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SEVENTH EDITION

LINEAR

ALGEBRA WITH

APPLICATIONS

Instructor’s Solutions Manual

Steven J. Leon

PREFACE

This solutions manual is designed to accompany the seventh edition of Linear

Algebra with Applications by Steven J. Leon. The answers in this manual supple-

ment those given in the answer key of the textbook. In addition this manual contains

the complete solutions to all of the nonroutine exercises in the book.

At the end of each chapter of the textbook there are two chapter tests (A and

B) and a section of computer exercises to be solved using MATLAB. The questions

in each Chapter Test A are to be answered as either true or false. Although the true-

false answers are given in the Answer Section of the textbook, students are required

to explain or prove their answers. This manual includes explanations, proofs, and

counterexamples for all Chapter Test A questions. The chapter tests labelled B

contain workout problems. The answers to these problems are not given in the

Answers to Selected Exercises Section of the textbook, however, they are provided

in this manual. Complete solutions are given for all of the nonroutine Chapter Test

B exercises.

In the MATLAB exercises most of the computations are straightforward. Con-

sequently they have not been included in this solutions manual. On the other hand,

the text also includes questions related to the computations. The purpose of the

questions is to emphasize the significance of the computations. The solutions man-

ual does provide the answers to most of these questions. There are some questions

for which it is not possible to provide a single answer. For example, aome exercises

involve randomly generated matrices. In these cases the answers may depend on

the particular random matrices that were generated.

Steven J. Leon

sleon@umassd.edu

CHAPTER

SECTION 1

2. (d)

5. (a) 3x 1

• 2x 2

x 1

• 5x 2

(b) 5x 1 − 2 x 2 + x 3 = 3

2 x 1 + 3x 2 − 4 x 3 = 0

(c) 2x 1 + x 2 + 4x 3 = − 1

4 x 1 − 2 x 2 + 3x 3 = 4

5 x 1 + 2x 2 + 6x 2 = − 1

(d) 4x 1 − 3 x 2 + x 3 + 2x 4 = 4

3 x 1 + x 2 − 5 x 3 + 6x 4 = 5

x 1 + x 2 + 2x 3 + 4x 4 = 8

5 x 1 + x 2 + 3x 3 − 2 x 4 = 7

9. Given the system

−m 1

x 1

• x 2

= b 1

−m 2

x 1

• x 2

= b 2

one can eliminate the variable x 2 by subtracting the first row from the

second. One then obtains the equivalent system

−m 1

x 1

• x 2

= b 1

(m 1 − m 2 )x 1 = b 2 − b 1

1

2 CHAPTER 1

(a) If m 1 6 = m 2 , then one can solve the second equation for x 1

x 1

b 2 − b 1

m 1 − m 2

One can then plug this value of x 1

into the first equation and solve for

x 2

. Thus, if m 1

= m 2

, there will be a unique ordered pair (x 1

, x 2

) that

satisfies the two equations.

(b) If m 1

= m 2

, then the x 1

term drops out in the second equation

0 = b 2 − b 1

This is possible if and only if b 1 = b 2.

(c) If m 1

= m 2

, then the two equations represent lines in the plane with

different slopes. Two nonparallel lines intersect in a point. That point

will be the unique solution to the system. If m 1

= m 2

and b 1

= b 2

, then

both equations represent the same line and consequently every point on

that line will satisfy both equations. If m 1

= m 2

and b 1

= b 2

, then the

equations represent parallel lines. Since parallel lines do not intersect,

there is no point on both lines and hence no solution to the system.

10. The system must be consistent since (0, 0) is a solution.
11. A linear equation in 3 unknowns represents a plane in three space. The

solution set to a 3 × 3 linear system would be the set of all points that lie

on all three planes. If the planes are parallel or one plane is parallel to the

line of intersection of the other two, then the solution set will be empty. The

three equations could represent the same plane or the three planes could

all intersect in a line. In either case the solution set will contain infinitely

many points. If the three planes intersect in a point then the solution set

will contain only that point.

SECTION 2

2. (b) The system is consistent with a unique solution (4, −1).
3. (b) x 1

and x 3

are lead variables and x 2

is a free variable.

