 31.1: Tony said that is irrational because it is not the ratio of integer...
 31.2: Maria said that since the solution of the inequality 2x 5 , 3 can b...
 31.3: In 314, determine whether each of the numbers is rational or irrati...
 31.4: In 314, determine whether each of the numbers is rational or irrati...
 31.5: In 314, determine whether each of the numbers is rational or irrati...
 31.6: In 314, determine whether each of the numbers is rational or irrati...
 31.7: In 314, determine whether each of the numbers is rational or irrati...
 31.8: In 314, determine whether each of the numbers is rational or irrati...
 31.9: In 314, determine whether each of the numbers is rational or irrati...
 31.10: In 314, determine whether each of the numbers is rational or irrati...
 31.11: In 314, determine whether each of the numbers is rational or irrati...
 31.12: In 314, determine whether each of the numbers is rational or irrati...
 31.13: In 314, determine whether each of the numbers is rational or irrati...
 31.14: In 314, determine whether each of the numbers is rational or irrati...
 31.15: In 1526, find and graph the solution set of each inequality. x , 7
 31.16: In 1526, find and graph the solution set of each inequality. a 2 5 $ 3
 31.17: In 1526, find and graph the solution set of each inequality. . 2y 1...
 31.18: In 1526, find and graph the solution set of each inequality. 2 2 4b...
 31.19: In 1526, find and graph the solution set of each inequality. 5 2 a . 4
 31.20: In 1526, find and graph the solution set of each inequality. 9 2 3x...
 31.21: In 1526, find and graph the solution set of each inequality. 2 1 2 ...
 31.22: In 1526, find and graph the solution set of each inequality. 2x 1 2...
 31.23: In 1526, find and graph the solution set of each inequality. 2x 1 2...
 31.24: In 1526, find and graph the solution set of each inequality. x 1 2 ...
 31.25: In 1526, find and graph the solution set of each inequality. P 1 1 . 1
 31.26: In 1526, find and graph the solution set of each inequality. 2 2 1 ...
 31.27: The temperature on Mars roughly satisfies the inequality t 2 75 # 1...
 31.28: Elevation of land in the United States is given by the inequality h...
Solutions for Chapter 31: The Real Numbers And Absolute Value
Full solutions for Amsco's Algebra 2 and Trigonometry  1st Edition
ISBN: 9781567657029
Solutions for Chapter 31: The Real Numbers And Absolute Value
Get Full SolutionsAmsco's Algebra 2 and Trigonometry was written by and is associated to the ISBN: 9781567657029. Since 28 problems in chapter 31: The Real Numbers And Absolute Value have been answered, more than 30992 students have viewed full stepbystep solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Amsco's Algebra 2 and Trigonometry, edition: 1. Chapter 31: The Real Numbers And Absolute Value includes 28 full stepbystep solutions.

Circulant matrix C.
Constant diagonals wrap around as in cyclic shift S. Every circulant is Col + CIS + ... + Cn_lSn  l . Cx = convolution c * x. Eigenvectors in F.

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).

Cyclic shift
S. Permutation with S21 = 1, S32 = 1, ... , finally SIn = 1. Its eigenvalues are the nth roots e2lrik/n of 1; eigenvectors are columns of the Fourier matrix F.

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.

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].

Four Fundamental Subspaces C (A), N (A), C (AT), N (AT).
Use AT for complex A.

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.

Hilbert matrix hilb(n).
Entries HU = 1/(i + j 1) = Jd X i 1 xj1dx. Positive definite but extremely small Amin and large condition number: H is illconditioned.

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.

Inverse matrix AI.
Square matrix with AI A = I and AAl = I. No inverse if det A = 0 and rank(A) < n and Ax = 0 for a nonzero vector x. The inverses of AB and AT are B1 AI and (AI)T. Cofactor formula (Al)ij = Cji! detA.

Linearly dependent VI, ... , Vn.
A combination other than all Ci = 0 gives L Ci Vi = O.

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

Nullspace matrix N.
The columns of N are the n  r special solutions to As = O.

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

Plane (or hyperplane) in Rn.
Vectors x with aT x = O. Plane is perpendicular to a =1= O.

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

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.

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

Semidefinite matrix A.
(Positive) semidefinite: all x T Ax > 0, all A > 0; A = any RT R.

Symmetric matrix A.
The transpose is AT = A, and aU = a ji. AI is also symmetric.