). Since polynomial addition is equivalent to adding like coefficients, 0(x) must
have every coefficient be zero. Hence, 0(x) = 0 + 0x + 0x2 + 0x3 . Equivalently, 0(x) = 0.
(c) (2pts) Describe the additive inverse of a vector in the vector space.
Solution: The additive inverse of a polynomial p(x) = a0 +a1 x+a2 x2 +a3 x3 is the polynomial
−p(x) such that p(x) + (−p(x)) = 0(x). This happens when the coefficients cancel to zero
upon polynomial addition. Therefore, −p(x) = −a0 − a1 x − a2 x2 − a3 x3 .
7.
(a) (5pts) Define a subspace W of a vector space V .
Solution: A subset W of V is a subspace of V if it is closed under V ’s operations of vector
addition and scalar multiplication.
·
¸
a b
(b) (5pts) Determine if the set of all 2×2 matrices of the form
is a subspace of M2×2 .
c 0
Solution: To show that this is a subspace, one needs to show that the set is closed under
addition and scalar multiplication.
·
¸
·
¸
a1 a2
b1 b2
Let A =
and B =
be any two matrices in the set.
a3 0
b3 0
closed under addition
·
¸ ·
¸
a1 a2
b1 b2
A+B =
+
a3 0
b3 0
·
¸
a1 + b1 a2 + b2
=
a3 + b3
0
which is still a matrix of the form of the set. Therefore, the set is closed under addition.
closed under scalar multiplication
Let k be any real number. Then
·
¸
a1 a2
kA = k
a3 0
·
¸
ka1 ka2
=
ka3 0
which is still a matrix of the form of the set. Therefore, the set is closed scalar multiplication.
¸
·
a b
is a subspace of M2×2 .
Hence, the set of all 2 × 2 matrices of the form
c 0
8. (5pts) Determine if the following statement is true or false. For each part, you will receive
1 point for a correct answer, −1 point for an incorrect answer, and 0 points for no answer.
The lowest possible score on this problem is zero.
(a) It is possible to find a pair of 2-dimensional subspaces of R3 that only intersect at the
origin.
Solution: FALSE. We know that a 2-dimensional subspace is a plane. If two planes intersect, they have to intersect in a line. It is impossible to intersect at a single point.
(b) If W is a subspace of R2 , then W must contain the vector (0, 0).
Solution: TRUE. All subspaces of a vector space must contain the zero vector from the
original vector space. Since the zero vector of R2 is (0, 0), it must be in W .
(c) Two vectors u and v in a vector space V are linearly dependent if and only if they are
parallel.
Solution: TRUE. If two vectors are linearly dependent, then one has to be a scalar multiple
of the other. One vector being a scalar multiple of the other is definition of parallel vectors.
(d) The set S = {v1 , v2 , . . . , vk }, k ≥ 2 is called linearly independent if and only if at least
one of the vectors vj can be written as a linear combination of the other vectors.
Solution: FALSE. This is the definition of a linearly dependent set. (See the answer to
question 5(a).)
(e) If a set S spans a vector space V , then every vector in V can be written as a linear
combination of the vectors in S.
Solution: TRUE. A set S spans a vector space V is the same as span(S) = V . By the definition of a span, this means that every vector in V can be written as a linear combination
of the vectors in S.
(f) A vector space is called a “real” vector space if the vectors are comprised of real numbers.
Solution: FALSE. The term “real” in the phrase “a real vector space” is talking about the
set of scalars. Every scalar in a “real vector space” has to be a real number. (This is in
contrast to imaginary or complex numbers.)
(g) If W is a subspace of a vector space V , then W is a vector space.
Solution: TRUE. The definition of a subspace requires that it is a vector space.
Math 125 – Exam 3 – Version 2
November 29, 2006
60 points possible
1. Consider v = (2, 3, 7, −3).
(a) (2pts) In what Euclidean space does v reside? Clearly justify your answer.
Solution: Since the vector v is a 4-tuple of numbers, we know that it an element of R4 .
(b) (2pts) What are the standard basis vectors for the Euclidean space you determined in
part (a)?
Solution: e1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), e3 = (0, 0, 1, 0), and e4 = (0, 0, 0, 1).
(c) (2pts) Write v as a linear combination of the standard basis vectors.
Solution:
v = 2e1 + 3e2 + 7e3 − 3e4 .
2. Consider
3x + 5y − 6z = 4
5x + y + 2z = 12
2x − 2y + 4z = 8.
(a) (3pts) Clearly state why it is appropriate to use Cramer’s Rule to solve this system.
Solution: Note that the above system can be written as

   
3 5 −6
x
4
5 1
2  y  = 12
2 −2 4
z
8
Let A denote the coefficient matrix. Since the coefficient matrix is square and is invertible
(det (A) = 16), Cramer’s Rule can be used to solve the system.
(b) (7pts) Use Cramer’s Rule to solve the system. Be sure to show enough detail that
shows you are solving this problem using Cramer’s Rule.
Solution:


4 5 −6
det A1
64
2  and det A1 = 64. By Cramer’s rule, x =
Let A1 = 12 1
=
= 4.
det A
16
8 −2 4


