Solutions Manual to Accompany Applied Mathematics and Modeling for Chemical Engineers E-Book

Richard G. Rice

• Herausgeber: John Wiley & Sons
• Kategorie: Fachliteratur
• Sprache: Englisch

Seitenzahl: 325

Veröffentlichungsjahr: 2023

Beschreibung

This book is a Solutions Manual to accompany Applied Mathematics and Modeling for Chemical Engineers, Third Edition. There are many examples provided as homework in the original text and the solution manual provides detailed solutions of many of these problems that are in the parent book Applied Mathematics and Modeling for Chemical Engineers, Third Edition.

Leseprobe

Table of Contents

Cover

Title Page

Copyright

CHAPTER 1: Formulation of Physicochemical Problems

Problem 1.1

Problem 1.2

Problem 1.3

Problem 1.4

Problem 1.5

Problem 1.6

Problem 1.7

Problem 1.8

Problem 1.9

Problem 1.10

CHAPTER 2: Modelling with Linear Algebra and Matrices

Problem 2.1

Problem 2.2

Problem 2.3

Problem 2.4

Problem 2.5

Problem 2.6

Problem 2.7

Problem 2.8

Problem 2.9

Problem 2.10

Problem 2.11

Problem 2.12

Problem 2.13

Problem 2.14

Problem 2.15

Problem 2.16

CHAPTER 3: Solution Techniques for Models Yielding Ordinary Differential Equations

Problem 3.1

Problem 3.2

Problem 3.3

Problem 3.4

Problem 3.5

Problem 3.6

Problem 3.7

Problem 3.8

Problem 3.9

Problem 3.10

Problem 3.11

Problem 3.12

Problem 3.13

Problem 3.14

Problem 3.15

Problem 3.16

Problem 3.17

Problem 3.18

Problem 3.19

Problem 3.20

Problem 3.21

Problem 3.22

Problem 3.23

Problem 3.24

Problem 3.25

Problem 3.26

Problem 3.27

Problem 3.28

CHAPTER 4: Series Solution Methods and Special Functions

Problem 4.1

Problem 4.2

Problem 4.3

Problem 4.4

Problem 4.5

Problem 4.6

Problem 4.7

Problem 4.8

Problem 4.9

Problem 4.10

Problem 4.11

Problem 4.12

Problem 4.13

CHAPTER 5: Integral Functions

Problem 5.1

Problem 5.2

Problem 5.3

Problem 5.4

Problem 5.5

Problem 5.6

Problem 5.7

Problem 5.8

Problem 5.9

Problem 5.10

CHAPTER 6: Staged‐Process Models: The Calculus of Finite Differences

Problem 6.1

Problem 6.2

Problem 6.3

Problem 6.4

Problem 6.5

Problem 6.6

Problem 6.7

CHAPTER 7: Probability and Statistical Modelling

Problem 7.1

Problem 7.2

Problem 7.3

Problem 7.4

Problem 7.5

Problem 7.6

Problem 7.7

Problem 7.8

Problem 7.9

Problem 7.10

Problem 7.11

Problem 7.12

Problem 7.13

Problem 7.14

Problem 7.15

Problem 7.16

CHAPTER 8: Approximate Solution Methods for ODE: Perturbation Methods

Problem 8.1

Problem 8.2

Problem 8.3

Problem 8.4

Problem 8.5

Problem 8.6

Problem 8.7

Problem 8.8

Problem 8.9

Problem 8.10

Problem 8.11

Problem 8.12

Problem 8.13

Problem 8.14

CHAPTER 9: Numerical Solution Methods (Initial Value Problems)

Problem 9.1

Problem 9.2

Problem 9.3

Problem 9.4

Problem 9.5

Problem 9.6

Problem 9.7

Problems 9.8 and 9.9:

