 3.4.1: Match the recursive routine in the first column with the equation i...
 3.4.2: You can use the equation d = 24 45t to model the distance from a de...
 3.4.3: You can use the equation d = 4.7 + 2.8t to model a walk in which th...
 3.4.4: Undo the order of operations to find the xvalue in each equation. ...
 3.4.5: The equation y = 35 + 0.8x gives the distance a sports car is from ...
 3.4.6: APPLICATION Louis is beginning a new exercise workout. His trainer ...
 3.4.7: Jo mows lawns after school. She finds that she can use the equation...
 3.4.8: As part of a physics experiment, June threw an object off a cliff a...
 3.4.9: APPLICATION Manny has a parttime job as a waiter. He makes $45 per...
 3.4.10: APPLICATION Paula is crosstraining for a triathlon in which she cy...
 3.4.11: At a family picnic, your cousin tells you that he always has a hard...
 3.4.12: APPLICATION Carl has been keeping a record of his gas purchases for...
 3.4.13: Match each recursive routine to a graph below. Explain how you made...
 3.4.14: Bjarne is training for a bicycle race by riding on a stationary bic...
 3.4.15: Consider the expression . a. Find the value of the expression if y ...
Solutions for Chapter 3.4: Linear Equations and the Intercept Form
Full solutions for Discovering Algebra: An Investigative Approach  2nd Edition
ISBN: 9781559537636
Solutions for Chapter 3.4: Linear Equations and the Intercept Form
Get Full SolutionsThis expansive textbook survival guide covers the following chapters and their solutions. Discovering Algebra: An Investigative Approach was written by and is associated to the ISBN: 9781559537636. Since 15 problems in chapter 3.4: Linear Equations and the Intercept Form have been answered, more than 4671 students have viewed full stepbystep solutions from this chapter. Chapter 3.4: Linear Equations and the Intercept Form includes 15 full stepbystep solutions. This textbook survival guide was created for the textbook: Discovering Algebra: An Investigative Approach, edition: 2.

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

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 nl  An).

Complete solution x = x p + Xn to Ax = b.
(Particular x p) + (x n in nullspace).

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.

Fibonacci numbers
0,1,1,2,3,5, ... satisfy Fn = Fnl + Fn 2 = (A7 A~)I()q A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ } A].

Fourier matrix F.
Entries Fjk = e21Cijk/n give orthogonal columns FT F = nI. Then y = Fe is the (inverse) Discrete Fourier Transform Y j = L cke21Cijk/n.

Graph G.
Set of n nodes connected pairwise by m edges. A complete graph has all n(n  1)/2 edges between nodes. A tree has only n  1 edges and no closed loops.

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

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

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.

Orthogonal subspaces.
Every v in V is orthogonal to every w in W.

Positive definite matrix A.
Symmetric matrix with positive eigenvalues and positive pivots. Definition: x T Ax > 0 unless x = O. Then A = LDLT with diag(D» O.

Pseudoinverse A+ (MoorePenrose inverse).
The n by m matrix that "inverts" A from column space back to row space, with N(A+) = N(AT). A+ A and AA+ are the projection matrices onto the row space and column space. Rank(A +) = rank(A).

Reflection matrix (Householder) Q = I 2uuT.
Unit vector u is reflected to Qu = u. All x intheplanemirroruTx = o have Qx = x. Notice QT = Q1 = Q.

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.

Solvable system Ax = b.
The right side b is in the column space of A.

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

Transpose matrix AT.
Entries AL = Ajj. AT is n by In, AT A is square, symmetric, positive semidefinite. The transposes of AB and AI are BT AT and (AT)I.

Vector v in Rn.
Sequence of n real numbers v = (VI, ... , Vn) = point in Rn.