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Instructor Solutions Manual for

Introduction to Linear Algebra with Applications

Jim DeFranza St. Lawrence University

Contents 1 Systems of Linear Equations and Matrices 1 Exercise Set 1.1 Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Exercise Set 1.2 Matrices and Elementary Row Operations . . . . . . . . . . . . . . . . . . . . . . . 7 Exercise Set 1.3 Matrix Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Exercise Set 1.4 The Inverse of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Exercise Set 1.5 Matrix Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Exercise Set 1.6 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Exercise Set 1.7 Elementary Matrices and LU Factorization . . . . . . . . . . . . . . . . . . . . . . 27 Exercise Set 1.8 Applications of Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . 32 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 Chapter Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 2 Linear Combinations and Linear Independence Exercise Set 2.1 Vectors in Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercise Set 2.2 Linear Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercise Set 2.3 Linear Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

42 42 46 51 55 58

3 Vector Spaces Exercise Set 3.1 Definition of a Vector Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercise Set 3.2 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercise Set 3.3 Basis and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercise Set 3.4 Coordinates and Change of Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercise Set 3.5 Application: Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

60 60 64 71 77 81 82 86

4 Linear Transformations 88 Exercise Set 4.1 Linear Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 Exercise Set 4.2 The Null Space and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 Exercise Set 4.3 Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 Exercise Set 4.4 Matrix Transformation of a Linear Transformation . . . . . . . . . . . . . . . . . . 101 Exercise Set 4.5 Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 Exercise Set 4.6 Application: Computer Graphics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 Chapter Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 5 Eigenvalues Exercise Set Exercise Set Exercise Set Exercise Set

and Eigenvectors 118 5.1 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 5.2 Diagonalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 5.3 Application: Systems of Linear Differential Equations . . . . . . . . . . . . . . . . . 128 5.4 Application: Markov Chains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 i

ii

CONTENTS Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 Chapter Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

6 Inner Product Spaces 137 Exercise Set 6.1 The Dot Product on Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 Exercise Set 6.2 Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 Exercise Set 6.3 Orthonormal Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 Exercise Set 6.4 Orthogonal Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 Exercise Set 6.5 Application: Least Squares Approximation . . . . . . . . . . . . . . . . . . . . . . . 157 Exercise Set 6.6 Diagonalization of Symmetric Matrices . . . . . . . . . . . . . . . . . . . . . . . . . 161 Exercise Set 6.7 Application: Quadratic Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 Exercise Set 6.8 Application: Singular Value Decomposition . . . . . . . . . . . . . . . . . . . . . . 166 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 Chapter Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 A Preliminaries Exercise Set A.1 Exercise Set A.2 Exercise Set A.3 Exercise Set A.4

173 Algebra of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 Techniques of Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

1.1 Systems of Linear Equations

1

Solutions to All Exercises

1

Systems of Linear Equations and Matrices

Exercise Set 1.1 In Section 1.1 of the text, Gaussian Elimination is used to solve a linear system. This procedure utilizes three operations that when applied to a linear system result in a new system that is equivalent to the original. Equivalent means that the linear systems have the same solutions. The three operations are: • Interchange two equations. • Multiply any equation by a nonzero constant. • Add a multiple of one equation to another. When used judiciously these three operations allow us to reduce a linear system to a triangular linear system, which can be solved. A linear system is consistent if there is at least one solution and is inconsistent if there are no solutions. Every linear system has either a unique solution, infinitely many solutions or no solutions. For example, the triangular linear systems       + x3 = 1 =2 =2  x1 − x2 + x3  x1 − 2 x2 + x3 2x1 −x2 + 2x3 = −3 , x2 − 2x3 = −1 , x2 − x3 = 2       x3 = 2 0 =4

have a unique solution, infinitely many solutions, and no solutions, respectively. In the second linear system, the variable x3 is a free variable, and once assigned any real number the values of x1 and x2 are determined. In this way the linear system has infinitely many solutions. If a linear system has the same form as the second system, but also has the additional equation 0 = 0, then the linear system will still have free variables. The third system is inconsistent since the last equation 0 = 4 is impossible. In some cases, the conditions on the right hand side of a linear system are not specified. Consider for example, the linear system     =a =a −x1 − x2 −x1 − x2 . 2x1 + 2x2 + x3 = b which is equivalent to x 3 = b + 2a  −−−−−−−−−−−−−−−−→    2x3 = c 0 = c − 2b − 4a This linear system is consistent only for values a, b and c such that c − 2b − 4a = 0.

