Download Solutions - U.I.U.C. Math

Name:
Section Registered In:
Math 125 – Exam 3 – Version 1
April 24, 2006
60 total points possible
1. (5pts) Use Cramer’s Rule to solve
−3x + 4y = 30
−x − 2y = 8.
Be sure to show enough detail that shows you are
using Cramer’s Rule.
 solving this
 problem
 
Solution: This system as a matrix equation is 

Here A = 

−3
4
−3
4
−1 −2

x
y
=
30
8
.
 and det(A) = 10.
−1 −2


30
4
 and det A1 = −92. By Cramer’s rule, x = det A1 = −92 = −46 .
Let A1 = 
det A
10
5
8 −2


−3 30
 and det A2 = 6. By Cramer’s rule, y = det A2 = 6 = 3 .
Let A2 = 
det A
10
5
−1 8
The solution point is (− 46
, 3 ).
5 5
2. (1pt each) Answer the following either True or False and justify your answer.
(a) If {v1 , v2 , v3 } is a linearly independent set, then so is the set {kv1 , kv2 , kv3 } for every
nonzero scalar k
Solution: TRUE. Since {v1 , v2 , v3 } is linearly independent, the only way to write the zero
vector as a linear combination of v1 , v2 , and v3 is
0v1 + 0v2 + 0v3 = 0.
Consider writing the zero vector as a linear combination of {kv1 , kv2 , kv3 }. That is, what
c1 , c2 , and c3 satisfy
c1 kv1 + c2 kv2 + c3 kv3 = 0.
Dividing both sides of this equation by k results in
c1 v1 + c2 v2 + c3 v3 = 0.
Since the set {v1 , v2 , v3 } is linearly independent, we know that c1 = c2 = c3 = 0. Hence the
set {kv1 , kv2 , kv3 } is linearly independent.
(b) If A~x = ~b does not have any solutions, then ~b is not in the column space of A.
Solution: TRUE. By theorem, a solution to A~x = ~b exists if and only if the vector ~b can be
written as a linear combination of the columns of A. An equivalent statement of the theorem is that a solution to A~x = ~b exists if and only if the vector ~b lies in the column space of A.
(c) There are three linearly independent vectors in R2 .
Solution: FALSE. One solution is to note that any basis of R2 has only two vectors since
the dimension of R2 is 2. Therefore, any set of three vectors can not be a basis and the set
has to be linearly dependent.
Another solution would be to remember that the homogeneous system ([v~1 v~2 v~3 | ~0]) formed
from three vectors in R2 that has only two rows. A homogeneous system of less equations
than unknowns has to have parametric solutions. Therefore the set of vectors is linearly
dependent.
Another solution would to draw a picture similar to Example 1(c) in Lesson 4.5.
(d) Every nonzero subspace of Rn has a unique basis.
Solution: FALSE. Bases are not unique The number of vectors in a basis of a subspace is
unique, but the vectors chosen for the basis are not unique. If the subspace is n-dimensional,
any selection of n linearly independent vectors from the subspace will span the subspace.


A
A12 A13
 11



(e) Let A be a 3 × 3 matrix. Then the adjoint of A is A21 A22 A23 , where Aij is


A31 A32 A33
the (i, j)−cofactor.


A11 A21 A31




Solution: FALSE. The definition of the adjoint of A is adj A = A12 A22 A32 .


A13 A23 A33
3.
(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.
(b) (5pts) Show that the set S of all vectors of the form (a, a + b, b) for all real numbers
a and b is a subspace of R3 .
Solution: One way to show that this is a subspace is to show that the set of vectors (a, a+b, b)
is the span of a set of vectors.
(a, a + b, b) = (a, a, 0) + (0, b, b) for all a and b
= a(1, 1, 0) + b(0, 1, 1) for all a and b
= span {(1, 1, 0), (0, 1, 1)} by def’n of a span
Since every span is a subspace, the set S is a subspace.
Another way to show that this is a subspace is to show that the set is closed under
addition and scalar multiplication.
Let u = (a1 , a1 + b1 , b1 ) and v = (a2 , a2 + b2 , b2 ) be two different vectors in R2 .
closed under addition
u + v = (a1 + a2 , a1 + b1 + a2 + b2 , b1 + b2 )
= (a1 + a2 , a1 + a2 + b1 + b2 , b1 + b2 )
= ((a1 + a2 ), (a1 + a2 ) + (b1 + b2 ), (b1 + b2 ))
which is still a vector of the form of the set. Therefore, the set S is closed under addition.
closed under scalar multiplication
Let k be any real number. Then
ku = k(a1 , a1 + b1 , b1 )
= (ka1 , k(a1 + b1 ), kb1 )
= (ka1 , ka1 + kb1 , kb1 )
which is still a vector of the form of the set. Therefore, the set S is closed scalar multiplication.
4.
(a) (3pts) Define a basis for a vector space V .
Solution: A set of vectors is a basis for a vector space V if the set of vectors is linearly
independent and the span of the set of vectors is all of V .
(b) (5pts) Use a dependency table and find a

