- 6.6.1: Rewrite each expression using only positive exponents. a. 23 b. 52 ...
- 6.6.2: Insert the appropriate symbol (<, =, or >) between each pair of num...
- 6.6.3: Find the exponent of 10 that you need to write each expression in s...
- 6.6.4: The population of a town is currently 45,647. It has been growing a...
- 6.6.5: Juan says that 63 is the same as 63.Write an explanation of how Jua...
- 6.6.6: Use the properties of exponents to rewrite each expression without ...
- 6.6.7: APPLICATION Suppose the annual rate of inflation is about 4%. This ...
- 6.6.8: APPLICATION The population of Japan in 2004 was about 1.3 108. Japa...
- 6.6.9: Decide whether each statement is true or false. Use expanded form t...
- 6.6.10: A large ball of string originally held 1 mile of string. Abigail cu...
- 6.6.11: Suppose 36(1 + 0.5)4 represents the number of bacteria cells in a s...
- 6.6.12: APPLICATION Camila received a $1,200 prize for one of her essays. S...
- 6.6.13: Mini-Investigation In the last few lessons, you have worked with eq...
- 6.6.14: APPLICATION A capacitor is charged with a nine-volt battery. The eq...
- 6.6.15: Set your calculator in scientific notation mode for this problem. a...
Solutions for Chapter 6.6: Zero and Negative Exponents
Full solutions for Discovering Algebra: An Investigative Approach | 2nd Edition
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).
Cramer's Rule for Ax = b.
B j has b replacing column j of A; x j = det B j I det A
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.
Dimension of vector space
dim(V) = number of vectors in any basis for V.
Dot product = Inner product x T y = XI Y 1 + ... + Xn Yn.
Complex dot product is x T Y . Perpendicular vectors have x T y = O. (AB)ij = (row i of A)T(column j of B).
Four Fundamental Subspaces C (A), N (A), C (AT), N (AT).
Use AT for complex A.
Gram-Schmidt orthogonalization A = QR.
Independent columns in A, orthonormal columns in Q. Each column q j of Q is a combination of the first j columns of A (and conversely, so R is upper triangular). Convention: diag(R) > o.
Hessenberg matrix H.
Triangular matrix with one extra nonzero adjacent diagonal.
Hilbert matrix hilb(n).
Entries HU = 1/(i + j -1) = Jd X i- 1 xj-1dx. Positive definite but extremely small Amin and large condition number: H is ill-conditioned.
Incidence matrix of a directed graph.
The m by n edge-node incidence matrix has a row for each edge (node i to node j), with entries -1 and 1 in columns i and j .
A symmetric matrix with eigenvalues of both signs (+ and - ).
lA-II = l/lAI and IATI = IAI.
The big formula for det(A) has a sum of n! terms, the cofactor formula uses determinants of size n - 1, volume of box = I det( A) I.
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.
Markov matrix M.
All mij > 0 and each column sum is 1. Largest eigenvalue A = 1. If mij > 0, the columns of Mk approach the steady state eigenvector M s = s > O.
The diagonal entry (first nonzero) at the time when a row is used in elimination.
Schur complement S, D - C A -} B.
Appears in block elimination on [~ g ].
Standard basis for Rn.
Columns of n by n identity matrix (written i ,j ,k in R3).
Trace of A
= sum of diagonal entries = sum of eigenvalues of A. Tr AB = Tr BA.
v + w = (VI + WI, ... , Vn + Wn ) = diagonal of parallelogram.
Vector v in Rn.
Sequence of n real numbers v = (VI, ... , Vn) = point in Rn.