Solutions to Exercises 1. Applying the given operations we obtain the equivalent triangular system    x1 − x2 − 2 x3 −x1 + 2x2 + 3x3   2 x1 − 2 x2 − 2 x3

   x1 − x2 − 2 x3 x2 + x3 E1 + E2 → E2 −−−−−−−−−−→   = −2 2x1 − 2x2 − 2x3

=3 =1

=3 =4 = −2

(−2)E1 + E3 → E3 −−−−−−−−−−−−−−→

2

Chapter 1 Systems of Linear Equations and Matrices    x1 − x2 − 2 x3 x2 + x3   2x3

=3 =4

. Using back substitution, the linear system has the unique solution

= −8 x1 = 3, x2 = 8, x3 = −4.

2. Applying the given operations we obtain the equivalent triangular system   2x1 − 2x2 − x3 x1 − 3 x2 + x3   x1 − 2 x2    x1 − 3 x2 + x3

  = −3 = −2  x1 − 3 x2 + x3 = −2 E1 ↔ E2 2x1 − 2x2 − x3 = −3 (−2)E1 + E2 → E2 −−−−−−→  −−−−−−−−−−−−−−→  =2 x1 − 2 x2 =2   = −2 = −2 x1 − 3x2 + x3 4x2 − 3x3 = 1 E2 ↔ E3 4x2 − 3x3 = 1 (−1)E1 + E3 → E3  −−−−−−−−−−−−−−→  −−−−−−→   x2 − x3 =4 x1 − 2 x2 =2     = −2  x1 − 3 x2 + x3 x1 − 3x2 + x3 = −2 . x2 − x3 = 4 (−4)E2 + E3 → E3 x2 − x3 = 4  −−−−−−−−−−−−−−→    4x2 − 3x3 = 1 x3 = −15

Using back substitution, the linear system has the unique solution x1 = −20, x2 = −11, x3 = −15. 3. Applying the given operations we obtain the equivalent triangular system  x1 + 3x 4    x + x + 4x 4 1 2  2x + x 3 + 8x4  1   x 1 + x 2 + x 3 + 6x 4

=2 =3

  x1   

+ 3x 4 + x4

=2 =1

x2 (−1)E1 + E2 → E2 = 3 −−−−−−−−−−−−−−→  2x + x3 + 8x4 = 3 1    =2 x 1 + x 2 + x 3 + 6x 4 = 2  x1 + 3x 4 =2     x2 + x4 =1 (−1)E1 + E4 → E4 (−2)E1 + E3 → E3 −−−−−−−−−−−−−−→  +x + 2x = −1 −−−−−−−−−−−−−−→ 3 4    x 1 + x 2 + x 3 + 6x 4 = 2   + 3x 4 = 2 x1 + 3x 4 = 2  x1        x2 + x4 = 1 x2 + x4 = 1 (−1)E2 + E4 → E4   − − − − − − − − − − − − − − → +x + 2x = −1 x + 2x 4 = −1 3 4 3       x 2 + x 3 + 3x 4 = 0 x 3 + 2x 4 = −1  x1 + 3x 4 = 2     x2 + x4 = 1 . (−1)E3 + E4 → E4 −−−−−−−−−−−−−−→  x + 2x 4 = −1 3    0 =0

The final triangular linear system has more variables than equations, that is, there is a free variable. As a result there are infinitely many solutions. Specifically, using back substitution, the solutions are given by x1 = 2 − 3x4 , x2 = 1 − x4 , x3 = −1 − 2x4 , x4 ∈ R. 4. Applying the given operations we obtain the equivalent triangular system   + x3  x1 x 1 + x 2 + 4x 3   2x1 + 2x 3 + x 4

  = −2 x1 = −1 (−1)E1 + E2 → E2 −−−−−−−−−−−−−−→   = −1 2x1

+ x3 x 2 + 3x 3 + 2x 3 + x 4

= −2 = 1 (−2)E1 + E3 → E3 −−−−−−−−−−−−−−→ = −1

1.1 Systems of Linear Equations    x1

3

+ x3 x 2 + 3x 3

= −2 = 1 . Using back substitution, the set of solutions is given by x1 = −2 − x3 , x2 = 1 −   x4 = 3 3x3 , x4 = 3, x3 ∈ R. 5. The second equation gives immediately that x = 0. Substituting the value x = 0 into the first equation, we have that y = − 23 . Hence, the linear system has the unique solution x = 0, y = − 23 . 6. From the second equation y = 1 and substituting this value in the first equation gives x = −4. 7. The first equation gives x = 1 and substituting this value in the second equation, we have y = 0. Hence, the linear system has the unique solution x = 1, y = 0.