1 −3 4


A = 2 −6 9

2 −6 9
basis for the column space of

−2 5


−1 8 .

−1 9
Solution: First, we create the initial dependency table using matrix A:
e~1
e~2
e~3
a~
 1
1


 2

2
a~2
a~3
a~4
−3
4
−2
−6
9
−1
−6
9
−1
a~5

5


8 .

9
Moving the matrix into reduced row echelon form, we get:
a~1
a~3
a~5
a~
 1
1


 0

0
a~2
a~3
a~4
−3
0
−14
0
1
3
0
0
0
a~5

0


0 .

1
We see that the set {a~1 , a~3 , a~5 } form a basis for column space.
(c) (2pts) What is the dimension of the column space of A? Explain your answer.
Solution: Since there are three vectors in the basis, we know that the dimension of the
columns space of A is three.
5. (5pts) Let u and v be vectors in a vector space. Under what conditions will it be true
that span(u) = span(v).
Solution: We know that the span of a single vector forms a line. If span(u) = span(v), then
the spans define the same line. In other words, the line ~x = tu for all t has to be the same
line as x~0 = sv for all s. Therefore, u and v have to be scalar multiples of each other.
Another decent answer would be to discuss Example 1(e) is Lesson 4.5.
6. Let A be a 4 × 4 matrix with det A = 2.
(a) (2pts) Does A−1 exist? Explain your answer.
Solution: Since the determinant of A is nonzero, we know that A−1 exists.
(b) (3pts) Can the linear system A~x = ~b have more than one solution? Explain your
answer.
Solution: Since A is invertible, we know that the reduced row echelon form of A is the
identity matrix I. Therefore [A| ~b] has the reduced row echelon form of [I| p~] where p~ is the
unique solution point to the linear system.
7.
(a) (5pts) Define what it means for a vector to be the additive identity element of the vector
space.
Solution: The additive identity element is the vector of a vector space that satisfies property
4 of the definition of a vector space. That is, it is the vector 0 such that for any other vector
v in the vector space,
v + 0 = 0 + v = v.
(b)
(5pts) Show that it is not possible for a vector space to have two different zero
vectors. That is, is it possible to have two different vectors 0~1 and 0~2 such that these vectors
both satisfy the fourth property a vector space? Explain your reasoning.
Solution: This is Homework 6, Problem 4.
Assume that there are two different zero vectors 0~1 and 0~2 . By the definition of a zero vector,
for any ~v in the vector space ~v + 0 = ~v . For 0~1 , this amounts to
~v + 0~1 = ~v .
Let ~v be the other zero. Then
0~2 + 0~1 = 0~2 .
Similarly, for 0~2 , this amounts to
~v + 0~2 = ~v .
Let ~v be the other zero. Then
0~1 + 0~2 = 0~1 .
Since addition is commutative for the vector space, 0~2 + 0~1 = 0~1 + 0~2 . Therefore, 0~1 = 0~2 ;
showing that every vector space has exactly one zero element.
8.
(a) (4pts) What are the standard basis vectors for R4 ?
Solution: e1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), e3 = (0, 0, 1, 0), and e4 = (0, 0, 0, 1).
(b) (3pts) Write the vector (3, −5, 9, 2) as a linear combination of the standard basis
vectors.
Solution: (3, −5, 9, 2) = 3e1 − 5e2 + 9e3 + 2e4
(c) (3pts) Explain why the standard basis vectors form a basis for R4 .
