 3.2.1: Construction For Exercises 15, construct the figures using only a c...
 3.2.2: Construction For Exercises 15, construct the figures using only a c...
 3.2.3: Construction For Exercises 15, construct the figures using only a c...
 3.2.4: Construction For Exercises 15, construct the figures using only a c...
 3.2.5: Construction For Exercises 15, construct the figures using only a c...
 3.2.6: Construction Do Exercises 15 using patty paper.
 3.2.7: Construction For Exercises 710, you have your choice of constructio...
 3.2.8: Construction For Exercises 710, you have your choice of constructio...
 3.2.9: Construction For Exercises 710, you have your choice of constructio...
 3.2.10: Construction For Exercises 710, you have your choice of constructio...
 3.2.11: The island shown at right has two post offices. The postal service ...
 3.2.12: Copy parallelogram FLAT onto your paper. Construct the perpendicula...
 3.2.13: Technology Use geometry software to construct a triangle. Construct...
 3.2.14: Construction Construct a very large triangle on a piece of cardboar...
 3.2.15: In Exercises 1520, match the term with its figure below. Scalene ac...
 3.2.16: In Exercises 1520, match the term with its figure below. Isosceles ...
 3.2.17: In Exercises 1520, match the term with its figure below. Isosceles ...
 3.2.18: In Exercises 1520, match the term with its figure below. Isosceles ...
 3.2.19: In Exercises 1520, match the term with its figure below. Scalene ob...
 3.2.20: In Exercises 1520, match the term with its figure below. Scalene ri...
 3.2.21: List the letters from the alphabet below that have a horizontal lin...
 3.2.22: Use your ruler and protractor to draw a triangle with angle measure...
Solutions for Chapter 3.2: Constructing Perpendicular Bisectors
Full solutions for Discovering Geometry: An Investigative Approach  4th Edition
ISBN: 9781559538824
Solutions for Chapter 3.2: Constructing Perpendicular Bisectors
Get Full SolutionsThis textbook survival guide was created for the textbook: Discovering Geometry: An Investigative Approach, edition: 4. Discovering Geometry: An Investigative Approach was written by and is associated to the ISBN: 9781559538824. Since 22 problems in chapter 3.2: Constructing Perpendicular Bisectors have been answered, more than 23337 students have viewed full stepbystep solutions from this chapter. Chapter 3.2: Constructing Perpendicular Bisectors includes 22 full stepbystep solutions. This expansive textbook survival guide covers the following chapters and their solutions.

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

Cramer's Rule for Ax = b.
B j has b replacing column j of A; x j = det B j I det A

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.

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.

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

Factorization
A = L U. If elimination takes A to U without row exchanges, then the lower triangular L with multipliers eij (and eii = 1) brings U back to A.

Fast Fourier Transform (FFT).
A factorization of the Fourier matrix Fn into e = log2 n matrices Si times a permutation. Each Si needs only nl2 multiplications, so Fnx and Fn1c can be computed with ne/2 multiplications. Revolutionary.

Iterative method.
A sequence of steps intended to approach the desired solution.

Krylov subspace Kj(A, b).
The subspace spanned by b, Ab, ... , AjIb. Numerical methods approximate A I b by x j with residual b  Ax j in this subspace. A good basis for K j requires only multiplication by A at each step.

Left nullspace N (AT).
Nullspace of AT = "left nullspace" of A because y T A = OT.

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 •

Pivot columns of A.
Columns that contain pivots after row reduction. These are not combinations of earlier columns. The pivot columns are a basis for the column space.

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.

Pseudoinverse A+ (MoorePenrose inverse).
The n by m matrix that "inverts" A from column space back to row space, with N(A+) = N(AT). A+ A and AA+ are the projection matrices onto the row space and column space. Rank(A +) = rank(A).

Random matrix rand(n) or randn(n).
MATLAB creates a matrix with random entries, uniformly distributed on [0 1] for rand and standard normal distribution for randn.

Singular matrix A.
A square matrix that has no inverse: det(A) = o.

Spanning set.
Combinations of VI, ... ,Vm fill the space. The columns of A span C (A)!

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

Toeplitz matrix.
Constant down each diagonal = timeinvariant (shiftinvariant) filter.

Triangle inequality II u + v II < II u II + II v II.
For matrix norms II A + B II < II A II + II B II·