Problem 9.10

Problem 9.11

Problem 9.12

CHAPTER 10: Approximate Methods for Boundary Value Problems: Weighted Residuals

Problem 10.1

Problem 10.2

Problem 10.3

Problem 10.4

Problem 10.5

Problem 10.6

Problem 10.7

Problem 10.8

Problem 10.9

Problem 10.10

Problem 10.11

Problem 10.12

Problem 10.13

Problem 10.14

Problem 10.15

Problem 10.16

Problem 10.17

Problem 10.18

Problem 10.19

Problem 10.20

CHAPTER 11: Introduction to Complex Variables and Laplace Transforms

Problem 11.1

Problem 11.2

Problem 11.3

Problem 11.4

Problem 11.5

Problem 11.6

Problem 11.7

Problem 11.8

Problem 11.9

Problem 11.10

Problem 11.11

Problem 11.12

Problem 11.13

Problem 11.14

Problem 11.15

Problem 11.16

CHAPTER 12: Solution Techniques for Models Producing PDEs

Problem 12.1

Problem 12.2

Problem 12.3

Problem 12.4 (Dialysis of Blood)

Problem 12.5

Problem 12.6

Problem 12.7

Problem 12.8

Problem 12.9

Problem 12.10

Problem 12.11

Problem 12.12

Problem 12.13

Problem 12.14

Problem 12.15

Problem 12.16

Problem 12.17

Problem 12.18

Problem 12.19

Problem 12.20

Problem 12.21

Problem 12.22

Problem 12.23

Problem 12.24

Problem 12.25

CHAPTER 13: Transform Methods for Linear PDEs

Problem 13.1

Problem 13.2

Problem 13.3

Problem 13.4

Problem 13.5

Problem 13.6

Problem 13.7

Problem 13.8

CHAPTER 14: Approximate and Numerical Solution Methods for PDEs

Problem 14.1

Problem 14.2

Problem 14.3

Problem 14.4

Problem 14.5

Problem 14.6

Problem 14.7

Problem 14.8

Problem 14.9

Problem 14.10

Problem 14.11

Problem 14.12

Problem 14.13

Problem 14.14

Problem 14.15

Problem 14.16

End User License Agreement

List of Tables

Chapter 7

Table 7.11.1 Approximate 95% confidence interval on the coefficient of the L...

Table 7.13.1 ANOVA table for fit of a line to the data

Table 7.13.2 ANOVA table for the fit of a parabola to the data

List of Illustrations

Chapter 7

Figure 7.11.1 Summaries of linear‐model fits of data for the different appro...

Figure 7.11.2 Comparison of fits of the different approaches to parameter es...

Figure 7.12.1 Fit of the model to data in both the linear form and the origi...

Figure 7.13.1 Fit of a linear and parabolic calibration curve to the given d...

Figure 7.14.1 Comparison of the fit of the activated‐rate model in to the da...

Guide

Cover

Table of Contents

Title Page

Copyright

Begin Reading

End User License Agreement

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Solutions Manual to Accompany

Applied Mathematics and Modeling for Chemical Engineers

Third Edition

RICHARD G. RICE

Emeritus Professor (LSU)

Punta Gorda, Florida, USA

DUONG D. DO

Emeritus Professor (University of Queensland)

Middle Park, Queensland, Australia

JAMES E.MANEVAL

Professor, Department of Chemical Engineering Bucknell University

Lewisburg, Pennsylvania, USA

This edition first published 2023.

© 2023 John Wiley & Sons, Inc.

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, except as permitted by law. Advice on how to obtain permission to reuse material from this title is available at http://www.wiley.com/go/permissions.

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Library of Congress Cataloging‐in‐Publication Data applied for

[PB ISBN: 9781119833888]

Cover Design: Wiley

CHAPTER 1Formulation of Physicochemical Problems

Problem 1.1

Length Required for Cooling Coil

Model 1 – Plug Flow, in section 1.2 applies:

The heat transfer coefficient is obtained from the correlation:

(1.1.1)

where

(1.1.2)

Substitution in Eq. 1.17 (in the text), with T(z) = TL, gives:

(1.1.3)

Solving the above equation for L, we get

(1.1.4)

Problem 1.2

Cooling of Fluids in Tube Flow: Locally Varying h

Model 1 (Section 1.2) applies up to equation 1.8:

(1.2.1)

Substitution of h for gives:

(1.2.2)

Lumping parameters and changing the dependent variable, we get:

(1.2.3)

where:

(1.2.4)

Separation of variables and integration using the boundary condition at z = 0

(1.2.5)

yields:

(1.2.6)

Problem 1.3

Dissolution of Benzoic Acid

Part (a):

Carrying out material balance around the shell at x with a thickness of Δx gives:

