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Solutions for Chapter 49: Dividing by a Decimal Number

Saxon Math, Course 1 | 1st Edition | ISBN: 9781591417835 | Authors: Stephan Hake

Full solutions for Saxon Math, Course 1 | 1st Edition

ISBN: 9781591417835

Saxon Math, Course 1 | 1st Edition | ISBN: 9781591417835 | Authors: Stephan Hake

Solutions for Chapter 49: Dividing by a Decimal Number

Solutions for Chapter 49
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Textbook: Saxon Math, Course 1
Edition: 1
Author: Stephan Hake
ISBN: 9781591417835

This expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Saxon Math, Course 1, edition: 1. Since 30 problems in chapter 49: Dividing by a Decimal Number have been answered, more than 35518 students have viewed full step-by-step solutions from this chapter. Chapter 49: Dividing by a Decimal Number includes 30 full step-by-step solutions. Saxon Math, Course 1 was written by and is associated to the ISBN: 9781591417835.

Key Math Terms and definitions covered in this textbook
  • Column picture of Ax = b.

    The vector b becomes a combination of the columns of A. The system is solvable only when b is in the column space C (A).

  • Companion matrix.

    Put CI, ... ,Cn in row n and put n - 1 ones just above the main diagonal. Then det(A - AI) = ±(CI + c2A + C3A 2 + .•. + cnA n-l - An).

  • Covariance matrix:E.

    When random variables Xi have mean = average value = 0, their covariances "'£ ij are the averages of XiX j. With means Xi, the matrix :E = mean of (x - x) (x - x) T is positive (semi)definite; :E is diagonal if the Xi are independent.

  • Cross product u xv in R3:

    Vector perpendicular to u and v, length Ilullllvlll sin el = area of parallelogram, u x v = "determinant" of [i j k; UI U2 U3; VI V2 V3].

  • Eigenvalue A and eigenvector x.

    Ax = AX with x#-O so det(A - AI) = o.

  • Elimination.

    A sequence of row operations that reduces A to an upper triangular U or to the reduced form R = rref(A). Then A = LU with multipliers eO in L, or P A = L U with row exchanges in P, or E A = R with an invertible E.

  • Ellipse (or ellipsoid) x T Ax = 1.

    A must be positive definite; the axes of the ellipse are eigenvectors of A, with lengths 1/.JI. (For IIx II = 1 the vectors y = Ax lie on the ellipse IIA-1 yll2 = Y T(AAT)-1 Y = 1 displayed by eigshow; axis lengths ad

  • Factorization

    A = L U. If elimination takes A to U without row exchanges, then the lower triangular L with multipliers eij (and eii = 1) brings U back to A.

  • Hypercube matrix pl.

    Row n + 1 counts corners, edges, faces, ... of a cube in Rn.

  • Least squares solution X.

    The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b - Ax is orthogonal to all columns of A.

  • Linearly dependent VI, ... , Vn.

    A combination other than all Ci = 0 gives L Ci Vi = O.

  • 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.

  • Minimal polynomial of A.

    The lowest degree polynomial with meA) = zero matrix. This is peA) = det(A - AI) if no eigenvalues are repeated; always meA) divides peA).

  • Multiplier eij.

    The pivot row j is multiplied by eij and subtracted from row i to eliminate the i, j entry: eij = (entry to eliminate) / (jth pivot).

  • Partial pivoting.

    In each column, choose the largest available pivot to control roundoff; all multipliers have leij I < 1. See condition number.

  • Reduced row echelon form R = rref(A).

    Pivots = 1; zeros above and below pivots; the r nonzero rows of R give a basis for the row space of A.

  • Reflection matrix (Householder) Q = I -2uuT.

    Unit vector u is reflected to Qu = -u. All x intheplanemirroruTx = o have Qx = x. Notice QT = Q-1 = Q.

  • Schwarz inequality

    Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.

  • Singular Value Decomposition

    (SVD) A = U:E VT = (orthogonal) ( diag)( orthogonal) First r columns of U and V are orthonormal bases of C (A) and C (AT), AVi = O'iUi with singular value O'i > O. Last columns are orthonormal bases of nullspaces.

  • Tridiagonal matrix T: tij = 0 if Ii - j I > 1.

    T- 1 has rank 1 above and below diagonal.

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