8. The operation 3E2 + E1 → E1 gives 5x = −1, so x = − 15 . Substitution in equation two, then gives y = − 15 .

9. Notice that the first equation is three times the second and hence, the equations have the same solutions. Since each many solutions the linear system has infinitely many solutions with solution   has infinitely   equation 2t+4  t∈R . set S = , t 3 10. Since the first equation is −3 times the second, the equations describe the same line and hence, there are infinitely many solutions, given by x = 35 y + 13 , y ∈ R. 11. The operations E1 ↔ E3 , E1 + E2 → E2 , 3E1 + E3 → E3 to the equivalent triangular system    x − 2y + z −5y + 2z   9 z 5

and − 58 E2 + E3 → E3 , reduce the linear system = −2 = −5 . =0

The unique solution is x = 0, y = 1, z = 0. 12. Reducing the linear system gives   

  x + 3y + z = 2  to −2x + 2y − 4z = −1 reduces −−−−−−−→     −y + 3z = 1

x + 3y + z 8y − 2z −y + 3z

  =2  to = 3 reduces −−−−−−−→   =1

x + 3y + z 8y − 2z z

=2 =3 . = 12

So the unique solution is x = 0, y = 21 , z = 12 . 13. The operations E1 ↔ E2 , 2E1 + E2 → E2 , −3E1 + E3 → E3 and E2 + E3 → E3 , reduce the linear system to the equivalent triangular system   + 5z = −1  x −2y + 12z = −1 .   0 =0    The linear system has infinitely many solutions with solution set S = −1 − 5t, 6t + 12 , t  t ∈ R . 14. Reducing the linear system gives   −x + y + 4z 3x − y + 2z   2x − 2y − 8z

    = −1 −x + y + 4z = −1 −x + y + 4z to reduces to = 2 reduces 2y + 14 z = − 1 2y + 14 z = −1 . −−−−−−−→  −−−−−−−→    0 =0 =2 2x − 2y − 8z =2    There are infinitely many solutions with solution set S = −3t + 12 , −7t − 21, t | t ∈ R . 15. Adding the two equations yields 6x1 + 6x3 = 4, so that x1 = 23 − x3 . Substituting this value in the x2 = − 21 . The linear system has infinitely many solutions with solution set  gives    first equation 2 , −1  t ∈ R . , t S = −t + 3 2 = −1

16. Reducing the linear system gives

4

Chapter 1 Systems of Linear Equations and Matrices ( −2x1 + x2 3x1 − x2 + 2x3

( =2 =2 . reduces to −2x1 + x2 x 2 + 4x 3 = 8 = 1 −−−−−−−→

There are infinitely many solutions with solution set S = {(−2t + 3, −4t + 8, t) | t ∈ R} . 17. The operation 2E1 +E2 → E2 gives the equation −3x2 −3x3 −4x = −9. Hence, the linear system has two 4  free variables, x3 and x4 . The two parameter set of solutions is S = 3 − 53 t, −s − 34 t + 3, s, t  s, t ∈ R .

18. The form.  linear system is in reduced   The solution set is a two parameter family given by S = 21 s − 3t + 25, 3t − 2, s, t | s, t ∈ R .

19. The operation −2E1 + E2 → E2 gives x = b − 2a. Then y = a + 2x = a + 2(b − 2a) = 2b − 3a, so that the unique solution is x = −2a + b, y = −3a + 2b. 20. The linear system ( 2x + 3y x +y

( 2x + 3y =a reduces to y = b −−−−−−−→

=a , = a − 2b

so the solution is x = −a + 3b, y = a − 2b. 21. The linear system is equivalent to the triangular linear system    −x  

y

−z z

=b , = a + 3b = c − 7b − 2a

which has the unique solution x = 2a + 6b − c, y = a + 3b, z = −2a − 7b + c. 22. The linear system     −y − 2z  −3x + 2y + z = a x − y − z = b −E2 + E3 → E3 , E1 + 3E2 → E1 reduces to x − y − z  −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→    −z x − y − 2z = c

= a + 3b , =b = −b + c

so the solution is x = −a − 3b + c, y = −a − 5b + 2c, z = b − c. 23. Since the operation 2E1 + E2 → E2 gives the equation 0 = 2a + 2, then the linear system is consistent for a = −1. 24. Since ( −x + 3y 2x − 6y

( −x + 3y =a reduces to 0 = 3 −−−−−−−→

=a , = 3 + 2a

the linear system is consistent if a = − 32 . 25. Since the operation 2E1 + E2 → E2 gives the equation 0 = a + b, then the linear system is consistent for b = −a. 26. Since (

6x − 3y −2x + y

the linear system is consistent if b = − 13 a.