(1.3.1)

where

(1.3.2)

Dividing Eq. (1.3.1) by πDΔx gives:

(1.3.3)

Taking the limit as Δx → 0, we get:

(1.3.4)

Part (b):

Defining θ = C − C* and substituting into Eq. (1.3.4), we obtain:

(1.3.5)

Separating the variables and integrating gives:

(1.3.6)

Part (c):

Using the boundary condition at x = 0, θ = −C*, gives

Thus, the final solution for C(x) is:

(1.3.7)

Part (d):

The concentration of benzoic acid leaving the tube at x = L is:

(1.3.8)

The total flow rate through the tube is:

(1.3.9a)

The rate of benzoic acid leaving the tube is:

(1.3.9b)

The weight change of the tube is:

(1.3.9c)

Substituting QT, QB and C(L) in the expression for ΔW, we get:

(1.3.10)

Part (e):

Rearranging Eq. (1.3.10), we get:

(1.3.11)

Approximation of the exponential term:

(1.3.12)

Substitution the above equation into Eq. (1.3.11), rearrangement and simplification gives:

(1.3.13)

Part (f):

Assumptions of the model are:

plug flow,

only resistance to dissolution of benzoic acid is due to liquid phase film resistance,

no axial or radial diffusion,

perfect radial mixing, and

constant wall diameter.

The maximum error can be estimated by the following equation (absolute value assures that all errors are additives, hence maximum errors):

(1.3.14)

where:

(1.3.15a)

(1.3.15b)

(1.3.15c)

Substitute the values of the partial derivatives using the previous equations and the errors, εΔW, εΔt and εD, from the experimental data, and simplifying:

(1.3.16)

Problem 1.4

Lumped Thermal Model for Thermocouple

Heat balance equation is

(1.4.1)

where

(1.4.2a)

(1.4.2b)

Substituting Eqs. (1.4.2) into Eq. (1.4.1), we get:

(1.4.3)

Part (b):

Defining θ = Tf − T and substituting into Eq. (1.4.3), we have:

(1.4.4)

where:

(1.4.5)

Part (c):

Separating variables and integrating Eq. (1.4.4) with the boundary condition T(0) = T0, we get:

(1.4.6)

Multiplying both sides by −1, adding 1 to each side, and rearranging, we obtain:

(1.4.7)

Evaluating for t = τ, (T0 − T)(T0 − Tf) = 0.63, therefore:

(1.4.8)

Problem 1.5

Distributed Thermal Model for Thermocouple

Heat Balance around the shell at r with a thickness of Δr is:

(1.5.1)

where:

(1.5.2a)

(1.5.2b)

Substituting Q, dividing throughout by 4πΔr and taking the limits as Δr → 0, we get:

(1.5.3)

Substituting qr from Eq. (1.5.2a) into the above equation, and rearranging, we get:

(1.5.4)

Part (b):

Heat balance at the surface of the sphere:

Heat Flux within Sphere by Conduction = Heat Flux to fluid by convection

Mathematically this continuity of heat flux is:

(1.5.5)

where:

(1.5.6a)

(1.5.6b)

Substituting qcond and qconv into Eq. (1.5.5), we get the boundary condition at r = R:

(1.5.7)

Problem 1.6

Modelling of Piston with Restraining Spring

Force Balance is:

(1.6.1)

Total Force (FT) = Pressure Force (FP) – Spring Force (FS) – Friction Force (Fμ)

where:

(1.6.2)

Substitution of Eq. (1.6.2) into the force balance equation (1.6.1) gives:

(1.6.3)

Part (b):

Dividing by K and defining new variables τ, ζ and f(t), Eq. (1.6.3) becomes:

(1.6.4)

where:

Time constant:

Damping coefficient:

Effective Force:

f

(

t

) =

AP

(

t

)/

K

Problem 1.7

Mass Transfer in Bubble Column

Part (a):

Steady state Material Balance for Dissolved Oxygen (Liquid Phase) is:

(1.7.1)

Dividing the above equation by AΔx and taking the limits as Δx → 0, we get:

(1.7.2)

Substituting J = − DedC/dx into the mass balance equation, we obtain:

(1.7.3)

Part (b):

Defining the following new variables:

