 0.5.1: Estimate the length of each segment in centimeters. Then measure an...
 0.5.2: Use a ruler to draw a segment that fits each description. a. oneth...
 0.5.3: Mark two points on your paper. Label them A and B. Draw a segment b...
 0.5.4: Do these calculations. Check your results with a calculator. a. 2 +...
 0.5.5: Youll need the program in Calculator Note 0E for Exercises 58. ]Dra...
 0.5.6: Youll need the program in Calculator Note 0E for Exercises 58. ] Dr...
 0.5.7: Youll need the program in Calculator Note 0E for Exercises 58. ]Exp...
 0.5.8: Youll need the program in Calculator Note 0E for Exercises 58. ]Sup...
 0.5.9: Draw a segment that is 12 cm in length. Find and label a point that...
 0.5.10: Use a calculator to investigate the behavior of the expressions bel...
 0.5.11: Look at the fractal below. The Stage 0 figure has a length of 1.
Solutions for Chapter 0.5: Out of Chaos
Full solutions for Discovering Algebra: An Investigative Approach  2nd Edition
ISBN: 9781559537636
Solutions for Chapter 0.5: Out of Chaos
Get Full SolutionsDiscovering Algebra: An Investigative Approach was written by and is associated to the ISBN: 9781559537636. This expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Discovering Algebra: An Investigative Approach, edition: 2. Chapter 0.5: Out of Chaos includes 11 full stepbystep solutions. Since 11 problems in chapter 0.5: Out of Chaos have been answered, more than 7675 students have viewed full stepbystep solutions from this chapter.

Cofactor Cij.
Remove row i and column j; multiply the determinant by (I)i + j •

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).

Echelon matrix U.
The first nonzero entry (the pivot) in each row comes in a later column than the pivot in the previous row. All zero rows come last.

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.

Fourier matrix F.
Entries Fjk = e21Cijk/n give orthogonal columns FT F = nI. Then y = Fe is the (inverse) Discrete Fourier Transform Y j = L cke21Cijk/n.

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.

Matrix multiplication AB.
The i, j entry of AB is (row i of A)·(column j of B) = L aikbkj. By columns: Column j of AB = A times column j of B. By rows: row i of A multiplies B. Columns times rows: AB = sum of (column k)(row k). All these equivalent definitions come from the rule that A B times x equals A times B x .

Minimal polynomial of A.
The lowest degree polynomial with meA) = zero matrix. This is peA) = det(A  AI) if no eigenvalues are repeated; always meA) divides peA).

Norm
IIA II. The ".e 2 norm" of A is the maximum ratio II Ax II/l1x II = O"max· Then II Ax II < IIAllllxll and IIABII < IIAIIIIBII and IIA + BII < IIAII + IIBII. Frobenius norm IIAII} = L La~. The.e 1 and.e oo norms are largest column and row sums of laij I.

Permutation matrix P.
There are n! orders of 1, ... , n. The n! P 's have the rows of I in those orders. P A puts the rows of A in the same order. P is even or odd (det P = 1 or 1) based on the number of row exchanges to reach I.

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

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

Special solutions to As = O.
One free variable is Si = 1, other free variables = o.

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

Stiffness matrix
If x gives the movements of the nodes, K x gives the internal forces. K = ATe A where C has spring constants from Hooke's Law and Ax = stretching.

Sum V + W of subs paces.
Space of all (v in V) + (w in W). Direct sum: V n W = to}.

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

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.

Vandermonde matrix V.
V c = b gives coefficients of p(x) = Co + ... + Cn_IXn 1 with P(Xi) = bi. Vij = (Xi)jI and det V = product of (Xk  Xi) for k > i.