Solution: The standard basis vectors are a linearly independent set. The subspace of R4
that is spanned by the standard basis vectors is four dimensional since there are four vectors
in the set. Since R4 itself is four dimensional, then the span of the standard basis vectors
must be all of R4 .
Name:
Section Registered In:
Math 125 – Exam 3 – Version 2
April 24, 2006
60 total points possible
1. (5pts) Determine a basis for the subspace 3x − 2y + 5z = 0 of R3 .
Solution: To find a basis, we need to convert the this equation in to a vector equation.
Solve the equation for x.
2
5
x = y − z.
3
3
Now let y and z be parameters. Letting y = s and z = t, we find that any vector (x, y, z)
2
5
that lies in the plane can be written as the vector equation (x, y, z) = ( s − t, s, t) for all
3
3
5
2
s and t. Separating the parameters, we find that (x, y, z) = s( , 1, 0) + t(− , 0, 1) for all s
3
3
2
5
and t. Therefore, the set {( , 1, 0), (− , 0, 1)} is a basis for the plane.
3
3
2. (1pt each) Answer the following either True or False and justify your answer.
(a) If S is a finite set of vectors in a vector space V , then span S must be closed under
vector addition and scalar multiplication.
Solution: TRUE. The span of a set of vectors is a subspace. By definition, a subspace is
closed under vector addition and scalar multiplication.
(b) If span(S1 ) = span(S2 ) then S1 = S2 .
Solution: FALSE. The spanning sets describe the same subspace but that does not mean
that the exact same set of vectors are are in each spanning set. For example, if S1 and S2 were
bases (which they don’t have to be in this question), we know that even bases are not unique.
(c) If {v1 , v2 } is a linearly dependent set of nonzero vectors, then each vector is a scalar
multiple of the other.
Solution: TRUE. By the definition of a linear dependent set, there exist constants c1 and c2
such that c1 v1 + c2 v2 = 0. This shows that either vector can be written as a scalar multiple
c2
of the other. For example, v1 = − v2 .
c1
(d) No set of two vectors can span R3 .
Solution: TRUE. Since R3 is 3-dimensional, a set of vectors that span all of R3 has to have
at least 3 vectors.
(e) If A, B, and C are square matrices, then det ABC = det A det B det C.
Solution: TRUE. Recall that we have the theorem det (AB) = det A det B To show this
equation is true, we need to apply the theorem twice.
det ABC = det (AB)C
= det (AB) det C
= det A det B det C
3.
(a) (3pts) Define what it means for a vector v to be a linear combination of a set of vectors.
Solution: A vector v is a linear combination of a set of vectors if v can be written as the
sum of scalar multiples of the set. In other words, if v is a linear combination of the set
{v1 , v2 , . . . , vk } then v = c1 v1 + c2 v2 + . . . + ck vk for some constants ci .
(b) (5pts) In R3 , determine if b = (5, 1, −1) can be written as a linear combination of
the vectors (1, 9, 1), (−1, 3, 1), and (1, 1, 1). If so, write the combination.
Solution: We are determining if there are constants c1 , c2 and c3 such that c1 (1, 9, 1) +
c2 (−1, 3, 1) + c3 (1, 1, 1) = (5, 1, −1). Forming the augmented matrix,