(d) x 1

and x 3

are lead variables and x 2

and x 4

are free variables.

(f) x 2

and x 3

are lead variables and x 1

is a free variable.

5. (l) The solution is (0, − 1. 5 , − 3 .5).
6. (c) The solution set consists of all ordered triples of the form (0, −α, α).
7. A homogeneous linear equation in 3 unknowns corresponds to a plane that

passes through the origin in 3-space. Two such equations would correspond

to two planes through the origin. If one equation is a multiple of the other,

then both represent the same plane through the origin and every point on

that plane will be a solution to the system. If one equation is not a multiple of

the other, then we have two distinct planes that intersect in a line through the

origin. Every point on the line of intersection will be a solution to the linear

system. So in either case the system must have infinitely many solutions.

4 CHAPTER 1

(g)

2. (d)

5. (a) 5A =

2 A + 3A =

(b) 6A =

3(2A) = 3

(c) A

T

(A

T )

T

T

= A

6. (a) A + B =

 =^ B^ +^ A

(b) 3(A + B) = 3

3 A + 3B =

(c) (A + B)

T

T

A

T

• B

T

7. (a) 3(AB) = 3

(3A)B =

Section 3 5

A(3B) =

(b) (AB)

T

T

B

T A

T

8. (a) (A + B) + C =

A + (B + C) =

(b) (AB)C =

A(BC) =

(c) A(B + C) =

AB + AC =

(d) (A + B)C =

AC + BC =

9. Let

D = (AB)C =

a 11

b 11

• a 12

b 21

a 11

b 12

• a 12

b 22

a 21

b 11

• a 22

b 21

a 21

b 12

• a 22

b 22

c 11

c 12

c 21

c 22

It follows that

d 11 = (a 11 b 11 + a 12 b 21 )c 11 + (a 11 b 12 + a 12 b 22 )c 21

= a 11 b 11 c 11 + a 12 b 21 c 11 + a 11 b 12 c 21 + a 12 b 22 c 21

d 12 = (a 11 b 11 + a 12 b 21 )c 12 + (a 11 b 12 + a 12 b 22 )c 22

= a 11 b 11 c 12 + a 12 b 21 c 12 + a 11 b 12 c 22 + a 12 b 22 c 22

d 21 = (a 21 b 11 + a 22 b 21 )c 11 + (a 21 b 12 + a 22 b 22 )c 21

= a 21

b 11

c 11

• a 22

b 21

c 11

• a 21

b 12

c 21

• a 22

b 22

c 21

d 22

= (a 21

b 11

• a 22

b 21

)c 12

• (a 21

b 12

• a 22

b 22

)c 22

= a 21

b 11

c 12

• a 22

b 21

c 12

• a 21

b 12

c 22

• a 22

b 22

c 22

If we set

E = A(BC) =

a 11 a 12

a 21 a 22

b 11 c 11 + b 12 c 21 b 11 c 12 + b 12 c 22

b 21 c 11 + b 22 c 21 b 21 c 12 + b 22 c 22

Section 3 7

16. Since

A

− 1 A = AA

− 1 = I

it follows from the definition that A

− 1 is nonsingular and its inverse is A.

17. Since

A

T (A

− 1 )

T = (A

− 1 A)

T = I

(A

− 1 )

T A

T = (AA

− 1 )

T = I

it follows that

(A

− 1 )

T = (A

T )

− 1

18. If Ax = Ay and x 6 = y, then A must be singular, for if A were nonsingular

then we could multiply by A

− 1 and get

A

− 1 Ax = A

− 1 Ay

x = y

19. For m = 1,

(A

1

)

− 1

= A

− 1

= (A

− 1

)

1

Assume the result holds in the case m = k, that is,

(A

k )

− 1 = (A

− 1 )

k

It follows that

(A

− 1 )

k+ A

k+ = A

− 1 (A

− 1 )

k A

k A = A

− 1 A = I

and

A

k+ (A

− 1 )

k+ = AA

k (A

− 1 )

k A

− 1 = AA

− 1 = I

Therefore

(A

− 1 )

k+ = (A

k+ )

− 1

and the result follows by mathematical induction.