(1.7.4)

the derivatives written in terms of these variables are:

(1.7.5a)

(1.7.5b)

Substituting Eqs. (1.7.5) into the dissolved oxygen equation (Eq. 1.7.3), we get:

(1.7.6)

Lumping the parameters in the second order term:

(1.7.7a)

where

(1.7.7b)

Part (c):

Material Balance at the Entrance of the Column gives (see the above figure):

(1.7.8)

Dividing the above equation by AΔx, taking the limit as Δx → 0 and substituting the diffusive flux J = −DedC/dx, we finally get:

(1.7.9)

The second boundary condition, dC/dx = 0 @ x = L, indicates that at the exit of the column there is no net flux of the dissolved oxygen, that is, at the end of the column the concentration of dissolved oxygen is constant with respect to axial position, because the liquid phase ends there (the top liquid interface).

Problem 1.8

Dissolution and Reaction of Gas in Liquids

The reaction rate and the diffusion flux equations are:

(1.8.1)

(1.8.2)

Material Balance for Dissolved Oxygen around the shell at z with a thickness of Δz is:

(1.8.3)

Dividing the above equation by AΔz, taking the limit as Δz → 0, and substituting Fick's Law of diffusion (Eq. 1.8.2) and the rate expression (Eq. 1.8.1), we get:

(1.8.4)

Problem 1.9

Modelling of a Catalytic Chemical Reactor

Part (a):

Material Balance for Toxic Gas:

(1.9.1)

This equation is written mathematically as follows:

(1.9.2)

Dividing the above equation by AΔz and taking the limit as z → 0, we get:

(1.9.3)

Part (b):

Separating the variables in Eq. (1.9.3) and integrating the result with the following boundary condition:

(1.9.4)

we get:

(1.9.5)

Part (c):

From Eq. (1.9.5), we derive the concentration at the exit of the reactor:

(1.9.6)

Part (d):

Multiplying and dividing the argument of the exponential term by the cross‐sectional area, A, we get:

(1.9.7)

However:

(1.9.8a)

(1.9.8b)

The exit concentration (Eq. (1.9.7)) becomes:

(1.9.9)

Part (e):

For 95 % conversion:

(1.9.10)

Simplifying and rearranging the above equation, we get:

(1.9.11)

Solving for the mass of catalyst, we finally obtain:

(1.9.12)

Part (f):

Accounting for axial diffusion, Eq. (1.9.2) can be rewritten as follows:

(1.9.13)

Dividing the above equation by AΔz and taking the limit as Δz → 0, we get:

(1.9.14)

Substituting the equation for the axial dispersion, J = −DdC/dz, the above equation becomes:

(1.9.15)

Part (g):

Defining the following variables and parameters:

(1.9.16)

The derivatives can be written in terms of the new variables as follows:

(1.9.17)

Substituting dC/dz and d2C/dz2 in the material balance equation (Eq. 1.9.15):

(1.9.18)

Multiplying by L/C0u0 and using W = (1 − ε)ρPLA and F = u0A, the above equation becomes:

(1.9.19)

From the definition of Pe and N, the final equation is:

(1.9.20)

For the boundary conditions:

(1.9.21)

which are written in terms of new variables as:

(1.9.22)

Therefore, for the two models:

(1.9.23)

(1.9.24)

It is seen that Model 2 reduces to Model 1 when Pe ≫ 1, as the second order term becomes negligible.

Part (h):

Material Balance inside the catalyst gives the following equation:

(1.9.25)

Dividing the above equation by 4πΔr, we get:

(1.9.26)

Taking the limit as Δr → 0 and substituting Fick's Law of diffusion, J = − DedCP/dr, we obtain:

(1.9.27)

Part (i):

The boundary conditions state that at the surface of the particle, (r= R), the concentration of the toxic gas in the catalyst is the same as that in the flowing gas. In the interior of the catalyst particle the concentration profile is a continuous function of r.

