 0.3.1: Evaluate each expression. Write your answer as a fraction and as a ...
 0.3.2: The fractal from the example is shown below. How much longer is the...
 0.3.3: At what stage does the figure above first exceed a length of 10?
 0.3.4: Evaluate each expression and check your results with a calculator.
 0.3.5: The Stage 0 figure below has a length of 1. At Stage 1, each segmen...
 0.3.6: The Stage 0 figure below has a length of 1. a. Complete a table lik...
 0.3.7: The Stage 0 figure below has a length of 1. a. Complete a table lik...
 0.3.8: The figures below look a little more complicated than others you ha...
 0.3.9: Write as a decimal.
 0.3.10: What is ?
 0.3.11: Look at the fractal cross pattern below. At each stage, new line se...
Solutions for Chapter 0.3: Shorter yet Longer
Full solutions for Discovering Algebra: An Investigative Approach  2nd Edition
ISBN: 9781559537636
Solutions for Chapter 0.3: Shorter yet Longer
Get Full SolutionsChapter 0.3: Shorter yet Longer includes 11 full stepbystep solutions. Since 11 problems in chapter 0.3: Shorter yet Longer have been answered, more than 3088 students have viewed full stepbystep solutions from this chapter. 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. This textbook survival guide was created for the textbook: Discovering Algebra: An Investigative Approach, edition: 2.

Affine transformation
Tv = Av + Vo = linear transformation plus shift.

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).

Diagonal matrix D.
dij = 0 if i # j. Blockdiagonal: zero outside square blocks Du.

Diagonalization
A = S1 AS. A = eigenvalue matrix and S = eigenvector matrix of A. A must have n independent eigenvectors to make S invertible. All Ak = SA k SI.

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 IIA1 yll2 = Y T(AAT)1 Y = 1 displayed by eigshow; axis lengths ad

Exponential eAt = I + At + (At)2 12! + ...
has derivative AeAt; eAt u(O) solves u' = Au.

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].

GaussJordan method.
Invert A by row operations on [A I] to reach [I AI].

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.

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.

Kirchhoff's Laws.
Current Law: net current (in minus out) is zero at each node. Voltage Law: Potential differences (voltage drops) add to zero around any closed loop.

Multiplicities AM and G M.
The algebraic multiplicity A M of A is the number of times A appears as a root of det(A  AI) = O. The geometric multiplicity GM is the number of independent eigenvectors for A (= dimension of the eigenspace).

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

Outer product uv T
= column times row = rank one matrix.

Projection matrix P onto subspace S.
Projection p = P b is the closest point to b in S, error e = b  Pb is perpendicularto S. p 2 = P = pT, eigenvalues are 1 or 0, eigenvectors are in S or S...L. If columns of A = basis for S then P = A (AT A) 1 AT.

Row picture of Ax = b.
Each equation gives a plane in Rn; the planes intersect at x.

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.
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