 11.4.1: For Exercises 113, use your new conjectures. All measurements are i...
 11.4.2: For Exercises 113, use your new conjectures. All measurements are i...
 11.4.3: For Exercises 113, use your new conjectures. All measurements are i...
 11.4.4: For Exercises 113, use your new conjectures. All measurements are i...
 11.4.5: For Exercises 113, use your new conjectures. All measurements are i...
 11.4.6: For Exercises 113, use your new conjectures. All measurements are i...
 11.4.7: For Exercises 113, use your new conjectures. All measurements are i...
 11.4.8: For Exercises 113, use your new conjectures. All measurements are i...
 11.4.9: For Exercises 113, use your new conjectures. All measurements are i...
 11.4.10: For Exercises 113, use your new conjectures. All measurements are i...
 11.4.11: For Exercises 113, use your new conjectures. All measurements are i...
 11.4.12: For Exercises 113, use your new conjectures. All measurements are i...
 11.4.13: For Exercises 113, use your new conjectures. All measurements are i...
 11.4.14: Triangle PQR is a dilated image of ABC. Find the coordinates of B a...
 11.4.15: Aunt Florence has willed to her two nephews a plot of land in the s...
 11.4.16: Developing Proof Prove that corresponding angle bisectors of simila...
 11.4.17: Use algebra to show that if then
 11.4.18: A rectangle is divided into four rectangles, each similar to the or...
 11.4.19: Developing Proof In Chapter 5, you discovered that when you constru...
 11.4.20: A rectangle has sides a and b. For what values of a and b is anothe...
 11.4.21: Assume = Find AB and BC.
Solutions for Chapter 11.4: Corresponding Parts of Similar Triangles
Full solutions for Discovering Geometry: An Investigative Approach  4th Edition
ISBN: 9781559538824
Solutions for Chapter 11.4: Corresponding Parts of Similar Triangles
Get Full SolutionsThis expansive textbook survival guide covers the following chapters and their solutions. Chapter 11.4: Corresponding Parts of Similar Triangles includes 21 full stepbystep solutions. Discovering Geometry: An Investigative Approach was written by and is associated to the ISBN: 9781559538824. Since 21 problems in chapter 11.4: Corresponding Parts of Similar Triangles have been answered, more than 23771 students have viewed full stepbystep solutions from this chapter. This textbook survival guide was created for the textbook: Discovering Geometry: An Investigative Approach, edition: 4.

Back substitution.
Upper triangular systems are solved in reverse order Xn to Xl.

Big formula for n by n determinants.
Det(A) is a sum of n! terms. For each term: Multiply one entry from each row and column of A: rows in order 1, ... , nand column order given by a permutation P. Each of the n! P 's has a + or  sign.

Covariance matrix:E.
When random variables Xi have mean = average value = 0, their covariances "'£ ij are the averages of XiX j. With means Xi, the matrix :E = mean of (x  x) (x  x) T is positive (semi)definite; :E is diagonal if the Xi are independent.

Determinant IAI = det(A).
Defined by det I = 1, sign reversal for row exchange, and linearity in each row. Then IAI = 0 when A is singular. Also IABI = IAIIBI and

Incidence matrix of a directed graph.
The m by n edgenode incidence matrix has a row for each edge (node i to node j), with entries 1 and 1 in columns i and j .

Kronecker product (tensor product) A ® B.
Blocks aij B, eigenvalues Ap(A)Aq(B).

lAII = 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.

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).

Nilpotent matrix N.
Some power of N is the zero matrix, N k = o. The only eigenvalue is A = 0 (repeated n times). Examples: triangular matrices with zero diagonal.

Nullspace N (A)
= All solutions to Ax = O. Dimension n  r = (# columns)  rank.

Orthogonal subspaces.
Every v in V is orthogonal to every w in W.

Orthonormal vectors q 1 , ... , q n·
Dot products are q T q j = 0 if i =1= j and q T q i = 1. The matrix Q with these orthonormal columns has Q T Q = I. If m = n then Q T = Q 1 and q 1 ' ... , q n is an orthonormal basis for Rn : every v = L (v T q j )q j •

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.

Projection p = a(aTblaTa) onto the line through a.
P = aaT laTa has rank l.

Rayleigh quotient q (x) = X T Ax I x T x for symmetric A: Amin < q (x) < Amax.
Those extremes are reached at the eigenvectors x for Amin(A) and Amax(A).

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.

Row space C (AT) = all combinations of rows of A.
Column vectors by convention.

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

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

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.