Material Balance Excluding Catalyst Volume gives:

(1.9.28)

Mathematically, the above equation can be written as:

(1.9.29)

where we note that J(R) is a negative flux and:

(1.9.30)

Dividing Eq. (1.9.29) by AΔz, taking the limit as Δz → 0 and substituting the expression for a and J(R) = −DedCp/dr|r = R, we obtain:

(1.9.31)

Problem 1.10

Dimensional Analysis of a Correlation

Part (a):

Solve

2 = 3x + y

1 = 3x

∴

x = 1/3 and y = 4/3

Given c = 0.35, show that

Take ε = void fraction

and given δ = d (eddy scale)

(1.10.1)

CHAPTER 2Modelling with Linear Algebra and Matrices

Problem 2.1

Properties of matrices

Part (a):

ATA is the product of an (M × N) matrix and an (N × M) matrix. The inner dimensions match (both are N) and so the product is defined. The result is an (M × M) matrix. By similar reasoning, the product AAT is defined (inner dimensions are both M). The result is an (N × N) matrix.

Part (b):

Using the results of Part a, the sum B = ATA + AAT is not defined since the dimensions of ATA and AAT are not the same. If A were and N × N matrix, all products and sums would be defined.

Part (c):

Both AB and BA are assumed to be defined. Expanding the square of their difference,

(2.1.1)

which is clearly not A2 − 2AB + B2 due to the non‐commutative nature of matrix multiplication. If in fact AB = −BA, the expression would be correct.

Part (d):

From the definition of the inverse of a (square) matrix, AA−1 = I. Taking the determinant of this definition gives

(2.1.2)

Problem 2.2

Properties of matrices

All matrices in this problem are (N × N). The matrices C and D are diagonal.

Part (a):

The product B = AC is computed as

(2.2.1)

Only the elements on the diagonal of C are non‐zero and so the sum reduces to a single expression for each i (each row) leading to

That is, the ith element of the diagonal of C multiplies the elements in column i of A.

To see this visually, consider the 3‐by‐3 case:

The same steps can be used for the product CA to give

(2.2.2)

which is the ith row of A multiplied by the ith element of the diagonal of C. In visual form,

Part (b):

Using the logic presented in Part (a), the product B = CD is computed as

(2.2.3)

But the elements of D are non‐zero only when i = j and the result is a diagonal matrix with elements bii = cidi. The product DC can be developed similarly leading to the result bii = dici which is the same as the product CD because scalar multiplication is commutative. Hence CD = DC and so matrix multiplication for square diagonal matrices is commutative.

Visually, we have

Problem 2.3

Cramer's Rule

Using Cramer's Rule, the elements of the solution vector for the problem Ax = b are computed according to

(2.3.1)

where Di is the determinant computed from the matrix A whose the ith column has been replaced by the vector b.

Parts (a) and (b):

All details will be laid out for the first problem. The second and third problems will be done in brief.

Problem

Matrix elements are

For this problem, det(A) = 4 − 1 = 3. Compute the elements of x one at a time from Eq. (2.3.1).

Check the result:

Problem

det(A) = 3(12 − 15) − (−4)(8 + 15) + 2(−10 − 15) = 33. Compute the elements of x from Eq. (2.3.1).

Check the result

Problem

det(A) = 3(−4 + 3) − 4(−4 − 3) + (−2)(2 + 2) = 17. Compute the elements of x from Eq. (2.3.1).

Check the result

Part (c):

To compute the inverse, set up the problem as AX = I and then solve the set of N equations (for an (N × N) system) for the columns of X, as

where ei (the unit vectors) are the columns of the identity matrix. Then use Eq. (2.3.1) to compute the components of the ith column of the inverse. Details will be given for the first problem and summaries for the balance of the problems.

Hence, we have the inverse of A

The calculations for the two 3‐by‐3 problems can be summarized by providing a listing of the i‐jth element of the inverse computed, as above, according to

where Di, j is the determinant of A with column i replaced, by ej. For example, in the second problem, D2, 3 is computed by replacing the second column (i = 2) with e3 = [ 0 0 1 ]T (j = 3):

The inserted unit vector is in bold.

The results for the remaining problems are

Accuracy can be verified by multiplying A by the computed inverse.

Problem 2.4

Matrix algebra

All matrices are square with N = 2. Hence all sums and products are defined, and the results are also square matrices with N = 2. Details are provided for at least one of the operations.

Part (a):

where is the 2‐by‐2 identity matrix

(2.4.1)

(2.4.2)

Part (b):

The matrices A and C do not commute because, comparing Eqs. (2.4.1) and (2.4.2), R = AC ≠ CA = T.

