Solutions Manual to accompany Fundamentals of Matrix Analysis with Applications

 
 
Wiley (Verlag)
  • erschienen am 20. Mai 2016
  • |
  • 268 Seiten
 
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978-1-118-99634-8 (ISBN)
 
Solutions Manual to accompany Fundamentals of Matrix Analysis with Applications - an accessible and clear introduction to linear algebra with a focus on matrices and engineering applications
1. Auflage
  • Englisch
  • New York
  • |
  • USA
  • Für Beruf und Forschung
  • 2,69 MB
978-1-118-99634-8 (9781118996348)
weitere Ausgaben werden ermittelt
1 Systems of Linear Algebraic Equations 1
1.1 Linear Algebraic Equations 1
1.2 Matrix Representation of Linear Systems and the Gauss-Jordan Algorithm 12
1.3 The Complete Gauss Elimination Algorithm 17
1.4 Echelon Form and Rank 26
1.5 Computational Considerations 34
2 Matrix Algebra 39
2.1 Matrix Multiplication 39
2.2 Some Useful Applications of Matrix Operators 48
2.3 The Inverse and the Transpose 54
2.4 Determinants 61
2.5 Three Important Determinant Rules 68
PART I REVIEW PROBLEMS FOR PART I 79
3 Vector Spaces 89
3.1 General Spaces, Subspaces, and Spans 89
3.2 Linear Dependence 93
3.3 Bases, Dimension, and Rank 97
4 Orthogonality 105
4.1 Orthogonal Vectors and the Gram-Schmidt Algorithm 105
4.2 Orthogonal Matrices 115
4.3 Least Squares 123
4.4 Function Spaces 133
PART II REVIEW PROBLEMS FOR PART II 140
5 Eigenvectors and Eigenvalues 144
5.1 Eigenvector Basics 144
5.2 Calculating Eigenvalues and Eigenvectors 155
5.3 Symmetric and Hermitian Matrices 168
6 Similarity 181
6.1 Similarity Transformations and Diagonalizability 181
6.2 Principal Axes and Normal Modes 189
6.3 Schur Decomposition and Its Implications 198
6.4 The Singular Value Decomposition 212
6.5 The Power Method and the QR Algorithm 217
7 Linear Systems of Differential Equations 221
7.1 First-Order Linear Systems 221
7.2 The Matrix Exponential Function 230
7.3 The Jordan Normal Form 236
7.4 Matrix Exponentiation via Generalized Eigenvectors 246
PART III REVIEW PROBLEMS FOR PART III 255

1
SYSTEMS OF LINEAR ALGEBRAIC EQUATIONS


1.1 LINEAR ALGEBRAIC EQUATIONS


    1. Solving the last equation for x4 yields . We substitute this value of x4 into the third equation and solve for x3 to get

      With these values of x3, the second equation becomes

      For x1, we substitute the values of x2 and x4 into the first equation.

      Therefore, the answer is: , , , and .

    2. Solving the first equation for x1, we obtain . We now substitute this value of x1 into the second equation and solve for x2.

      The third equation then becomes

      Therefore, the answer is , , and .

    3. From the third equation, we immediately get . This value, when substituted into the first equation, yields

      From the fourth equation, we obtain

      Finally, from the second equation, we conclude that

      So, the solution is , , , and .

    4. The third equation implies that . Then, the first equation says

      We now substitute these values of x1 and x2 into the second equation to get

      The solution to this problem is .

  1. To eliminate x1 from the first and second equations, we subtract from the first equation the third equation multiplied by 3 and subtract from the second equation the third equation multiplied by 2, resp. Thus we get an equivalent system

    From the first equation, we get . Making the back substitutions into the second and third equations yields

    The solution is , , and .

  2. We eliminate x1 and x4 from the first and the the last equations by subtracting from them the second equation. We also eliminate x1 from the third equation by subtracting from it twice the second equation. The new system is

    Next, we eliminate x2 from the third and fourth by subtracting from them the first equation multiplied by 2. This gives

    We now can go with the back substitution. The last equation gives . With this value, we find x1 and x4 from the first and third equations, resp.

    Finally, the second equation says . so, the answer is , , , and .

  3. To eliminate x1 from the second equation, we multiply the first equation by 0.987/0.123 and subtract the result from the second one. Thus, we get

    From the second equation, we get . Substituting this value into the first equation, we get

    The solution, rounded to three decimal places, is , . The number of arithmetic operations required is 3 divisions, 3 multiplications, and 3 additions.

