- 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
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).
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).
Diagonal matrix D.
dij = 0 if i #- j. Block-diagonal: zero outside square blocks Du.
A = S-1 AS. A = eigenvalue matrix and S = eigenvector matrix of A. A must have n independent eigenvectors to make S invertible. All Ak = SA k S-I.
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 IIA-1 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.
0,1,1,2,3,5, ... satisfy Fn = Fn-l + Fn- 2 = (A7 -A~)I()q -A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ } A].
Invert A by row operations on [A I] to reach [I A-I].
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 A-I.
Square matrix with A-I A = I and AA-l = 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 B-1 A-I and (A-I)T. Cofactor formula (A-l)ij = Cji! detA.
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).
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 A-I are BT AT and (AT)-I.
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