Part (c):

The matrix C is singular because its determinant is zero. All other matrices here are non‐singular.

Part (d):

Since these matrices are 2‐by‐2, the formula in Eq. (2.60) (Chapter 2, Section 2.5.2) is easiest to apply.

(2.4.3)

Part (e):

Since A is non‐singular, the solution to AZ = B is Z = A−1B. From Eq. (2.4.3),

Part (f):

Since D is non‐singular, there is only one solution to the problem Dy = b. Using the computed inverse in Eq. (2.4.3), the solution is

Problem 2.5

Solving systems of equations

Problem

:

Solution Method 1: Use the analytical expression for the inverse of A and the find the solution by multiplication.

Solution Method 2: Use Gauss Elimination on the augmented matrix [ A |  b ].

Step 1: Multiply row 1 by

−0.5

and add to row 2

Step 2: Use back‐substitution

The balance of the problems are best solved using Gauss Elimination on the augmented matrix [ A |  b ] since solvability aspects will appear in the solution process.

Problem

:

Multiply row 1 by 1/3

Eliminate entries down the first column. Add row 1 to row 2; multiply row 1 by 7 and add to row 3.

Eliminating down the balance of column 2 by dividing row 2 by 13/3 and then multiplying row 2 by −13/3 and adding to row 3 leads to the final row‐reduced form of the system

(2.5.1)

This matrix is singular (i.e., rank(A) = 2 when it needs to be 3 to be non‐singular) since there are two rows with non‐zero elements. The system is consistent since rank(A) = rank ([A| b]) = 2. Hence there is an infinite number of solutions possible with N − r = 3 − 2 = 1 free parameter. Choose x3 = α with α an arbitrary parameter. From the second equation in Eq. (2.5.1) we have

Using the values of x2 and x3 in the first equation then gives

and so all vectors of the form

are solutions to the problem.

Problem

:

Multiply row 1 of the augmented matrix by 0.25 to start:

Eliminate entries down the first column. Multiply row 1 by −3 and add to row 2; subtract row 1 by 7 from row 3.

The coefficients in the third row are the negatives of those in the second row and so we will find that the matrix is singular. Adding row 2 to row 3:

(2.5.2)

As observed, the matrix is singular (rank(A) = 2) since there are only 2 rows with non‐zero elements in the first 3 columns of the augmented matrix. The zeros in the third row do not extend to the third element and so that rank([A| b]) = 3. We therefore have the finding 2 = rank (A) < rank ([A| b]) = 3 so that the system of equations is inconsistent, and no solution is possible.

Problem:

Divide row 1 of the augmented matrix by 3 to start:

Eliminate entries down the first column. Multiply row 1 by −2 and add to row 2; Multiply row 1 by −5 and add to row 3:

Multiply row 2 by 3/17 to place a 1 on the diagonal. Then multiply row 2 by (−5/3) to eliminate the remaining row in the second column.

The matrix is non‐singular since the rank is 3 (all of the rows of the augmented matrix have non‐zero elements in the first 3 columns). Use back‐substitution to solve for the elements of z. From the third row, z3 = 3. Using this result in the second row gives z2 = (13/17)(3) − 5/17 = 2 and from the first row, z1 = (4/3)(2) − (2/3)(3) + (1/3) = 1. Hence, the solution vector is

Problem 2.6

Properties of eigenvalues and eigenvectors

The starting point is the eigensystem of A which is found from the solutions to Ax = λx.

Part (a):

A is non‐singular and so it has an inverse and none of its eigenvalues are zero. Multiply the eigenvalue problem by A−1 to find

(2.6.1)

Divide through by λ to give

(2.6.2)

Equation (2.6.2) is the definition of the eigenvalue problem for the inverse of the matrix A and clearly the eigenvalues of A−1 are the reciprocals of the eigenvalues of A.

Part (b):

From the form of the problem developed in Part (a), it is clear that given the eigensystem of A{λi, xi} for i = 1, 2, …, N, the eigensystem of A−1 is for i = 1, 2, …, N. The eigenvalues of the two matrices are reciprocals but the eigenvectors are the same.

