Solutions for Chapter 4.9: SOLVING SYSTEMS OF LINEAR DES BY ELIMINATION

A First Course in Differential Equations with Modeling Applications | 10th Edition | ISBN: 9781111827052 | Authors: Dennis G. Zill

Full solutions for A First Course in Differential Equations with Modeling Applications | 10th Edition

ISBN: 9781111827052

A First Course in Differential Equations with Modeling Applications | 10th Edition | ISBN: 9781111827052 | Authors: Dennis G. Zill

Solutions for Chapter 4.9: SOLVING SYSTEMS OF LINEAR DES BY ELIMINATION

Solutions for Chapter 4.9
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Textbook: A First Course in Differential Equations with Modeling Applications
Edition: 10th
Author: Dennis G. Zill
ISBN: 9781111827052

This textbook survival guide was created for the textbook: A First Course in Differential Equations with Modeling Applications, edition: 10th. Chapter 4.9: SOLVING SYSTEMS OF LINEAR DES BY ELIMINATION includes 27 full step-by-step solutions. Since 27 problems in chapter 4.9: SOLVING SYSTEMS OF LINEAR DES BY ELIMINATION have been answered, more than 23260 students have viewed full step-by-step solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions. A First Course in Differential Equations with Modeling Applications was written by and is associated to the ISBN: 9781111827052.

Key Math Terms and definitions covered in this textbook
  • Change of basis matrix M.

    The old basis vectors v j are combinations L mij Wi of the new basis vectors. The coordinates of CI VI + ... + cnvn = dl wI + ... + dn Wn are related by d = M c. (For n = 2 set VI = mll WI +m21 W2, V2 = m12WI +m22w2.)

  • Cholesky factorization

    A = CTC = (L.J]))(L.J]))T for positive definite A.

  • Cramer's Rule for Ax = b.

    B j has b replacing column j of A; x j = det B j I det A

  • Diagonalization

    A = S-1 AS. A = eigenvalue matrix and S = eigenvector matrix of A. A must have n independent eigenvectors to make S invertible. All Ak = SA k S-I.

  • Dot product = Inner product x T y = XI Y 1 + ... + Xn Yn.

    Complex dot product is x T Y . Perpendicular vectors have x T y = O. (AB)ij = (row i of A)T(column j of B).

  • Fast Fourier Transform (FFT).

    A factorization of the Fourier matrix Fn into e = log2 n matrices Si times a permutation. Each Si needs only nl2 multiplications, so Fnx and Fn-1c can be computed with ne/2 multiplications. Revolutionary.

  • Free variable Xi.

    Column i has no pivot in elimination. We can give the n - r free variables any values, then Ax = b determines the r pivot variables (if solvable!).

  • Gram-Schmidt orthogonalization A = QR.

    Independent columns in A, orthonormal columns in Q. Each column q j of Q is a combination of the first j columns of A (and conversely, so R is upper triangular). Convention: diag(R) > o.

  • Incidence matrix of a directed graph.

    The m by n edge-node incidence matrix has a row for each edge (node i to node j), with entries -1 and 1 in columns i and j .

  • Indefinite matrix.

    A symmetric matrix with eigenvalues of both signs (+ and - ).

  • Iterative method.

    A sequence of steps intended to approach the desired solution.

  • Left inverse A+.

    If A has full column rank n, then A+ = (AT A)-I AT has A+ A = In.

  • Lucas numbers

    Ln = 2,J, 3, 4, ... satisfy Ln = L n- l +Ln- 2 = A1 +A~, with AI, A2 = (1 ± -/5)/2 from the Fibonacci matrix U~]' Compare Lo = 2 with Fo = O.

  • Markov matrix M.

    All mij > 0 and each column sum is 1. Largest eigenvalue A = 1. If mij > 0, the columns of Mk approach the steady state eigenvector M s = s > O.

  • Multiplication Ax

    = Xl (column 1) + ... + xn(column n) = combination of columns.

  • Orthogonal matrix Q.

    Square matrix with orthonormal columns, so QT = Q-l. Preserves length and angles, IIQxll = IIxll and (QX)T(Qy) = xTy. AlllAI = 1, with orthogonal eigenvectors. Examples: Rotation, reflection, permutation.

  • Row space C (AT) = all combinations of rows of A.

    Column vectors by convention.

  • Simplex method for linear programming.

    The minimum cost vector x * is found by moving from comer to lower cost comer along the edges of the feasible set (where the constraints Ax = b and x > 0 are satisfied). Minimum cost at a comer!

  • Spanning set.

    Combinations of VI, ... ,Vm fill the space. The columns of A span C (A)!

  • Volume of box.

    The rows (or the columns) of A generate a box with volume I det(A) I.

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