5
1 −1 1




 9
3 1
1 .


−1
1
1 1
The reduced row echelon form of this matrix is

1 0 0


 0 1 0

0 0 1

1


−3  .

1
Therefore, the constants are c1 = 1, c2 = −3 and c3 = 1 and b can be written as the combination b = (1, 9, 1) − 3(−1, 3, 1) + (1, 1, 1).
(c) (2pts)
Without
 doing any computations, explain if b = (5, 1, −1) lies in the column

1 −1 1




space of 9 3 1.


1 1 1


1 −1 1




Solution: The column space of 9 3 1 is the span {(1, 9, 1), (−1, 3, 1), (1, 1, 1)}. Above


1 1 1
we have shown that b is a linear combination of these vectors. Therefore, b lies in the
column space.
4.
(a) (5pts) List 4 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 four 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
(b) (5pts) Under the normal operations of addition and scalar multiplication, show that
the set of integers is not a real vector space.
Solution: The set of integers {. . . , −3, −2, −1, 0, 1, 2, 3, . . .} are not closed under scalar multiplication. Let n be any integer. Then kn does not have to be an integer. For example, if
k = π then πn is not an integer.
5. (5pts) Under what conditions will two vectors in R3 span a plane? A line? Clearly
explain your answer.
Solution: Let v1 and v2 be vectors in R3 . If the two vectors are linearly independent then
the set {v1 , v2 } form a basis for the subspace spanned by the set {v1 , v2 }. Two basis vectors
mean that the subspace is 2-dimensional. Hence, if the two vectors are linearly independent
the span of them forms a plane.
If v1 and v2 are linearly dependent, then span {v1 , v2 } = span {v1 } = span {v2 } since
v1 = kv2 for some constant k. The span of a single vector forms a line.
6. (5pts) Show that it is not possible for a vector u in a vector space to have two different
negatives. That is, for the vector u, it is not possible to have two different vectors (−u)1
and (−u)2 . Clearly justify your answer.
Solution: This is Homework 6, Problem 5.
Assume u does have two different additive inverses. By the definition of inverses, we have
the two equations
u + (−u)1 = 0 and u + (−u)2 = 0.
Therefore u + (−u)1 = u + (−u)2 . On the left-hand side of both sides of the equation, add
one of u’s negatives. (It doesn’t matter which.) Then
(−u)1 + u + (−u)1 = (−u)1 + u + (−u)2 .
[(−u)1 + u] + (−u)1 = [(−u)1 + u] + (−u)2 by associativity.
[0] + (−u)1 = [0] + (−u)2 by the properties of negatives.
(−u)1 = (−u)2 by the properties of zeros.
Therefore every element can have only one unique additive inverse.
7. Let v1 = (−2, 0, 1), v2 = (3, 2, 5), v3 = (6, −1, 1), and v4 = (7, 0, −2).
(a) (3pts) What vector space are these vectors elements of? Explain your answer.
Solution: Since every vector is an 3-tuple of numbers, the vectors are elements of R3 .
(b) (4pts) Show that the set {v1 , v2 , v3 , v4 } is linearly dependent.
Solution: One solution would be to site the theorem that says that since we have more
vectors (4) than dimensions of the the vector space (3), the four vectors have to be linearly
dependent. This is a fine answer to part (b) but won’t help us with (c).
I will show dependency using the definition. 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 the linear system [v1 v2 v3 v4 |0]. Forming the augmented matrix,


−2 3
6
7
0




 0 2 −1
0 .
0


1 5
1 −2
0
The reduced row echelon form of this

1 0


 0 1

0 0
matrix is
0 −79/29
0
3/29
1
6/29

0


0 .

0
Since the system has parametric solutions, the set is linearly dependent.
(c) (3pts) Find a dependency equation.
79
Solution: Solving for the coefficients in the above reduced row echelon form, we get c1 = t,
29
3
6
c2 = − t, c3 = − t, and c4 = t. Therefore,
29
29
3
6
79
0 = tv1 − tv2 − tv3 + tv4 for all t.
29
29
29
To find a dependency equation, the easiest way is to let t = 1. This yields the dependency
equation,
0=
79
3
6
v1 − v2 − v3 + v4 .
29
29
29
8. Let A be a 4 × 4 matrix with det A = 2.
(a) (3pts) If A is transformed into the reduced row echelon form matrix B, what is B?
Explain your answer.
Solution: Since the determinant is nonzero, we know that the matrix A is invertible. Since A
is invertible, we know that the reduced row echelon form of the matrix is the identity matrix.
(b) (2pts) Describe the set of all solutions to the homogeneous system A~x = ~0.
Solution: Since the matrix is invertible, we know that the system has a unique solution.
The only unique solution to the null space problem is the trivial solution (0, 0, 0, 0).
9. (5pts) Sink Inc. manufactures sinks from a steel alloy that is 76% iron, 16% nickel,
and 8% chromium and counter tops from an alloy that is 70% iron, 20% nickel, and 10%
chromium. They have located two supplies of scrap metal and can save substantially on
the production costs if their products can be blended from these scraps. One scrap metal
mixture contains 80% iron, 15% nickel, and 5% chromium, and the other scrap metal mixture contains 60% iron, 20% nickel, and 20% chromium. Determine if the products can be
manufactured from a blend of these scrap mixtures; if so, find the blend.
Solution: Interpreting the metal composition from the sources and the desired alloys as
vectors, we can write the above information in the following form:
S~1 = (80, 15, 5), S~2 = (60, 20, 20), A~1 = (76, 16, 8), and A~2 = (70, 20, 10).
To determine if the alloys can be made from the sources, we look at the matrix [S~1 S~2 | A~1 A~2 ]
Putting this matrix into reduced row echelon form, we get


.8 0
1 0




 0 1
.2 0  .


0 1
0 0
Note that only the Alloy 1 vector is consistent. Therefore, from these two sources, we can
only make Alloy 1.
The conclusion: To make Alloy 1 (and subsequently the sinks), use a combination of
metal that consists of 80% of the metal being from Source 1 and 20% from Source 2.