Part (c):

Multiply the eigenvalue problem for A by the matrix A to find

(2.6.3)

demonstrating the eigenvalues of A2 are the squares of the eigenvalues of A. Repeated multiplications of the eigenvalue problem by A (always from the left) then leads to

(2.6.4)

for n = 1, 2, … and so the eigenvalues of An are λn.

Part (d):

The eigenvalues of AT are computed from det(AT − λI) = 0. Since (AT − λI) = (A − λI)T and because det(B) = det (BT), the same characteristic polynomial will result for both A and AT so that the eigenvalues of A and AT are the same. It is not true that the eigenvectors ofAandATare the same so, generally, the eigensystems of A and ATare not the same.

Problem 2.7

Computing eigensystems

General approach: Expand the expression det(A − λI) = 0 to find the characteristic polynomial for the matrix. The eigenvalues are the roots of this polynomial. For each eigenvalue, λi, solve the problem (A − λi)xi = 0 to find the associated eigenvector, xi. In the solutions here, the arbitrary constant is left in place rather than be determined by normalization or by any other requirement on an eigenvector.

Problem

:

Characteristic polynomial equation is λ2 − tr(A)λ + det (A) = λ2 + 7λ + 12 = (λ + 4)(λ + 3) = 0 and so the eigenvalues are λi = −4, −3.

λ1 = −3:

Solve for x1 in . Let x1, 1 = a and so the associated eigenvector is .

λ2 = −4:

Solve for x2 in . Let x2, 1 = b and so the associated eigenvector is .

Problem

:

Characteristic polynomial equation is λ2 − 5λ = λ(λ − 5) = 0 and so the eigenvalues are λi = 0, 5.

λ1 = 0:

Solve for x1 in . Let x1, 1 = a and so the associated eigenvector is . Note that a zero eigenvalue leads to a non‐zero eigenvector.

λ2 = 5:

Solve for x2 in . Let x2, 1 = b and so the associated eigenvector is .

Problem

:

The determinant was computed by expanding in minors over the second row to take advantage of the number of zeros in that row. Note that the matrix is clearly singular (row 3 is a multiple of row 1) and we should expect a zero eigenvalue.

Characteristic polynomial is λ2(1 + λ) = 0 and so the eigenvalues are λi = 0, 0, − 1. Note that 0 is a repeated root. It will not be certain that we will be able to find two eigenvectors associated with this eigenvalue.

λ1, λ2 = 0:

Solve for x1 in . Add row 1 to row 3 to obtain the Gauss‐eliminated form of the problem as

Any triple of numbers works for the third equation so use that fact to set x1, 3 = a. The second equation sets x1, 2 = 0 and the first equation then reads x1, 1 + x1, 3 = 0 so x1, 1 = − a. The associated eigenvector is then .

We see that there is no way (at least as far as the presentation in the text is concerned) to find a second eigenvector associated with λ = 0 and so we do not have a simple eigensystem for this problem.

λ3 = −1:

Solve for x3 in . The second equation is satisfied by any triple of numbers and so use that fact to set x3, 2 = b. From the third equation we have −x3, 1 − 2x3, 2 = 0 and so x3, 1 = −2b. Using the last two results in the first equation then gives 2x3, 1 + 2x3, 2 + x3, 3 = 0 so that x3, 3 = 2b. The associated eigenvector is then .

Problem

:

The computation of the determinant takes advantage of the fact that the determinant of a triangular matrix is just the product of the diagonal elements of that matrix (see Problem 2.8).

Characteristic polynomial is (1 − λ)2(2 − λ) = 0 and so the eigenvalues are λi = 1, 1, 2. Note that 1 is a repeated root. It will not be certain that we will be able to find two eigenvectors associated with this eigenvalue.

λ1, λ2 = 1:

Solve for x1 in . The second and third equations give, respectively, x1, 2 = 0 and x1, 3 = 0. With these results, the first equation is satisfied by any value of x1, 1 and so the associated eigenvector is . A second eigenvector is not available given the techniques in this text.

λ3 = 2:

Solve for x3 in . The third equation is satisfied by any triple and so use that fact to set x3, 3 = a. From the second equation, we have −x3, 2 + x3, 3 = 0 and so x3, 2 = a. Finally, the first equation gives −x3, 1 − x3, 2 = 0 and so x3, 1 = −a