3 4 −6
det A2
−64
Let A2 = 5 12 2  and det A2 = −64. By Cramer’s rule, y =
=
= −4.
det A
16
2 8 4


3 5 4
det A1
−32
Let A3 = 5 1 12 and det A3 = −32. By Cramer’s rule, z =
=
= −2.
det A
16
2 −2 8
The solution point is (4, −4, −2).
3. Let a1 = (3, −6, 3), a2 = (−3, 5, 6), and a3 = (3, −7, 12).
(a) (3pts) Define span{a1 , a2 , a3 }.
Solution: The span of {a1 , a2 , a3 } is the set of all linear combinations of the set of vectors.
Alternatively,
span{a1 , a2 , a3 } = {ra1 + sa2 + ta3 for all values of r, s, and t}.
(b) (4pts) Find the general form of a vector in span{a1 , a2 , a3 }.
Solution: If ~x = (x, y, z) is in span{a1 , a2 , a3 } then we can solve for the coefficients c1 , c2 ,
and c3 such that ~x = c1 a1 + c2 a2 + c3 a3 .
One method of solving this problem is using the linear system [a1 a2 a3 |~x] and requiring
the system to be consistent.


3 −3
3
x
 −6
5 −7
y 
augmented
3
6 12
z 

x
3 −3
3
R20 = R2 + 2R1

0
−1
−1
y
+
2x 
R30 = R3 − R1
0
9
9
z−x 

3 −3
3
x

 0 −1 −1
y + 2x
R30 = R3 + 9R2
0
0
0
z − x + 9(y + 2x)
This system will be consistent only when the bottom equation is satisfied. That is
0 = z − x + 9(y + 2x). Simplifying this expression, we see that any vector (x, y, z) in the
span satisfies the equation 17x + 9y + z = 0.
Alternatively, we could have used the technique we
with

3 −3
3
1 0
 −6
5 −7
0 1
3
6 12
0 0
and using the calculator, we get

1
 0
0
learned in Homework 7. Starting

0
0 
1
the reduced row echelon form

0 −2/17 5/51
0 2
1 1
0
1/17 2/17  .
1
9/17 1/17
0 0
Requiring consistency, the bottom row yields the equation
x+
1
9
y + z = 0.
17
17
This is equivalent to the previous answer.
(c) (1pt) What is the Euclidean equation form of the span{a1 , a2 , a3 }.
This is exactly what the equation we found in part (b) represented. The plane 17x +
9y + z = 0 is the span{a1 , a2 , a3 }.
4. (6pts) The Peoria Bottle Company manufactures two bottle types from mixtures of recycled white, brown, and green glass. The first bottle type is made from 74% white glass, 22%
brown glass, and 4% green glass; the second type is made from 81% white glass, 13% brown
glass, and 6% green glass. Scout troops in the North End and on the West Side have bottle
collection days, and the bottles collected in the North End are consistently in the mixture
92% white glass, 6% brown glass, and 2% green glass., while the bottles collected on the
West Side are consistently in the mixture 70% white glass, 20% brown glass, and 10% green
glass. Determine if the products can be blended from these glass sources; if so, find the blend.
Solution: We begin by representing the data as vectors. (I will choose to represent the
percentages as whole numbers.) Let B1 = (74, 22, 4) be the glass composition of the first
bottle, B2 = (81, 13, 6) be the composition of the second bottle, N = (92, 6, 2) be the
composition of the recycled glass from North End, and W = (70, 20, 10) be the composition
from West Side. The company wants to make the bottles from the recycled glass. To
determine if the bottles can be made from the recycling sources, we look at the matrix
[N W |B1 B2 ] Putting this matrix into reduced row echelon form, we get


1 0
0 1/2
 0 1
0 1/2  .
0 0
1
0
Note that only the Bottle 2 vector is consistent. Therefore, from these two sources, we can
only make Bottle 2. As a linear combination of vectors,
1
1
B2 = N + W.
2
2
The conclusion: To make the second bottle type, the company should mix equal parts of
the recycled glass from each source.
5.
(a) (3pts) Define what it means for a set of vectors to be linearly independent.
Solution: A set of vectors {v1 , v2 , . . . , vk } are linearly independent if the only linear combination c1 v1 + c2 v2 + . . . + ck vk of the set that equals the zero vector is the linear combination
where c1 = c2 = . . . = Ck = 0.
In other words, no one vector in the set is a linear combination other vectors in the set.
(b) (2pts) Clearly show that (2, 3, −5, −2), (4, −4, −11, −2), and (−4, −2, 6, 5) are linearly
independent.
Solution: We want to show that the only linear combination of the set that forms the zero
vector is when all the coefficients are zero. Solving for the coefficients equates to solving the
linear system


−2
4 −4
0
 3 −4 −2
0 

.
 −5 −11
6
0 
−2 −2
5
0
The reduced row echelon form of this matrix is

1 0 0
 0 1 0

 0 0 1
0 0 0

0
0 
.
0 
0
The only solution to this system is c1 = 0, c2 = 0, and c3 = 0. Hence, the three vectors are
linearly independent.
6.
(a) (3pts) Define the column space of a matrix A.
Solution: The column space is the span of the column vectors a~1 , a~2 , a~3 ,. . ., a~k . (Similarly,
all linear combinations of the column vectors a~1 , a~2 , a~3 ,. . ., a~k .)


