 4.6.1: APPLICATION This table shows that the traveling distances between s...
 4.6.2: APPLICATION Let x represent total fat in grams, and let y represent...
 4.6.3: Give the coordinates of the Qpoints for each data set.
 4.6.4: The table gives the winning times for the Olympic mens 10,000meter...
 4.6.5: Create a data set that has Qpoints at (4, 28) and (12, 47) so that...
 4.6.6: Which linear equation below best fits the data at right? Explain yo...
 4.6.7: At 2:00 P.M., elevator A passes the second floor of the Empire Stat...
 4.6.8: At 2:00 P.M., elevator B passes the 94th floor of the same building...
 4.6.9: Think about the elevators in Exercises 7 and 8. a. Estimate when el...
 4.6.10: A car is traveling from Sioux Falls, South Dakota, to Mt. Rushmore,...
 4.6.11: A 4 oz bottle of mustard costs $0.88, a 7.5 oz bottle costs $1.65, ...
 4.6.12: Imagine that a classmate has been out of school for the past few da...
Solutions for Chapter 4.6: More on Modeling
Full solutions for Discovering Algebra: An Investigative Approach  2nd Edition
ISBN: 9781559537636
Solutions for Chapter 4.6: More on Modeling
Get Full SolutionsThis textbook survival guide was created for the textbook: Discovering Algebra: An Investigative Approach, edition: 2. This expansive textbook survival guide covers the following chapters and their solutions. Discovering Algebra: An Investigative Approach was written by Patricia and is associated to the ISBN: 9781559537636. Chapter 4.6: More on Modeling includes 12 full stepbystep solutions. Since 12 problems in chapter 4.6: More on Modeling have been answered, more than 2880 students have viewed full stepbystep solutions from this chapter.

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.

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 IIA1 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!).

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

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.

Krylov subspace Kj(A, b).
The subspace spanned by b, Ab, ... , AjIb. Numerical methods approximate A I b by x j with residual b  Ax j in this subspace. A good basis for K j requires only multiplication by A at each step.

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.

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

Left nullspace N (AT).
Nullspace of AT = "left nullspace" of A because y T A = OT.

Nilpotent matrix N.
Some power of N is the zero matrix, N k = o. The only eigenvalue is A = 0 (repeated n times). Examples: triangular matrices with zero diagonal.

Normal matrix.
If N NT = NT N, then N has orthonormal (complex) eigenvectors.

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 •

Pascal matrix
Ps = pascal(n) = the symmetric matrix with binomial entries (i1~;2). Ps = PL Pu all contain Pascal's triangle with det = 1 (see Pascal in the index).

Pivot columns of A.
Columns that contain pivots after row reduction. These are not combinations of earlier columns. The pivot columns are a basis for the column space.

Polar decomposition A = Q H.
Orthogonal Q times positive (semi)definite H.

Right inverse A+.
If A has full row rank m, then A+ = AT(AAT)l has AA+ = 1m.

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.

Spectral Theorem A = QAQT.
Real symmetric A has real A'S and orthonormal q's.

Spectrum of A = the set of eigenvalues {A I, ... , An}.
Spectral radius = max of IAi I.

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