20. (a) (A+B)

2 = (A+B)(A+B) = (A+B)A+(A+B)B = A

2 +BA+AB+B

2

In the case of real numbers ab + ba = 2ab, however, with matrices

AB + BA is generally not equal to 2AB.

(b)

(A + B)(A − B) = (A + B)(A − B)

= (A + B)A − (A + B)B

= A

2

• BA − AB − B

2

In the case of real numbers ab−ba = 0, however, with matrices AB−BA

is generally not equal to O.

21. If we replace a by A and b by the identity matrix, I, then both rules will

work, since

(A + I)

2 = A

2

• IA + AI + B

2 = A

2

• AI + AI + B

2 = A

2

• 2AI + B

2

and

(A + I)(A − I) = A

2

• IA − AI − I

2 = A

2

• A − A − I

2 = A

2 − I

2

8 CHAPTER 1

22. There are many possible choices for A and B. For example, one could choose

A =

 and^ B^ =

More generally if

A =

a b

ca cb

 B^ =

db eb

−da −ea

then AB = O for any choice of the scalars a, b, c, d, e.

23. To construct nonzero matrices A, B, C with the desired properties, first find

nonzero matrices C and D such that DC = O (see Exercise 22). Next, for

any nonzero matrix A, set B = A + D. It follows that

BC = (A + D)C = AC + DC = AC + O = AC

24. A 2 × 2 symmetric matrix is one of the form

A =

a b

b c

Thus

A

2

a

2

• b

2 ab + bc

ab + bc b

2

• c

2

If A

2 = O, then its diagonal entries must be 0.

a

2

• b

2 = 0 and b

2

• c

2 = 0

Thus a = b = c = 0 and hence A = O.

25. For most pairs of symmetric matrices A and B the product AB will not be

symmetric. For example

See Exercise 27 for a characterization of the conditions under which the

product will be symmetric.

26. (a) A

T is an n × m matrix. Since A

T has m columns and A has m rows,

the multiplication A

T A is possible. The multiplication AA

T is possible

since A has n columns and A

T has n rows.

(b) (A

T A)

T = A

T (A

T )

T = A

T A

(AA

T )

T = (A

T )

T A

T = AA

T

27. Let A and B be symmetric n × n matrices. If (AB)

T = AB then

BA = B

T A

T = (AB)

T = AB

Conversely if BA = AB then

(AB)

T = B

T A

T = BA = AB

28. If A is skew-symmetric then A

T = −A. Since the (j, j) entry of A

T is ajj

and the (j, j) entry of −A is −a jj

, it follows that is a jj

= −a jj

for each j

and hence the diagonal entries of A must all be 0.

10 CHAPTER 1

The product L = E

− 1

1

E

− 1

2

E

− 1

3

is lower triangular.

L =

7. A can be reduced to the identity matrix using three row operations

The elementary matrices corresponding to the three row operations are

E 1 =

 ,^ E 2 =

 ,^ E 3 =

1

2

So

E 3 E 2 E 1 A = I

and hence

A = E

− 1

1

E

− 1

3

E

− 1

3

and A

− 1 = E 3

E

2

E

1

8. (b)

(d)

9. (a)

10. (e)

12. (b) XA + B = C

X = (C − B)A

− 1

(d) XA + C = X

XA − XI = −C

X(A − I) = −C

X = −C(A − I)

− 1

Section 4 11

13. (a) If E is an elementary matrix of type I or type II then E is symmetric.

Thus E

T = E is an elementary matrix of the same type. If E is the

elementary matrix of type III formed by adding α times the ith row of

the identity matrix to the jth row, then E

T is the elementary matrix

of type III formed from the identity matrix by adding α times the jth

row to the ith row.

(b) In general the product of two elementary matrices will not be an ele-

mentary matrix. Generally the product of two elementary matrices will

be a matrix formed from the identity matrix by the performance of two

row operations. For example, if

E

1

and E 2

then E 1

and E 2

are elementary matrices, but

E

1

E

2

is not an elementary matrix.