1 2 −1
2
−1
9
−9 . Use a dependency table and find a basis
(b) (3pts) Let A =  5 11 −7
−2 3 −12 −12 −28
for the column space of A.
Solution: First, we create the initial dependency table using matrix A:
e~1
e~2
e~3
 a~1

1
5
−2
a~2
2
11
3
a~3
−1
−7
−12
a~4
2
9
−12
a~5 
−1
−9  .
−28
Moving the matrix into reduced row echelon form, we get:
a~1
a~2
a~4
a~1
1
 0
0
a~2
0
1
0
a~3
3
−2
0
a~4
0
0
1
a~5 
−1
−2  .
2
We see that the set {a~1 , a~2 , a~4 } form a basis for column space.
(c) (2pts) What is the dimension of the column space. Justify your answer.
Solution: Since there are three vectors in the basis of the column space of A, we know that
the dimension is three.
(d) (2pts) Explain why the columns of A span all of R3 .
Solution: Since each column vector is from the Euclidean space R3 , we know that the column
space of A is a subspace of R3 . We also know that the column space of A is a 3-dimensional
vector space by our answer in part (c). Therefore, column space of A must be equal to all
of R3 .
7.
(a) (5pts) List 5 of the 10 properties in the definition of a real vector space.
Solution: Let V be the vector space, u and v vectors in V and c and d be any real numbers.
Then any five of the following will work:
1. V is closed under vector addition
2. addition is commutative
3. addition is associative
4. V has an additive identity element
5. every element has an additive inverse
6. V is closed under scalar multiplication
7. c(u + v) = cu + cv
8. (c + d)u = cu + du
9. c(du) = (cd)u
10. 1u = u
·
¸
a b
(b) (5pts) Explain why the set of all 2 × 2 matrices of the form
together with the
c 1
usual matrix operations does not form a real vector space.
Solution: There are many reasons. The most obvious being that this set is not closed under
addition or scalar multiplication.
¸
·
¸
·
b1 b2
a1 a2
and B =
be any two matrices in the set.
Let A =
a3 1
b3 1
closed under addition
¸ ·
¸
b1 b2
a1 a2
+
A+B =
a3 1
b3 1
¸
·
a1 + b1 a2 + b2
=
a3 + b3
2
·
which is not a matrix of the form of the set. Therefore, the set is not closed under addition.
closed under scalar multiplication
Let k be any real number. Then
¸
·
a1 a2
kA = k
a3 1
¸
·
ka1 ka2
=
ka3 k
which is not a matrix of the form of the set. Therefore, the set is not closed scalar multiplication.
Another ·intuitive
¸ answer is that this set does not contain a zero vector. The only candi0 0
.
date is 0 =
0 1
·
¸ ·
¸
a1 a2
0 0
A+0 =
+
a3 1
0 1
·
¸
a1 a2
=
a3 2
6= A
Hence, there is no zero vector for this set. This shows that this is not a vector space.
There are many more answers.
8. (5pts) Determine if the following statement is true or false. For each part, you will receive
1 point for a correct answer, −1 point for an incorrect answer, and 0 points for no answer.
The lowest possible score on this problem is zero.
(a) For a nonzero scalar c, the vector cv is c times as long as v and has the same direction
as v if c > 0 and the opposite direction if c < 0.
Solution: FALSE. The phrase “the vector cv is c times as long as v” does not make any
sense if c is a negative number. Length always has to be a positive quantity. This problem
was removed from the exam. Everyone received a point for it.
(b) It is sufficient to show that a set is not a vector space by showing that just one axiom
is not satisfied.
Solution: TRUE. If any one of the 10 properties in the definition of a vector space fails to
hold for a set with given operations, then it is not a vector space.
(c) The zero vector of a vector space always forms a subspace.
Solution: TRUE. The zero vector is closed under addition (~0 + ~0 = ~0) and scalar multiplication (k~0 = ~0 for any k). Therefore it is a subspace.
(d) A set of vectors S = {v1 , v2 , . . . , vk } is called linearly dependent if the vector equation
c1 v~1 + c2 v~2 + . . . + ck v~k = 0 has only the trivial solution.
Solution: FALSE. This is the definition of a linearly independent set. (See the answer to
question 5(a).)
(e) A system A~x = ~b is consistent if and only if ~b can be written as a linear combination of
the columns of the coefficient matrix A.
Solution: TRUE. This is the exact statement of one of the versions of the Consistency theorem.
(f) Every subspace can be written as the span of a set of vectors.
Solution: TRUE. By theorem, every span of a set of vectors is a subspace and every subspace can be written as the span of a set of vectors.
(g) For a square matrix A, the adjoint matrix adj(A) exists only if A is invertible.
Solution: FALSE. The adjoint matrix always exists. It is just a matrix of cofactors of A.