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Solutions for Chapter 5.1: Solving Systems of Equations

Discovering Algebra: An Investigative Approach | 2nd Edition | ISBN: 9781559537636 | Authors: Jerald Murdock, Ellen Kamischke, Eric Kamischke

Full solutions for Discovering Algebra: An Investigative Approach | 2nd Edition

ISBN: 9781559537636

Discovering Algebra: An Investigative Approach | 2nd Edition | ISBN: 9781559537636 | Authors: Jerald Murdock, Ellen Kamischke, Eric Kamischke

Solutions for Chapter 5.1: Solving Systems of Equations

This textbook survival guide was created for the textbook: Discovering Algebra: An Investigative Approach, edition: 2. Chapter 5.1: Solving Systems of Equations includes 16 full step-by-step solutions. This expansive textbook survival guide covers the following chapters and their solutions. Since 16 problems in chapter 5.1: Solving Systems of Equations have been answered, more than 8479 students have viewed full step-by-step solutions from this chapter. Discovering Algebra: An Investigative Approach was written by and is associated to the ISBN: 9781559537636.

Key Math Terms and definitions covered in this textbook
  • Affine transformation

    Tv = Av + Vo = linear transformation plus shift.

  • Augmented matrix [A b].

    Ax = b is solvable when b is in the column space of A; then [A b] has the same rank as A. Elimination on [A b] keeps equations correct.

  • Big formula for n by n determinants.

    Det(A) is a sum of n! terms. For each term: Multiply one entry from each row and column of A: rows in order 1, ... , nand column order given by a permutation P. Each of the n! P 's has a + or - sign.

  • 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

  • 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!).

  • Hankel matrix H.

    Constant along each antidiagonal; hij depends on i + j.

  • Independent vectors VI, .. " vk.

    No combination cl VI + ... + qVk = zero vector unless all ci = O. If the v's are the columns of A, the only solution to Ax = 0 is x = o.

  • Left nullspace N (AT).

    Nullspace of AT = "left nullspace" of A because y T A = OT.

  • 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.

  • 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).

  • Network.

    A directed graph that has constants Cl, ... , Cm associated with the edges.

  • Normal equation AT Ax = ATb.

    Gives the least squares solution to Ax = b if A has full rank n (independent columns). The equation says that (columns of A)·(b - Ax) = o.

  • Nullspace N (A)

    = All solutions to Ax = O. Dimension n - r = (# columns) - rank.

  • Orthonormal vectors q 1 , ... , q n·

    Dot products are q T q j = 0 if i =1= j and q T q i = 1. The matrix Q with these orthonormal columns has Q T Q = I. If m = n then Q T = Q -1 and q 1 ' ... , q n is an orthonormal basis for Rn : every v = L (v T q j )q j •

  • 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.

  • Saddle point of I(x}, ... ,xn ).

    A point where the first derivatives of I are zero and the second derivative matrix (a2 II aXi ax j = Hessian matrix) is indefinite.

  • Singular matrix A.

    A square matrix that has no inverse: det(A) = o.

  • Spanning set.

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

  • Trace of A

    = sum of diagonal entries = sum of eigenvalues of A. Tr AB = Tr BA.

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