14. If T = U R, then

tij =

n ∑

k=

uikrkj

Since U and R are upper triangular

u i 1

= u i 2

= · · · = u i,i− 1

r j+1,j

= r j+2,j

= · · · − r nj

If i > j, then

t ij

j ∑

k=

u ik

r kj

n ∑

k=j+

u ik

r kj

j ∑

k=

0 r kj

n ∑

k=j+

u ik

Therefore T is upper triangular.

If i = j, then

t jj

= t ij

i− 1 ∑

k=

u ik

r kj

• u jj

r jj

n ∑

k=j+

u ik

r kj

i− 1 ∑

k=

0 rkj + ujj rjj +

n ∑

k=j+

uik 0

= u jj

r jj

Section 4 13

21. (a) If the diagonal entries of D 1 are α 1 , α 2 ,... , αn and the diagonal entries

of D 2

are β 1

, β 2

,... , β n

, then D 1

D

2

will be a diagonal matrix with diag-

onal entries α 1

β 1

, α 2

β 2

,... , α n

β n

and D 2

D

1

will be a diagonal matrix

with diagonal entries β 1

α 1

, β 2

α 2

,.. ., β n

α n

. Since the two have the same

diagonal entries it follows that D 1

D

2

= D

2

D

1

(b)

AB = A(a 0 I + a 1 A + · · · + ak A

k )

= a 0 A + a 1 A

2

• · · · + ak A

k+

= (a 0 I + a 1 A + · · · + ak A

k )A

= BA

22. If A is symmetric and nonsingular, then

(A

− 1

)

T

= (A

− 1

)

T

(AA

− 1

) = ((A

− 1

)

T

A

T

)A

− 1

= A

− 1

23. If A is row equivalent to B then there exist elementary matrices E 1

, E

2

,... , E

k

such that

A = EkEk− 1 · · · E 1 B

Each of the Ei’s is invertible and E

− 1

i

is also an elementary matrix (Theorem

1.4.1). Thus

B = E

− 1

1

E

− 1

2

· · · E

− 1

k

A

and hence B is row equivalent to A.

24. (a) If A is row equivalent to B, then there exist elementary matrices E 1 , E 2 ,... , Ek

such that

A = EkEk− 1 · · · E 1 B

Since B is row equivalent to C, there exist elementary matrices H 1

, H

2

,.. ., H

j

such that

B = Hj Hj− 1 · · · H 1 C

Thus

A = EkEk− 1 · · · E 1 Hj Hj− 1 · · · H 1 C

and hence A is row equivalent to C.

(b) If A and B are nonsingular n × n matrices then A and B are row

equivalent to I. Since A is row equivalent to I and I is row equivalent

to B it follows from part (a) that A is row equivalent to B.

25. If U is any row echelon form of A then A can be reduced to U using row

operations, so A is row equivalent to U. If B is row equivalent to A then it

follows from the result in Exercise 24(a) that B is row equivalent to U.

26. If B is row equivalent to A, then there exist elementary matrices E 1 , E 2 ,.. ., Ek

such that

B = EkEk− 1 · · · E 1 A

Let M = E k

E

k− 1

· · · E

1

. The matrix M is nonsingular since each of the E i

’s

is nonsingular.

14 CHAPTER 1

Conversely suppose there exists a nonsingular matrix M such that

B = M A. Since M is nonsingular it is row equivalent to I. Thus there exist

elementary matrices E 1

, E

2

,... , E

k

such that

M = EkEk− 1 · · · E 1 I

It follows that

B = M A = EkEk− 1 · · · E 1 A

Therefore B is row equivalent to A.