( 6x − 3y = a =a reduces to , = b −−−−−−−→ 0 = 13 a + b

1.1 Systems of Linear Equations

5

27. The linear system is equivalent to the triangular linear system    x − 2y + 4z = a 5y − 9z = −2a + b   0 = c−a−b

and hence, is consistent for all a, b, and c such that c − a − b = 0. 28. Since    x − y + 2z 2x + 4y − 3z   4x + 2y + z

  =a  x reduces to = b −−−−−−−→   =c

− y + 2z 6y − 7z 0

=a , = b − 2a = c − 2a − b

the linear system is consistent if c − 2a − b = 0. 29. The operation −2E1 + E2 → E2 gives the equivalent linear system ( x+y = −2 . (a − 2)y = 7 Hence, if a = 2, the linear system is inconsistent. 30. Since ( 2x − y ax + 3y

( 2x − y =4 =4 reduces to , = 2 −−−−−−−→ 3 + 21 a = 2 − 2a

the linear system is inconsistent if 3 + 12 a = 0, that is a = −6. Notice that if a = −6, then 2 − 2a 6= 0. 31. The operation −3E1 + E2 → E2 gives the equivalent linear system ( x−y =2 . 0 =a−6 Hence, the linear system is inconsistent for all a 6= 6. 32. Since ( 2x − y 6x − 3y

( 2x − y =a reduces to = a −−−−−−−→ 0

=a = −2a

the linear system is inconsistent for a 6= 0. 33. To find the parabola y = ax2 + bx + c that passes through the specified points we solve the linear system   c = 0.25  a + b + c = −1.75 .   a − b + c = 4.25 The unique solution is a = 1, b = −3, and c =   vertex of the parabola is the point 23 , −2 .

1 , 4

so the parabola is y = x2 − 3x +

1 4

2  = x − 32 − 2. The

34. To find the parabola y = ax2 + bx + c that passes through the specified points we solve the linear system   

c =2 9a − 3b + c = −1 .   0.25a + 0.5b + c = 0.75

6

Chapter 1 Systems of Linear Equations and Matrices

The unique solution is a = −1, b = −2 and c = 2, so the parabola is y = −x2 − 2x + 2 = −(x + 1)2 + 3. The vertex of the parabola is (−1, 3). 35. To find the parabola y = ax2 + bx + c that passes through the specified points we solve the linear system  2   (−0.5) a − (0.5)b + c = −3.25 . a+ b+c =2   (2.3)2 a + (2.3)b + c = 2.91

The unique solution is a = −1, b = −4, and c = −1, so the parabola is y = −x2 + 4x − 1 = −(x − 2)2 + 3. The vertex of the parabola is the point (2, 3). 36. To find the parabola y = ax2 + bx + c that passes through the specified points we solve the linear system   

c

a+b+c   9a + 3b + c

= −2875

= −5675 . = 5525

The unique solution is a = 2800, b = −5600, and c = −2875, so the parabola is y = 2800x2 − 5600x − 2875. b The x coordinate of the vertex is given by x = − 2a , so x = 1. The y coordinate of the vertex is y = −5675. 37. a. The point of intersection of the three lines can be b. found by solving the linear system y  5  =1 −x + y −6x + 5y = 3 .   12x + 5y = 39 25

5

x

5

x

This linear system has the unique solution (2, 3).

25

38. a. The point of intersection of the three lines can be b. found by solving the linear system  2x + y = 0    x + y = −1 . 3x + y = 1    4x + y = 2 This linear system has the unique solution (1, −2).

y 5

25

25

( x+y =2 39. a. The linear system has the unique solution x = 1 and y = 1. Notice that the two lines x−y =0 have different slopes. ( x+y =1 b. The linear system has infinitely many solutions given by the one parameter set S = 2x + 2y = 2 {(1 − t, t) | t ∈ R}. Notice that the second equation is twice the first and the equations represent the same line. ( x+y =2 c. The linear system is inconsistent. 3x + 3y = −6 40. Using the operations dE1 → E1 , bE2 → E2 , followed by (−1)E2 + E1 → E1 gives (ad − bc)x = dx1 − bx2 . 2 −cx 1 1 −bx2 . In a similar way, we have that y = axad−bc Since ad − bc is not zero, we have that x = dxad−bc . 41. a. S = {(3 − 2s − t, 2 + s − 2t, s, t) | s, t ∈ R} b. S = {(7 − 2s − 5t, s, −2 + s + 2t, t) | s, t ∈ R}

1.2 Matrices and Elementary Row Operations

7

42. a. Let x4 = s, and x5 = t, so that x3 = 2 + 2s − 3t, x2 = −1 + s + t, and x1 = −2 + 3t. b. Let x3 = s and x5 = t, so that x4 = −1 + 21s + 23 t, x2 = −2 + 12 s + 25 t, and x1 = −2 + 3t. 43. Applying kE1 → E1 , 9E2 → E2 , and −E1 + E2 → E2 gives the equivalent linear system ( 9kx + k 2 y = 9k . (9 − k 2 )y = −27 − 9k

Whether the linear system is consistent or inconsistent can now be determined by examining the second equation. a. If k = 3, the second equation becomes 0 = −54, so the linear ...