  4. Following the notations in Problem 8, the coefficients in Problem 7 are

    Applying the formulas given in Problem 8, we obtain

    The number of arithmetic operations required is 2 divisions, 6 multiplications, and 6 additions.

  5. To eliminate x1 from the second equation, we subtract from it the first equation multiplied by 0.987/0.123. Similarly, we eliminate x1 from the third equation using the factor of 0.333/0.123, and obtain

    We subtract from the third equation the second one multiplied by (1.790/3.005):

    Going from the third equation up, using the back substitution we find that

    The number of arithmetic operations required is 6 divisions, 11 multiplications, and 11 additions.

    1. From the third equation we find that (rounded to four decimal places). Substituting this value into the second equation, we get

      Finally, we use the values of x2 and x3 to find x1 from the first equation.

      Rounding the results to two decimal places, the answer is

      The total number of arithmetic operations required is 3 divisions, 3 multiplications, and 3 additions.

    2. We start from the fourth equation to find x4. Substituting its value into the third equation, we evaluate x3, and so on. These computations give

      Rounding the results to three decimal places yields

      The number of arithmetic operations required is 4 divisions, 6 multiplications, and 6 additions.

  6. No, the suggested procedure is inconsistent with the Gauss elimination rules. First of all, these rules require, on each step, the elimination of one of the variables from all but one equations. One can do some preliminary steps adding to an equation a multiple of another equation, but on each such step the new system must be considered.

    In the problem, let's follow steps. Subtracting the second equation from the first, we get a new system:

    Subtracting the third equation from the second, yields a new system:

    Performing the last step suggested in the problem, we obtain

    After these preliminary steps (that were not really necessary), we can now go with Gauss elimination procedure. Adding the second equation to the third yields

    We can now use the back substitution method to solve the system.

  7. In the "derivation", a standard mistake was made: the equation does not conclude that . The correct conclusion is . With the sign " ", we get the answer given; i.e. , . Choosing the sign " ", we get

    The figure below indicates these two solutions - the points of intersection of the graphs of the equations.

  8. The coordinates of both points must satisfy the equation . Thus, we have a system of two linear equations with three unknowns - a, b, and c:

    Since there is always a line passing through two given points, this system is consistent. If the triple (a0,?b0,?c0) is a solution, multiplying both equations by , we get an equivalent system

    Thus, the triple (ka0,?kb0,?kc0) also satisfies the conditions and, actually, determines the same line because

    Thus, the equation of a line is determined up to a non-zero constant multiple.

  9. Since the points (x0,?y0),?.,?(xn,?yn) are on the graph  , they must satisfy the equation. Substituting the coordinates of these points into the equation, yield the system of equations with unknowns - a0,?.,?an.
  10. Multiplying the second equation by we get

    We eliminate now x1 from the first equation by subtracting the above result from it (keeping the original second equation unchanged). Thus we get

    To eliminate x2 from the first equation, we multiply it by and then subtract from it the third equation multiplied by . Thus, we get

    so that we have a new system

    From the first equation, we find that . Substituting this value into the third equation yields . With zero values of x2 and x3, the second equation says that . Therefore, the solution is

  11. If x and y are integers, then 6x and 4y are even numbers. Thus, their difference, , is even, and so cannot be equal to an odd number 9.
  12. The first equation is equivalent to

    In the integers modulo 6, we have the sides as

    Thus, the equation is inconsistent in integers modulo 6, and so is the system.

    Concerning a solution in integers modulo 7, there are different ways to go. One of them is the following. First, we get a system that is equivalent to the given system by multiplying the first equation by 3 and the second equation by 2.

    Subtracting the second equation from the first one yields . Rewriting this equation in integers modulo 7 (since ) as , the back substitution into the second equation in the original system yields

    Solving for x, we get . Therefore, in integers modulo 7, the solution to the given system is

  13. To simplify computations, we note that, for large n, the total number of each arithmetic operation can be approximated with a good relative accuracy by

    Thus, the total number N(n) of arithmetic operations can be approximated as

    For the "Thermal stress", . Therefore, using the performance of computers, we get that computers require approximately

    to solve the system. Performing similar computations for other systems, we fill in the following table.

    Model n Typical PC Tianhe-2 Thermal Stress 104 133.333 American Sign Language 105 0.020 Chemical Plant Modeling 0.532 Mechanics of Composite...

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