27. (a) The system V c = y is given by

1 x 1 x

2

1

· · · x

n

1

1 x 2 x

2

2

· · · x

n

2

1 x n+

x

2

n+

· · · x

n

n+

c 1

c 2

c n+

y 1

y 2

y n+

Comparing the ith row of each side, we have

c 1

• c 2

x i

• · · · + c n+

x

n

i

= y i

Thus

p(x i

) = y i

i = 1, 2 ,.. ., n + 1

(b) If x 1

, x 2

,... , x n+

are distinct and V c = 0 , then we can apply part (a)

with y = 0. Thus if p(x) = c 1

• c 2

x + · · · + c n+

x

n , then

p(xi) = 0 i = 1, 2 ,.. ., n + 1

The polynomial p(x) has n + 1 roots. Since the degree of p(x) is less

than n + 1, p(x) must be the zero polynomial. Hence

c 1 = c 2 = · · · = cn+1 = 0

Since the system V c = 0 has only the trivial solution, the matrix V

must be nonsingular.

SECTION 5

2. B = A

T A =

a

T

1

a

T

2

a

T

n

(a 1

, a 2

,... , a n

a

T

1

a 1 a

T

1

a 2 · · · a

T

1

an

a

T

2

a 1 a

T

2

a 2 · · · a

T

2

an

a

T

n

a 1

a

T

n

a 2

· · · a

T

n

a n

5. (a)

 (1^2 3) =

(c) Let

A

11

3

5

4

5

4

5

3

5

A

12

A 21 = (0 0) A 22 = (1 0)

16 CHAPTER 1

(b) If we let X = U Σ, then

X = U 1 Σ 1 = (σ 1 u 1 , σ 2 u 2 ,.. ., σnun)

and it follows that

A = U ΣV

T = XV

T = σ 1 u 1 v

T

1

• σ 2 u 2 v

T

2

• · · · + σnunv

T

n

A

− 1

11

C

O A

− 1

22

A

11

A

12

O A

22

I A

− 1

11

A 12 + CA 22

O I

If

A

− 1

11

A

12

+ CA

22

= O

then

C = −A

− 1

11

A 12 A

− 1

22

Let

B =

A

− 1

11

−A

− 1

11

A 12 A

− 1

22

O A

− 1

22

Since AB = BA = I it follows that B = A

− 1 .

12. Let 0 denote the zero vector in R

n

. If A is singular then there exists a vector

x 1

= 0 such that Ax 1

= 0. If we set

x =

x 1

then

M x =

A C

O B

x 1

Ax 1 + C 0

Ox 1 + B 0

By Theorem 1.4.2, M must be singular. Similarly, if B is singular then there

exists a vector x 2 6 = 0 such that Bx 2 = 0. So if we set

x =

x 2

then x is a nonzero vector and M x is equal to the zero vector.

15. The block form of S

− 1 is given by

S

− 1

=

I −A

O I

It follows that

S

− 1 M S =

I −A

O I

AB O

B O

I A

O I

I −A

O I

AB ABA

B BA

Section 5 17

O O

B BA

16. The block multiplication of the two factors yields

I O

B I

A

11

A

12

O C

A

11

A

12

BA

11

BA

12

+ C

If we equate this matrix with the block form of A and solve for B and C we

get

B = A

21

A

− 1

11

and C = A 22

− A

21

A

− 1

11

A

12

To check that this works note that

BA

11

= A

21

A

− 1

11

A

11

= A

21

BA

12

+ C = A

21

A

− 1

11

A

12

+ A

22

− A

21

A

− 1

11

A

12

= A

22

and hence

I O

B I

A

11

A

12

O C

A

11

A

12

A

21

A

22

 =^ A

17. In order for the block multiplication to work we must have

XB = S and Y M = T

Since both B and M are nonsingular, we can satisfy these conditions by

choosing X = SB

− 1 and Y = T M

− 1 .

18. (a)

BC =

b 1

b 2

bn

(c) =

b 1

c

b 2

c

bnc

= cb

(b)

Ax = (a 1 , a 2 ,... , an)

x 1

x 2

xn

= a 1

(x 1

) + a 2

(x 2

) + · · · + a n

(x n

(c) It follows from parts (a) and (b) that

Ax = a 1 (x 1 ) + a 2 (x 2 ) + · · · + an(xn)

= x 1 a 1 + x 2 a 2 + · · · + xnan

19. If Ax = 0 for all x ∈ R

n , then

a j

= Ae j

= 0 for j = 1,... , n

and hence A must be the zero matrix.